McClass

Question

Does VR training lead to faster surgery times than traditional Box training? Please determine the $χ^{2}$ -statistic.

First determine the pooled success proportion, then the $z$ -statistic. Finally, raise $z$ to the square to obtain $χ^{2}$ .
The pooled proportion of successes is $p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$ .
The $z$ -statistic is $\frac{p_{VR} - p_{Box}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})}}$
The $χ$ -statistic is based on the $z$ -statistic.

Feedback

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Solution(s)

The system accepts the following correct response(s)

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{p_{VR} - p_{Box}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
Correctly identified the $z$ -statistic.
Correctly identified the $χ^{2}$ -statistic.

Incorrect responses

The system detects the following incorrect responses

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{p_{VR} - p_{Box}}{\sqrt{\underset{instead of p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})}{\underset{⏟}{p_{pool} 1 - p_{pool} 1 / n_{VR} + 1 / n_{Box}}}}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
Correctly identified the $z$ -statistic.
The results corresponds to the $z$ -statistic without the parentheses around the different elements in $p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})$ .
Correctly identified the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{p_{VR} - p_{Box}}{\sqrt{\underset{instead of p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})}{\underset{⏟}{p_{pool} 1 - p_{pool} 1 / n_{VR} + 1 / n_{Box}}}}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
Correctly identified the $z$ -statistic.
The results corresponds to the $z$ -statistic without the parentheses around the different elements in $p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})$ .
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{p_{VR} - p_{Box}}{\sqrt{\underset{instead of p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})}{\underset{⏟}{p_{pool} 1 - p_{pool} 1 / n_{VR} + 1 / n_{Box}}}}}$
$χ^{2} = {| z |}^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
Correctly identified the $z$ -statistic.
The results corresponds to the $z$ -statistic without the parentheses around the different elements in $p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})$ .
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{p_{VR} - p_{Box}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) 1 / n_{VR} + 1 / n_{Box}}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
Correctly identified the $z$ -statistic.
The results corresponds to the $z$ -statistic without the parentheses around $(\frac{1}{n_{VR}} + \frac{1}{n_{Box}})$ .
Correctly identified the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{p_{VR} - p_{Box}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) 1 / n_{VR} + 1 / n_{Box}}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
Correctly identified the $z$ -statistic.
The results corresponds to the $z$ -statistic without the parentheses around $(\frac{1}{n_{VR}} + \frac{1}{n_{Box}})$ .
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{p_{VR} - p_{Box}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) 1 / n_{VR} + 1 / n_{Box}}}$
$χ^{2} = {| z |}^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
Correctly identified the $z$ -statistic.
The results corresponds to the $z$ -statistic without the parentheses around $(\frac{1}{n_{VR}} + \frac{1}{n_{Box}})$ .
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = 1 / \frac{p_{VR} - p_{Box}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) \cdot \underset{instead of 1 / n_{VR} + 1 / n_{Box}}{\underset{⏟}{\frac{1}{n_{VR} + n_{Box}}}}}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The result corresponds to the reciprocal of the test statistic $(1 / z)$ . Please double check the nominator and denominator of the test statistic.
The addition between $\frac{1}{n_{VR}}$ and $\frac{1}{n_{Box}}$ was not correct. Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{Box}} \neq \frac{1}{n_{VR} + n_{Box}} .$
Correctly identified the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{p_{VR} - p_{Box}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
Correctly identified the $z$ -statistic.
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{p_{VR} - p_{Box}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})}}$
$χ^{2} = {| z |}^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
