McClass

Question

Compare the OSATS scores of both groups assuming homogeneity of variance and calculate the pooled variance $s_{pool}^{2} .$

The pooled variance $s_{pool}^{2}$ needs to be calculated.

Feedback

Waiting for response...

Solution(s)

The system accepts the following correct response(s)

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$

Correctly determined the pooled variance $s_{pool}^{2} .$

Incorrect responses

The system detects the following incorrect responses

$s_{pool}^{2} = \frac{(\underset{instead of n_{VR}}{\underset{⏟}{n_{BOX}}} - 1) \cdot \underset{instead of s_{VR}^{2}}{\underset{⏟}{s_{VR}}} + (\underset{instead of n_{BOX}}{\underset{⏟}{n_{VR}}} - 1) \cdot \underset{instead of s_{BOX}^{2}}{\underset{⏟}{s_{BOX}}}}{n_{VR} + n_{BOX} - 2}$

Correctly determined the pooled variance $s_{pool}^{2} .$
The result matches the expression for the pooled variance with swaped sample sizes. Please double check the sample sizes $n_{VR}$ and $n_{BOX}$ of both groups.
The result matches the expression for the pooled variance with the standard deviation instead of the standard variations. Please remember to use the squares of the standard deviations $s_{VR}$ and $s_{BOX}$ to calculate the pooled variance.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot \underset{instead of s_{VR}^{2}}{\underset{⏟}{s_{VR}}} + (n_{BOX} - 1) \cdot \underset{instead of s_{BOX}^{2}}{\underset{⏟}{s_{BOX}}}}{n_{VR} + n_{BOX} - 2}$

Correctly determined the pooled variance $s_{pool}^{2} .$
The result matches the expression for the pooled variance with the standard deviation instead of the standard variations. Please remember to use the squares of the standard deviations $s_{VR}$ and $s_{BOX}$ to calculate the pooled variance.

$s_{pool}^{2} = \frac{(\underset{instead of n_{VR}}{\underset{⏟}{n_{BOX}}} - 1) \cdot s_{VR}^{2} + (\underset{instead of n_{BOX}}{\underset{⏟}{n_{VR}}} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$

Correctly determined the pooled variance $s_{pool}^{2} .$
The result matches the expression for the pooled variance with swaped sample sizes. Please double check the sample sizes $n_{VR}$ and $n_{BOX}$ of both groups.

Traps

Avoid these traps

Do not forget to use the square of the standard deviations when calculating the pooled variance.
Do not swap the sample sizes in the formula for the pooled variance $s_{pool}^{2} .$

Question

Is VR training superior to traditional Box training? Please report the $t$ -ratio using the Box trainingas the control intervention.

This is a $t$ -test for independent samples.
The pooled variance $s_{pool}^{2}$ needs to be calculated.
The $t$ -statistic must be determined: $\frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}} .$

Feedback

Waiting for response...

Solution(s)

The system accepts the following correct response(s)

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}$

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly determined the $t$ -statistic.

Incorrect responses

The system detects the following incorrect responses

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot \frac{1}{n_{VR} + n_{BOX}}}$

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly determined the $t$ -statistic.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly determined the $t$ -statistic.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\bar{VR} - \bar{BOX}}{\underset{instead of \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}{\underset{⏟}{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}}$

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly determined the $t$ -statistic.
The result matches the expression for the $t$ -ratio for independent samples without the square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot \frac{1}{n_{VR} + n_{BOX}}}$

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly determined the $t$ -statistic.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}$

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly determined the $t$ -statistic.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}$

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly determined the $t$ -statistic.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\bar{VR} - \bar{BOX}}{\underset{instead of \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}{\underset{⏟}{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}}$

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly determined the $t$ -statistic.
The result matches the expression for the $t$ -ratio for independent samples without the square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly determined the $t$ -statistic.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}$

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly determined the $t$ -statistic.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\bar{VR} - \bar{BOX}}{\underset{instead of \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}{\underset{⏟}{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}}$

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly determined the $t$ -statistic.
The result matches the expression for the $t$ -ratio for independent samples without the square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\bar{VR} - \bar{BOX}}{\underset{instead of \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}{\underset{⏟}{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}}$

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly determined the $t$ -statistic.
The result matches the expression for the $t$ -ratio for independent samples without the square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of \bar{BOX}}{\underset{⏟}{\bar{VR}}}}{\underset{instead of \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}{\underset{⏟}{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}}$

