McClass

*Table 1.* Observed WRAT scores at Pretest, Posttest, and $D = Pre - Post$
WRAT	Pretest	Posttest	$\bar{D}$
Average	$79.4$	$82.0$	$- 2.5$
SD	$1.9$	$2.7$	$1.8$

Question

Does the Cogmed training lead to a relevant improvement (i.e., more than $μ = 2.8$ units) in mean WRAT scores between pre- and posttest? Please report the $t$ -ratio.

This is a $t$ -test for paired samples.
The $t$ -ratio is $\frac{\bar{D} - μ}{s_{D} / \sqrt{N}} .$

Feedback

Waiting for response...

Solution(s)

The system accepts the following correct response(s)

$t = \frac{\bar{D} - μ}{s_{D} / \sqrt{N}}$

Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the $t$ -ratio for paired samples.

Incorrect responses

The system detects the following incorrect responses

$t = \frac{\bar{D} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Pre}}} / N^{1 / 2}}$

The result matches the $t$ -ratio with the standard deviation for the pretest $s_{Pre}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the $t$ -ratio for paired samples.
The result matches the $t$ -ratio without the square root around $N$ . Please do not forget the square root around $N$ .

$t = \frac{\bar{D} - μ}{s_{D} / N^{1 / 2}}$

Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the $t$ -ratio for paired samples.
The result matches the $t$ -ratio without the square root around $N$ . Please do not forget the square root around $N$ .

$t = \frac{\bar{D} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Post}}} / N^{1 / 2}}$

The result matches the $t$ -ratio with the standard deviation for the posttest $s_{Post}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the $t$ -ratio for paired samples.
The result matches the $t$ -ratio without the square root around $N$ . Please do not forget the square root around $N$ .

$t = \frac{\bar{D} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Pre}}} / N^{1 / 2}}$

The result matches the $t$ -ratio with the standard deviation for the pretest $s_{Pre}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
The result matches the $t$ -ratio, when the null hypothesis $μ$ has been omitted. Please do not forget $μ$ in the $t$ -ratio.
The result matches the $t$ -ratio without the square root around $N$ . Please do not forget the square root around $N$ .

$t = \frac{\bar{D} - μ}{s_{D} / N^{1 / 2}}$

Correctly recognised the problem as a $t$ -test for paired samples.
The result matches the $t$ -ratio, when the null hypothesis $μ$ has been omitted. Please do not forget $μ$ in the $t$ -ratio.
The result matches the $t$ -ratio without the square root around $N$ . Please do not forget the square root around $N$ .

$t = \frac{\bar{D} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Post}}} / N^{1 / 2}}$

The result matches the $t$ -ratio with the standard deviation for the posttest $s_{Post}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
The result matches the $t$ -ratio, when the null hypothesis $μ$ has been omitted. Please do not forget $μ$ in the $t$ -ratio.
The result matches the $t$ -ratio without the square root around $N$ . Please do not forget the square root around $N$ .

$t = \frac{\bar{D} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Pre}}} / \sqrt{N}}$

The result matches the $t$ -ratio with the standard deviation for the pretest $s_{Pre}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the $t$ -ratio for paired samples.

$t = \frac{\bar{D} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Post}}} / \sqrt{N}}$

The result matches the $t$ -ratio with the standard deviation for the posttest $s_{Post}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the $t$ -ratio for paired samples.

$t = \underset{instead of a paired t -test}{\underset{⏟}{\frac{\bar{Pre} - \bar{Post}}{\sqrt{s_{pool}^{2} \cdot \frac{1}{2 N}}}}}$ , with $s_{pool}^{2} = \frac{(N - 1) \cdot s_{Pre}^{2} + (N - 1) \cdot s_{Post}^{2}}{N + N - 2}$ denoting the pooled variance

The problem was mistakenly identified as a $t$ -test for independent samples.
Correctly identified the $t$ -ratio for independent samples.
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{2 N} .$ Please keep in mind that $\frac{1}{N} + \frac{1}{N} \neq \frac{1}{2 N} .$

$t = \frac{\bar{D} - μ}{s_{D} / \sqrt{N}}$

Correctly recognised the problem as a $t$ -test for paired samples.
The result matches the $t$ -ratio, when the null hypothesis $μ$ has been omitted. Please do not forget $μ$ in the $t$ -ratio.