Correctly identified the $z$ -statistic.
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = 1 / \frac{p_{VR} - p_{Box}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) \cdot \underset{instead of 1 / n_{VR} + 1 / n_{Box}}{\underset{⏟}{\frac{1}{n_{VR} + n_{Box}}}}}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The result corresponds to the reciprocal of the test statistic $(1 / z)$ . Please double check the nominator and denominator of the test statistic.
The addition between $\frac{1}{n_{VR}}$ and $\frac{1}{n_{Box}}$ was not correct. Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{Box}} \neq \frac{1}{n_{VR} + n_{Box}} .$
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = 1 / \frac{p_{VR} - p_{Box}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) \cdot \underset{instead of 1 / n_{VR} + 1 / n_{Box}}{\underset{⏟}{\frac{1}{n_{VR} + n_{Box}}}}}}$
$χ^{2} = {| z |}^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The result corresponds to the reciprocal of the test statistic $(1 / z)$ . Please double check the nominator and denominator of the test statistic.
The addition between $\frac{1}{n_{VR}}$ and $\frac{1}{n_{Box}}$ was not correct. Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{Box}} \neq \frac{1}{n_{VR} + n_{Box}} .$
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{p_{VR} - p_{Box}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) \cdot \underset{instead of 1 / n_{VR} + 1 / n_{Box}}{\underset{⏟}{\frac{1}{n_{VR} + n_{Box}}}}}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
Correctly identified the $z$ -statistic.
The addition between $\frac{1}{n_{VR}}$ and $\frac{1}{n_{Box}}$ was not correct. Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{Box}} \neq \frac{1}{n_{VR} + n_{Box}} .$
Correctly identified the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{p_{VR} - p_{Box}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) \cdot \underset{instead of 1 / n_{VR} + 1 / n_{Box}}{\underset{⏟}{\frac{1}{n_{VR} + n_{Box}}}}}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
Correctly identified the $z$ -statistic.
The addition between $\frac{1}{n_{VR}}$ and $\frac{1}{n_{Box}}$ was not correct. Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{Box}} \neq \frac{1}{n_{VR} + n_{Box}} .$
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{p_{VR} - p_{Box}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) \cdot \underset{instead of 1 / n_{VR} + 1 / n_{Box}}{\underset{⏟}{\frac{1}{n_{VR} + n_{Box}}}}}}$
$χ^{2} = {| z |}^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
Correctly identified the $z$ -statistic.
The addition between $\frac{1}{n_{VR}}$ and $\frac{1}{n_{Box}}$ was not correct. Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{Box}} \neq \frac{1}{n_{VR} + n_{Box}} .$
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = 1 / \frac{p_{VR} - p_{Box}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The result corresponds to the reciprocal of the test statistic $(1 / z)$ . Please double check the nominator and denominator of the test statistic.
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = 1 / \frac{p_{VR} - p_{Box}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})}}$
$χ^{2} = {| z |}^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The result corresponds to the reciprocal of the test statistic $(1 / z)$ . Please double check the nominator and denominator of the test statistic.
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = 1 / \frac{p_{VR} - p_{Box}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The result corresponds to the reciprocal of the test statistic $(1 / z)$ . Please double check the nominator and denominator of the test statistic.
Correctly identified the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{\underset{instead of p_{VR} - p_{Box}}{\underset{⏟}{p_{VR} + p_{Box}}}}{\sqrt{\underset{instead of p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})}{\underset{⏟}{p_{pool} 1 - p_{pool} 1 / n_{VR} + 1 / n_{Box}}}}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The results corresponds to the $z$ -statistic where $p_{Box}$ was added to $p_{VR}$ , rather than subtracted from it.
The results corresponds to the $z$ -statistic without the parentheses around the different elements in $p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})$ .
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{\underset{instead of p_{VR} - p_{Box}}{\underset{⏟}{p_{VR} + p_{Box}}}}{\sqrt{\underset{instead of p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})}{\underset{⏟}{p_{pool} 1 - p_{pool} 1 / n_{VR} + 1 / n_{Box}}}}}$