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly determined the $t$ -statistic.
The result matches the expression for the $t$ -ratio for independent samples without the square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of \bar{BOX}}{\underset{⏟}{\bar{VR}}}}{\underset{instead of \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}{\underset{⏟}{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}}$

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly determined the $t$ -statistic.
The result matches the expression for the $t$ -ratio for independent samples without the square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of \bar{BOX}}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}$

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly determined the $t$ -statistic.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of \bar{BOX}}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}$

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly determined the $t$ -statistic.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of \bar{BOX}}{\underset{⏟}{\bar{VR}}}}{\underset{instead of \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}{\underset{⏟}{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}}$

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly determined the $t$ -statistic.
The result matches the expression for the $t$ -ratio for independent samples without the square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of \bar{BOX}}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly determined the $t$ -statistic.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of \bar{BOX}}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}$

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly determined the $t$ -statistic.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of \bar{BOX}}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}$

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly determined the $t$ -statistic.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of \bar{BOX}}{\underset{⏟}{\bar{VR}}}}{\underset{instead of \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}{\underset{⏟}{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}}$

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly determined the $t$ -statistic.
The result matches the expression for the $t$ -ratio for independent samples without the square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of \bar{BOX}}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot \frac{1}{n_{VR} + n_{BOX}}}$

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly determined the $t$ -statistic.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of \bar{BOX}}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly determined the $t$ -statistic.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of \bar{BOX}}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot \frac{1}{n_{VR} + n_{BOX}}}$

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly determined the $t$ -statistic.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

Traps

Avoid these traps

Do not forget that $\frac{1}{a} + \frac{1}{b} \neq \frac{1}{a + b} .$
Do not forget the square root around the denominator of the $t$ -ratio.
Do not forget to use the square of the standard deviations when calculating the pooled variance $s_{pool}^{2} .$
The control intervention $\bar{BOX}$ must be subtracted from the tested intervention $\bar{VR}$ in the numerator of the $t$ -ratio.
You need to take the square root of the whole denomiator in the formula for the $t$ -ratio.

Question

Determine the confidence interval for the difference between the two group means using the Box training as the control intervention.

This is a $t$ -test for independent samples.
The pooled variance $s_{pool}^{2}$ needs to be calculated.
The formula to calculate the lower and upper bound of the confidence interval for the difference of group means is $\bar{VR} - \bar{BOX} \pm T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df) \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} .$
Make sure to use the $1 - α / 2$ -quantileof the $t$ -distribution.

Feedback

Waiting for response...

Solution(s)

The system accepts the following correct response(s)

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df) \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
Correctly used the $1 - α / 2$ -quantileof the $t$ -distribution.

Incorrect responses

The system detects the following incorrect responses

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df) \cdot {[s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})]}^{1 / 2}$

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
Correctly used the $1 - α / 2$ -quantileof the $t$ -distribution.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df) \cdot {[s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})]}^{1 / 2}$

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
Correctly used the $1 - α / 2$ -quantileof the $t$ -distribution.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot {[s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})]}^{1 / 2}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot {[s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})]}^{1 / 2}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot {[s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})]}^{1 / 2}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot {[s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})]}^{1 / 2}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df) \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
Correctly used the $1 - α / 2$ -quantileof the $t$ -distribution.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df) \cdot \sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}$

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the the confidence interval for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$
Correctly used the $1 - α / 2$ -quantileof the $t$ -distribution.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df) \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})$

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
Correctly used the $1 - α / 2$ -quantileof the $t$ -distribution.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df) \cdot {[s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})]}^{1 / 2}$

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
Correctly used the $1 - α / 2$ -quantileof the $t$ -distribution.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df) \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
Correctly used the $1 - α / 2$ -quantileof the $t$ -distribution.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df) \cdot \sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}$

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the the confidence interval for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$
Correctly used the $1 - α / 2$ -quantileof the $t$ -distribution.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the the confidence interval for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the the confidence interval for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df) \cdot \sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}$

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the the confidence interval for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$
Correctly used the $1 - α / 2$ -quantileof the $t$ -distribution.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot {[s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})]}^{1 / 2}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df) \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})$