$t = \frac{\bar{D} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Pre}}} / \sqrt{N}}$

The result matches the $t$ -ratio with the standard deviation for the pretest $s_{Pre}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
The result matches the $t$ -ratio, when the null hypothesis $μ$ has been omitted. Please do not forget $μ$ in the $t$ -ratio.

$t = \frac{\bar{D} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Post}}} / \sqrt{N}}$

The result matches the $t$ -ratio with the standard deviation for the posttest $s_{Post}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
The result matches the $t$ -ratio, when the null hypothesis $μ$ has been omitted. Please do not forget $μ$ in the $t$ -ratio.

$t = \underset{instead of a paired t -test}{\underset{⏟}{\frac{\bar{Pre} - \bar{Post}}{\sqrt{s_{pool}^{2} \cdot (1 / N + 1 / N)}}}}$ , with $s_{pool}^{2} = \frac{(N - 1) \cdot s_{Pre}^{2} + (N - 1) \cdot s_{Post}^{2}}{N + N - 2}$ denoting the pooled variance

The problem was mistakenly identified as a $t$ -test for independent samples.
Correctly identified the $t$ -ratio for independent samples.

$t = \bar{D} - \frac{μ}{s_{D}} / \sqrt{N}$

Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the $t$ -ratio for paired samples.
The result matches the fraction without parentheses around the numerator and the denominator, $\frac{(\bar{D} - μ)}{(s_{D} / \sqrt{N})} .$ Please do not forget the parentheses around the numerator and the denominator of a fraction.

$t = \bar{D} - \frac{μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Pre}}}} / \sqrt{N}$

The result matches the $t$ -ratio with the standard deviation for the pretest $s_{Pre}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the $t$ -ratio for paired samples.
The result matches the fraction without parentheses around the numerator and the denominator, $\frac{(\bar{D} - μ)}{(s_{D} / \sqrt{N})} .$ Please do not forget the parentheses around the numerator and the denominator of a fraction.

$t = \bar{D} - \frac{μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Post}}}} / \sqrt{N}$

The result matches the $t$ -ratio with the standard deviation for the posttest $s_{Post}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the $t$ -ratio for paired samples.
The result matches the fraction without parentheses around the numerator and the denominator, $\frac{(\bar{D} - μ)}{(s_{D} / \sqrt{N})} .$ Please do not forget the parentheses around the numerator and the denominator of a fraction.

$t = \bar{D} - \frac{μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Pre}}}} / N^{1 / 2}$

The result matches the $t$ -ratio with the standard deviation for the pretest $s_{Pre}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the $t$ -ratio for paired samples.
The result matches the fraction without parentheses around the numerator and the denominator, $\frac{(\bar{D} - μ)}{(s_{D} / \sqrt{N})} .$ Please do not forget the parentheses around the numerator and the denominator of a fraction.
The result matches the $t$ -ratio without the square root around $N$ . Please do not forget the square root around $N$ .

$t = \bar{D} - \frac{μ}{s_{D}} / N^{1 / 2}$

Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the $t$ -ratio for paired samples.
The result matches the fraction without parentheses around the numerator and the denominator, $\frac{(\bar{D} - μ)}{(s_{D} / \sqrt{N})} .$ Please do not forget the parentheses around the numerator and the denominator of a fraction.
The result matches the $t$ -ratio without the square root around $N$ . Please do not forget the square root around $N$ .

$t = \bar{D} - \frac{μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Post}}}} / N^{1 / 2}$

The result matches the $t$ -ratio with the standard deviation for the posttest $s_{Post}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the $t$ -ratio for paired samples.
The result matches the fraction without parentheses around the numerator and the denominator, $\frac{(\bar{D} - μ)}{(s_{D} / \sqrt{N})} .$ Please do not forget the parentheses around the numerator and the denominator of a fraction.
The result matches the $t$ -ratio without the square root around $N$ . Please do not forget the square root around $N$ .