$χ^{2} = {| z |}^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The results corresponds to the $z$ -statistic where $p_{Box}$ was added to $p_{VR}$ , rather than subtracted from it.
The results corresponds to the $z$ -statistic without the parentheses around the different elements in $p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})$ .
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{\underset{instead of p_{VR} - p_{Box}}{\underset{⏟}{p_{VR} + p_{Box}}}}{\sqrt{\underset{instead of p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})}{\underset{⏟}{p_{pool} 1 - p_{pool} 1 / n_{VR} + 1 / n_{Box}}}}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The results corresponds to the $z$ -statistic where $p_{Box}$ was added to $p_{VR}$ , rather than subtracted from it.
The results corresponds to the $z$ -statistic without the parentheses around the different elements in $p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})$ .
Correctly identified the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = 1 / \frac{p_{VR} - p_{Box}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) 1 / n_{VR} + 1 / n_{Box}}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The result corresponds to the reciprocal of the test statistic $(1 / z)$ . Please double check the nominator and denominator of the test statistic.
The results corresponds to the $z$ -statistic without the parentheses around $(\frac{1}{n_{VR}} + \frac{1}{n_{Box}})$ .
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = 1 / \frac{p_{VR} - p_{Box}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) 1 / n_{VR} + 1 / n_{Box}}}$
$χ^{2} = {| z |}^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The result corresponds to the reciprocal of the test statistic $(1 / z)$ . Please double check the nominator and denominator of the test statistic.
The results corresponds to the $z$ -statistic without the parentheses around $(\frac{1}{n_{VR}} + \frac{1}{n_{Box}})$ .
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = 1 / \frac{p_{VR} - p_{Box}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) 1 / n_{VR} + 1 / n_{Box}}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The result corresponds to the reciprocal of the test statistic $(1 / z)$ . Please double check the nominator and denominator of the test statistic.
The results corresponds to the $z$ -statistic without the parentheses around $(\frac{1}{n_{VR}} + \frac{1}{n_{Box}})$ .
Correctly identified the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{\underset{instead of p_{VR} - p_{Box}}{\underset{⏟}{p_{VR} + p_{Box}}}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) 1 / n_{VR} + 1 / n_{Box}}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The results corresponds to the $z$ -statistic where $p_{Box}$ was added to $p_{VR}$ , rather than subtracted from it.
The results corresponds to the $z$ -statistic without the parentheses around $(\frac{1}{n_{VR}} + \frac{1}{n_{Box}})$ .
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{\underset{instead of p_{VR} - p_{Box}}{\underset{⏟}{p_{VR} + p_{Box}}}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) 1 / n_{VR} + 1 / n_{Box}}}$
$χ^{2} = {| z |}^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The results corresponds to the $z$ -statistic where $p_{Box}$ was added to $p_{VR}$ , rather than subtracted from it.
The results corresponds to the $z$ -statistic without the parentheses around $(\frac{1}{n_{VR}} + \frac{1}{n_{Box}})$ .
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = 1 / \frac{p_{VR} - p_{Box}}{\sqrt{\underset{instead of p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})}{\underset{⏟}{p_{pool} 1 - p_{pool} 1 / n_{VR} + 1 / n_{Box}}}}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The result corresponds to the reciprocal of the test statistic $(1 / z)$ . Please double check the nominator and denominator of the test statistic.
The results corresponds to the $z$ -statistic without the parentheses around the different elements in $p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})$ .
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = 1 / \frac{p_{VR} - p_{Box}}{\sqrt{\underset{instead of p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})}{\underset{⏟}{p_{pool} 1 - p_{pool} 1 / n_{VR} + 1 / n_{Box}}}}}$
$χ^{2} = {| z |}^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The result corresponds to the reciprocal of the test statistic $(1 / z)$ . Please double check the nominator and denominator of the test statistic.