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
Correctly used the $1 - α / 2$ -quantileof the $t$ -distribution.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot {[s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})]}^{1 / 2}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{{(s_{pool}^{2} \cdot \frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}^{1 / 2}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the $t$ -ratio without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})$ , with $t = \frac{\bar{VR} - \bar{BOX}}{{(s_{pool}^{2} \cdot \frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}^{1 / 2}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the $t$ -ratio without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot {[s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})]}^{1 / 2}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{{(s_{pool}^{2} \cdot \frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}^{1 / 2}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the $t$ -ratio without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{{(s_{pool}^{2} \cdot \frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}^{1 / 2}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the the confidence interval for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the $t$ -ratio without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{{(s_{pool}^{2} \cdot \frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}^{1 / 2}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the $t$ -ratio without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{{(s_{pool}^{2} \cdot \frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}^{1 / 2}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the $t$ -ratio without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})$ , with $t = \frac{\bar{VR} - \bar{BOX}}{{(s_{pool}^{2} \cdot \frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}^{1 / 2}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the $t$ -ratio without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the the confidence interval for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the the confidence interval for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{{(s_{pool}^{2} \cdot \frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}^{1 / 2}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the the confidence interval for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the $t$ -ratio without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot {[s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})]}^{1 / 2}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df) \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})$

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
Correctly used the $1 - α / 2$ -quantileof the $t$ -distribution.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df) \cdot {[s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})]}^{1 / 2}$

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
Correctly used the $1 - α / 2$ -quantileof the $t$ -distribution.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the the confidence interval for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the the confidence interval for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the the confidence interval for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the the confidence interval for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df) \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
Correctly used the $1 - α / 2$ -quantileof the $t$ -distribution.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot {[s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})]}^{1 / 2}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{{(s_{pool}^{2} \cdot \frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}^{1 / 2}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the the confidence interval for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the $t$ -ratio without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the the confidence interval for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the the confidence interval for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df) \cdot \sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}$

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the the confidence interval for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$
Correctly used the $1 - α / 2$ -quantileof the $t$ -distribution.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{{(s_{pool}^{2} \cdot \frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}^{1 / 2}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the $t$ -ratio without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{{(s_{pool}^{2} \cdot \frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}^{1 / 2}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the $t$ -ratio without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{{(s_{pool}^{2} \cdot \frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}^{1 / 2}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the $t$ -ratio without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df) \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})$

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
Correctly used the $1 - α / 2$ -quantileof the $t$ -distribution.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{{(s_{pool}^{2} \cdot \frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}^{1 / 2}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the the confidence interval for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the $t$ -ratio without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot {[s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})]}^{1 / 2}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{{(s_{pool}^{2} \cdot \frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}^{1 / 2}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the $t$ -ratio without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})$ , with $t = \frac{\underset{instead of \bar{VR}}{\underset{⏟}{\bar{BOX}}} - \underset{instead of boc}{\underset{⏟}{\bar{VR}}}}{{(s_{pool}^{2} \cdot \frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}^{1 / 2}}$ denoting the observed $t$ -statistic

The sign of the result matches the negative $t$ -ratio, with $\bar{VR}$ subtracted from $\bar{BOX} .$ Please keep in mind that the control intervention must be subtracted from the tested intervention.
Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the $t$ -ratio without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot {[s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})]}^{1 / 2}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{{(s_{pool}^{2} \cdot \frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}^{1 / 2}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the $t$ -ratio without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot {[s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})]}^{1 / 2}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the the confidence interval for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the the confidence interval for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot \sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot {[s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})]}^{1 / 2}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
The result matches the expression for the pooled variance without the square of $s_{VR}$ and $s_{BOX} .$ Please do not forget the square of $s_{VR}$ and $s_{BOX}$ when calculating the pooled variance.
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot {[s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})]}^{1 / 2}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot {[s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})]}^{1 / 2}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot \frac{1}{n_{VR} + n_{BOX}}}}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{n_{VR} + n_{BOX}} .$ Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{BOX}} \neq \frac{1}{n_{VR} + n_{BOX}} .$

$s_{pool}^{2} = \frac{(n_{VR} - 1) \cdot s_{VR}^{2} + (n_{BOX} - 1) \cdot s_{BOX}^{2}}{n_{VR} + n_{BOX} - 2}$
$ci = \bar{VR} - \bar{BOX} \pm t \cdot {[s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})]}^{1 / 2}$ , with $t = \frac{\bar{VR} - \bar{BOX}}{\sqrt{s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}})}$ denoting the observed $t$ -statistic