$t = \underset{instead of \bar{D}}{\underset{⏟}{\bar{Pre}}} - \frac{μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Pre}}}} / \sqrt{N}$

The result matches the $t$ -ratio with the average $\bar{Pre}$ instead of the average change score $\bar{D}$ . Please insert the average change score $\bar{D}$ into the $t$ -ratio.
The result matches the $t$ -ratio with the standard deviation for the pretest $s_{Pre}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the $t$ -ratio for paired samples.
The result matches the fraction without parentheses around the numerator and the denominator, $\frac{(\bar{D} - μ)}{(s_{D} / \sqrt{N})} .$ Please do not forget the parentheses around the numerator and the denominator of a fraction.

$t = \underset{instead of \bar{D}}{\underset{⏟}{\bar{Pre}}} - \frac{μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Pre}}}} / N^{1 / 2}$

The result matches the $t$ -ratio with the average $\bar{Pre}$ instead of the average change score $\bar{D}$ . Please insert the average change score $\bar{D}$ into the $t$ -ratio.
The result matches the $t$ -ratio with the standard deviation for the pretest $s_{Pre}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the $t$ -ratio for paired samples.
The result matches the fraction without parentheses around the numerator and the denominator, $\frac{(\bar{D} - μ)}{(s_{D} / \sqrt{N})} .$ Please do not forget the parentheses around the numerator and the denominator of a fraction.
The result matches the $t$ -ratio without the square root around $N$ . Please do not forget the square root around $N$ .

$t = \underset{instead of \bar{D}}{\underset{⏟}{\bar{Post}}} - \frac{μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Post}}}} / \sqrt{N}$

The result matches the $t$ -ratio with the posttest average $\bar{Post}$ instead of the average change score $\bar{D}$ . Please insert the average change score $\bar{D}$ into the $t$ -ratio.
The result matches the $t$ -ratio with the standard deviation for the posttest $s_{Post}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the $t$ -ratio for paired samples.
The result matches the fraction without parentheses around the numerator and the denominator, $\frac{(\bar{D} - μ)}{(s_{D} / \sqrt{N})} .$ Please do not forget the parentheses around the numerator and the denominator of a fraction.

$t = \underset{instead of \bar{D}}{\underset{⏟}{\bar{Post}}} - \frac{μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Post}}}} / N^{1 / 2}$

The result matches the $t$ -ratio with the posttest average $\bar{Post}$ instead of the average change score $\bar{D}$ . Please insert the average change score $\bar{D}$ into the $t$ -ratio.
The result matches the $t$ -ratio with the standard deviation for the posttest $s_{Post}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the $t$ -ratio for paired samples.
The result matches the fraction without parentheses around the numerator and the denominator, $\frac{(\bar{D} - μ)}{(s_{D} / \sqrt{N})} .$ Please do not forget the parentheses around the numerator and the denominator of a fraction.
The result matches the $t$ -ratio without the square root around $N$ . Please do not forget the square root around $N$ .

$t = \frac{\underset{instead of \bar{D}}{\underset{⏟}{\bar{Post}}} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Post}}} / \sqrt{N}}$

The result matches the $t$ -ratio with the posttest average $\bar{Post}$ instead of the average change score $\bar{D}$ . Please insert the average change score $\bar{D}$ into the $t$ -ratio.
The result matches the $t$ -ratio with the standard deviation for the posttest $s_{Post}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the $t$ -ratio for paired samples.

$t = \frac{\underset{instead of \bar{D}}{\underset{⏟}{\bar{Post}}} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Post}}} / \sqrt{N}}$

The result matches the $t$ -ratio with the posttest average $\bar{Post}$ instead of the average change score $\bar{D}$ . Please insert the average change score $\bar{D}$ into the $t$ -ratio.
The result matches the $t$ -ratio with the standard deviation for the posttest $s_{Post}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
The result matches the $t$ -ratio, when the null hypothesis $μ$ has been omitted. Please do not forget $μ$ in the $t$ -ratio.

$t = \frac{\underset{instead of \bar{D}}{\underset{⏟}{\bar{Pre}}} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Pre}}} / \sqrt{N}}$

The result matches the $t$ -ratio with the average $\bar{Pre}$ instead of the average change score $\bar{D}$ . Please insert the average change score $\bar{D}$ into the $t$ -ratio.
The result matches the $t$ -ratio with the standard deviation for the pretest $s_{Pre}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the $t$ -ratio for paired samples.

$t = \frac{\underset{instead of \bar{D}}{\underset{⏟}{\bar{Pre}}} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Pre}}} / \sqrt{N}}$

The result matches the $t$ -ratio with the average $\bar{Pre}$ instead of the average change score $\bar{D}$ . Please insert the average change score $\bar{D}$ into the $t$ -ratio.
The result matches the $t$ -ratio with the standard deviation for the pretest $s_{Pre}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
The result matches the $t$ -ratio, when the null hypothesis $μ$ has been omitted. Please do not forget $μ$ in the $t$ -ratio.

$t = \frac{\underset{instead of \bar{D}}{\underset{⏟}{\bar{Post}}} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Post}}} / N^{1 / 2}}$

The result matches the $t$ -ratio with the posttest average $\bar{Post}$ instead of the average change score $\bar{D}$ . Please insert the average change score $\bar{D}$ into the $t$ -ratio.
The result matches the $t$ -ratio with the standard deviation for the posttest $s_{Post}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the $t$ -ratio for paired samples.
The result matches the $t$ -ratio without the square root around $N$ . Please do not forget the square root around $N$ .

$t = \frac{\underset{instead of \bar{D}}{\underset{⏟}{\bar{Post}}} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Post}}} / N^{1 / 2}}$

The result matches the $t$ -ratio with the posttest average $\bar{Post}$ instead of the average change score $\bar{D}$ . Please insert the average change score $\bar{D}$ into the $t$ -ratio.
The result matches the $t$ -ratio with the standard deviation for the posttest $s_{Post}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
The result matches the $t$ -ratio, when the null hypothesis $μ$ has been omitted. Please do not forget $μ$ in the $t$ -ratio.
The result matches the $t$ -ratio without the square root around $N$ . Please do not forget the square root around $N$ .

$t = \frac{\underset{instead of \bar{D}}{\underset{⏟}{\bar{Pre}}} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Pre}}} / N^{1 / 2}}$

The result matches the $t$ -ratio with the average $\bar{Pre}$ instead of the average change score $\bar{D}$ . Please insert the average change score $\bar{D}$ into the $t$ -ratio.
The result matches the $t$ -ratio with the standard deviation for the pretest $s_{Pre}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the $t$ -ratio for paired samples.
The result matches the $t$ -ratio without the square root around $N$ . Please do not forget the square root around $N$ .

$t = \frac{\underset{instead of \bar{D}}{\underset{⏟}{\bar{Pre}}} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{Pre}}} / N^{1 / 2}}$

The result matches the $t$ -ratio with the average $\bar{Pre}$ instead of the average change score $\bar{D}$ . Please insert the average change score $\bar{D}$ into the $t$ -ratio.
The result matches the $t$ -ratio with the standard deviation for the pretest $s_{Pre}$ instead of the standard deviation of the change score $s_{D}$ . Please insert the standard deviation of the change score $s_{D}$ into the $t$ -ratio.
Correctly recognised the problem as a $t$ -test for paired samples.
The result matches the $t$ -ratio, when the null hypothesis $μ$ has been omitted. Please do not forget $μ$ in the $t$ -ratio.
The result matches the $t$ -ratio without the square root around $N$ . Please do not forget the square root around $N$ .

Traps

Avoid these traps

Do not calculate a $t$ -test for independent samples here.
Do not forget the parentheses around the numerator and the denominator of a fraction, $\frac{(\bar{D} - μ)}{(s / \sqrt{N})}$ .
Do not forget the square root around $N$ .
Do not insert the average $\bar{Pre}$ into the $t$ -ratio. Use the change scores instead.
Do not insert the posttest average $\bar{Post}$ into the $t$ -ratio. Use the change scores instead.
Do not insert the standard deviation for the posttest $s_{Post}$ into the $t$ -ratio. Use the change scores instead.
Do not insert the standard deviation for the pretest $s_{Pre}$ into the $t$ -ratio. Use the change scores instead.
Do not omit the null hypothesis $μ$ from the $t$ -ratio.