Correctly identified the problem as a $t$ -test for independent samples.
Correctly determined the pooled variance $s_{pool}^{2} .$
Correctly identified the formula for the upper and lower bound of the confidence interval for the difference of group means.
The result matches the confidence interval without square root around $s_{pool}^{2} \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget the square root around the denominator.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution $T_{1 - α / 2} (n_{VR} + n_{BOX} - 2 df)$ instead.
The result matches the expression for the $t$ -ratio with the square root stopping before $(\frac{1}{n_{VR}} + \frac{1}{n_{BOX}}) .$ Please do not forget to take the square root of the whole denominator.

Traps

Avoid these traps

The square root of the whole denomiator should be taken.
Do not forget that $\frac{1}{a} + \frac{1}{b} \neq \frac{1}{a + b} .$
Do not forget the square root around the denominator of the $t$ -ratio.
Do not forget to use the square of the standard deviations when calculating the pooled variance $s_{pool}^{2} .$
Do not insert the observed $t$ -statistic into the formula for the confidence interval. Use the quantile of the $t$ -distribution instead.
The control intervention $\bar{BOX}$ must be subtracted from the tested intervention $\bar{VR}$ in the numerator of the $t$ -ratio.

Training of surgical skills

Question

814.63...817.42

28.54...28.59

28.54...28.59

814.30...817.21

Question

-0.38...-0.37

-43.40...-36

-10.34...-9.81

-7.49...-7.20

-7.49...-7.20

-4.10...-3.92

-2.03...-1.98

-1.88...-1.84

-1.88...-1.84

-0.76...-0.74

-0.27...-0.25

-0.07...-0.06

0.06...0.07

0.25...0.27

0.37...0.38

0.74...0.76

1.84...1.88

1.84...1.88

1.98...2.03

3.92...4.10

7.20...7.49

7.20...7.49

9.81...10.34

36...43.40

Question

-13.63...-13.61to9.28...9.30

-68.42...-68.19to63.86...64.09

-64.09...-63.86to68.19...68.42

-60.28...-59.25to63.57...64.60

-22.99...-22.70to27.03...27.31

-10.41...-10.27to14.60...14.74

-9.90...-9.28to13.61...14.23

-9.30...-9.28to13.61...13.63

-9.00...-8.44to12.76...13.33

-8.64...-8.48to12.81...12.97

-7.90...-7.89to3.56...3.57

-4.48...-4.48to0.15...0.15

-4.48...-4.48to0.15...0.15

-4.31...-4.31to-0.02...-0.02

-3.57...-3.56to7.89...7.90

-3.42...-3.13to7.46...7.75

-3.24...-3.16to7.49...7.57

-3.24...-3.24to-1.09...-1.09

-2.62...-2.41to6.74...6.94

-2.60...-2.60to-1.73...-1.73

-2.26...-2.08to6.40...6.59

-2.19...-2.15to6.47...6.52

-2.10...-2.05to6.38...6.43

-2.02...-2.02to-2.31...-2.31

-1.99...-1.90to6.22...6.32

-1.80...-1.80to-2.53...-2.53

-1.73...-1.72to-2.61...-2.60

-1.73...-1.72to-2.60...-2.60

-1.68...-1.60to5.93...6.01

-1.43...-1.43to-2.90...-2.89

-1.40...-1.39to-2.94...-2.92

-1.31...-1.27to-3.06...-3.02

-1.29...-1.28to-3.04...-3.04

-1.09...-1.07to-3.26...-3.23

-1.09...-1.08to-3.25...-3.24

-0.28...-0.24to-4.09...-4.04

-0.20...-0.15to4.47...4.53

-0.15...-0.15to4.48...4.48

-0.15...-0.15to4.48...4.48

-0.09...0.02to4.31...4.42

-0.05...0.04to4.28...4.38

-0.04...0.05to-4.38...-4.28

-0.03...0.02to4.30...4.36

-0.02...0.03to-4.36...-4.30

-0.02...0.09to-4.42...-4.31

-0.02...0.01to4.32...4.35

-0.01...0.02to-4.35...-4.32

-0.01...0.01to4.32...4.34