Question

Does the Cogmed training lead to a relevant improvement (i.e., more than $μ = 2.8$ units) in mean WRAT scores between pre- and posttest? Please report the one-tailed $p$ -value

This is a $t$ -test for paired samples.
The $t$ -ratio is $\frac{\bar{D} - μ}{s_{D} / \sqrt{N}} .$
The one-tailed $p$ -value must be determined.

This is a $t$ -test for paired samples.
The one-tailed $p$ -value must be determined.
The $t$ -ratio is $\frac{\bar{D} - μ}{s_{D} / \sqrt{N}} .$

Feedback

Waiting for response...

Solution(s)

The system accepts the following correct response(s)

$t = \frac{\bar{D} - μ}{s_{D} / \sqrt{N}}$
$p = P (t; N - 1 df)$

Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the $t$ -ratio for paired samples.
Correctly determined the one-tailed $p$ -value.

Incorrect responses

The system detects the following incorrect responses

$t = \frac{\bar{D} - μ}{s_{D} / \sqrt{N}}$
$p = P (\underset{instead of T \leq t}{\underset{⏟}{T = t}}; N - 1 df)$

Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the $t$ -ratio for paired samples.
The result matches the density of the $t$ -distribution.

$t = \frac{\bar{D} - μ}{s_{D} / \sqrt{N}}$
$p = P (\underset{instead of T \leq t}{\underset{⏟}{| T | > | t |}}; N - 1 df)$

Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the $t$ -ratio for paired samples.
The result matches the two-sided $p$ -value.

Traps

Avoid these traps

Do not report the density of the $t$ -distribution.
Do not report the two-sided $p$ -value.

Question

Determine the confidence interval for the change in the children's WRAT scores.

This is a $t$ -test for paired samples.
The formula to calculate the upper bound of the confidence interval is $d + T_{1 - α} (n - 1 df) \cdot s_d / \sqrt{n} .$
Make sure to use the $α$ -quantile of the $t$ -distribution.

Feedback

Waiting for response...

Solution(s)

The system accepts the following correct response(s)

$- \infty to \bar{D} + T_{1 - α} (N - 1 df) \cdot s_{D} / \sqrt{N}$

Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the formula for the upper bound of the confidence interval for a mean value in a $t$ -test.
Correctly used the $α$ -quantile of the $t$ -distribution.

Incorrect responses

The system detects the following incorrect responses

$- \infty to \bar{D} + \underset{instead of T_{α} (N - 1 df)}{\underset{⏟}{t}} \cdot s_{D} / \sqrt{N}$ , with $t = \frac{\bar{D} - μ}{s_{D} / N^{1 / 2}}$ denoting the observed $t$ -statistic.

Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the formula for the upper bound of the confidence interval for a mean value in a $t$ -test.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution instead.
The result matches the confidence interval based on the observed $t$ -statistic without square root around $N .$ Please do not forget the square root around $N$ .

$- \infty to \bar{D} + \underset{instead of T_{α} (N - 1 df)}{\underset{⏟}{t}} \cdot s_{D} / N^{1 / 2}$ , with $t = \frac{\bar{D} - μ}{s_{D} / N^{1 / 2}}$ denoting the observed $t$ -statistic.

Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the formula for the upper bound of the confidence interval for a mean value in a $t$ -test.
The result matches the confidence interval without the square root around $N$ . Please do not forget the square root around $N$ .
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution instead.
The result matches the confidence interval based on the observed $t$ -statistic without square root around $N .$ Please do not forget the square root around $N$ .

$- \infty to \bar{D} + \underset{instead of T_{α} (N - 1 df)}{\underset{⏟}{t}} \cdot s_{D} / \sqrt{N}$ , with $t = \frac{\bar{D} - μ}{s_{D} / \sqrt{N}}$ denoting the observed $t$ -statistic.

Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the formula for the upper bound of the confidence interval for a mean value in a $t$ -test.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution instead.

$- \infty to \bar{D} + \underset{instead of T_{α} (N - 1 df)}{\underset{⏟}{t}} \cdot s_{D} / N^{1 / 2}$ , with $t = \frac{\bar{D} - μ}{s_{D} / \sqrt{N}}$ denoting the observed $t$ -statistic.

Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the formula for the upper bound of the confidence interval for a mean value in a $t$ -test.
The result matches the confidence interval without the square root around $N$ . Please do not forget the square root around $N$ .
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution instead.

$- \infty to \bar{D} + \underset{instead of T_{α} (N - 1 df)}{\underset{⏟}{Φ^{- 1} (α)}} \cdot s_{D} / \sqrt{N}$

Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the formula for the upper bound of the confidence interval for a mean value in a $t$ -test.
The result matches the confidence interval based on the standard normal distribution. Please insert the quantile of the $t$ -distribution $T_{α} (N - 1 df)$ into the formula for the confidence interval.

$- \infty to \bar{D} + \underset{instead of T_{α} (N - 1 df)}{\underset{⏟}{Φ^{- 1} (α)}} \cdot s_{D} / N^{1 / 2}$

Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the formula for the upper bound of the confidence interval for a mean value in a $t$ -test.
The result matches the confidence interval without the square root around $N$ . Please do not forget the square root around $N$ .
The result matches the confidence interval based on the standard normal distribution. Please insert the quantile of the $t$ -distribution $T_{α} (N - 1 df)$ into the formula for the confidence interval.

$- \infty to \bar{D} + T_{1 - α} (N - 1 df) \cdot s_{D} / N^{1 / 2}$

Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the formula for the upper bound of the confidence interval for a mean value in a $t$ -test.
The result matches the confidence interval without the square root around $N$ . Please do not forget the square root around $N$ .
Correctly used the $α$ -quantile of the $t$ -distribution.

$\bar{D} - μ \pm T_{1 - α} (N - 1 df) \cdot s_{D} / \sqrt{N}$

Correctly recognised the problem as a $t$ -test for paired samples.
The result matches the upper and lower bound calculated by SPSS. Please do not forget to add $μ$ to the upper and lower bound of the confidence interval, if you calculate it with SPSS.
Correctly used the $α$ -quantile of the $t$ -distribution.

$\bar{D} - μ \pm T_{1 - α} (N - 1 df) \cdot s_{D} / N^{1 / 2}$

Correctly recognised the problem as a $t$ -test for paired samples.
The result matches the upper and lower bound calculated by SPSS. Please do not forget to add $μ$ to the upper and lower bound of the confidence interval, if you calculate it with SPSS.
The result matches the confidence interval without the square root around $N$ . Please do not forget the square root around $N$ .
Correctly used the $α$ -quantile of the $t$ -distribution.

Traps

Avoid these traps

Do not forget the square root around $N$ .
Do not insert the observed $t$ -statistic into the formula for the confidence interval. Use the quantile of the $t$ -distribution instead.
Do not insert the quantile of the $z$ -distribution into the formula for the confidence interval. Use the quantile of the $t$ -distribution instead.
If you calculate the confindence intervall with SPSS, keep in mind that SPSS subtracts $μ$ from the two bounds of the CI (which must be undone).

Evaluation of working memory training

Question

-22.51...-20.38

-180.01...-132.50

-180.01...-132.50

-108...-88.33

-86.67...-62.50

-86.67...-62.50

-52...-41.66

-20...-18.92

-13.85...-13.24

-11.05...-10.54

-10.84...-9.61

-9.63...-8.92

-6.67...-6.25

-5.41...-5.27

-2.84...-2.72

-2.82...-2.71

-2.75...-2.64

-2.64...-2.53

-2.64...-2.53

-2.63...-2.52

79.23...79.24

79.41...79.42

81.83...81.84

81.95...81.96

197.92...203.03

204.92...210.21

273.71...283.89

283.71...294.26

1319.50...1583.60

1366.16...1639.60

1916...2555.01

1986...2648.34

Question

Result 1

Result 2

0.000...0.000

0.000...0.000

0.000...0.000

Question

-∞to-2.19...-2.07

-∞to-48.89...-34.68

-∞to-9.21...-7.10

-∞to-8.39...-7.45

-∞to-3.43...-3.21

-∞to-3.02...-2.90

-∞to-2.66...-2.56

-∞to-2.54...-2.44

-5.83...-5.71to-4.99...-4.87

-5.46...-5.36to-5.34...-5.24