The results corresponds to the $z$ -statistic without the parentheses around the different elements in $p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})$ .
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{\underset{instead of p_{VR} - p_{Box}}{\underset{⏟}{p_{VR} + p_{Box}}}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The results corresponds to the $z$ -statistic where $p_{Box}$ was added to $p_{VR}$ , rather than subtracted from it.
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{\underset{instead of p_{VR} - p_{Box}}{\underset{⏟}{p_{VR} + p_{Box}}}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})}}$
$χ^{2} = {| z |}^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The results corresponds to the $z$ -statistic where $p_{Box}$ was added to $p_{VR}$ , rather than subtracted from it.
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{\underset{instead of p_{VR} - p_{Box}}{\underset{⏟}{p_{VR} + p_{Box}}}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) \cdot \underset{instead of 1 / n_{VR} + 1 / n_{Box}}{\underset{⏟}{\frac{1}{n_{VR} + n_{Box}}}}}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The results corresponds to the $z$ -statistic where $p_{Box}$ was added to $p_{VR}$ , rather than subtracted from it.
The addition between $\frac{1}{n_{VR}}$ and $\frac{1}{n_{Box}}$ was not correct. Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{Box}} \neq \frac{1}{n_{VR} + n_{Box}} .$
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{\underset{instead of p_{VR} - p_{Box}}{\underset{⏟}{p_{VR} + p_{Box}}}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) \cdot \underset{instead of 1 / n_{VR} + 1 / n_{Box}}{\underset{⏟}{\frac{1}{n_{VR} + n_{Box}}}}}}$
$χ^{2} = {| z |}^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The results corresponds to the $z$ -statistic where $p_{Box}$ was added to $p_{VR}$ , rather than subtracted from it.
The addition between $\frac{1}{n_{VR}}$ and $\frac{1}{n_{Box}}$ was not correct. Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{Box}} \neq \frac{1}{n_{VR} + n_{Box}} .$
The result corresponds to the $z$ -statistic instead of the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{\underset{instead of p_{VR} - p_{Box}}{\underset{⏟}{p_{VR} + p_{Box}}}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) 1 / n_{VR} + 1 / n_{Box}}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The results corresponds to the $z$ -statistic where $p_{Box}$ was added to $p_{VR}$ , rather than subtracted from it.
The results corresponds to the $z$ -statistic without the parentheses around $(\frac{1}{n_{VR}} + \frac{1}{n_{Box}})$ .
Correctly identified the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = 1 / \frac{p_{VR} - p_{Box}}{\sqrt{\underset{instead of p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})}{\underset{⏟}{p_{pool} 1 - p_{pool} 1 / n_{VR} + 1 / n_{Box}}}}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The result corresponds to the reciprocal of the test statistic $(1 / z)$ . Please double check the nominator and denominator of the test statistic.
The results corresponds to the $z$ -statistic without the parentheses around the different elements in $p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})$ .
Correctly identified the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{\underset{instead of p_{VR} - p_{Box}}{\underset{⏟}{p_{VR} + p_{Box}}}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) \cdot (\frac{1}{n_{VR}} + \frac{1}{n_{Box}})}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The results corresponds to the $z$ -statistic where $p_{Box}$ was added to $p_{VR}$ , rather than subtracted from it.
Correctly identified the $χ^{2}$ -statistic.

$p_{pool} = \frac{s_{VR} + s_{Box}}{n_{VR} + n_{Box}}$
$z = \frac{\underset{instead of p_{VR} - p_{Box}}{\underset{⏟}{p_{VR} + p_{Box}}}}{\sqrt{p_{pool} \cdot (1 - p_{pool}) \cdot \underset{instead of 1 / n_{VR} + 1 / n_{Box}}{\underset{⏟}{\frac{1}{n_{VR} + n_{Box}}}}}}$
$χ^{2} = z^{2}$

Correctly identified the main steps of the calculation.
Correctly identified the pooled proportion of successes.
The results corresponds to the $z$ -statistic where $p_{Box}$ was added to $p_{VR}$ , rather than subtracted from it.
The addition between $\frac{1}{n_{VR}}$ and $\frac{1}{n_{Box}}$ was not correct. Please keep in mind that $\frac{1}{n_{VR}} + \frac{1}{n_{Box}} \neq \frac{1}{n_{VR} + n_{Box}} .$
Correctly identified the $χ^{2}$ -statistic.

Traps

Avoid these traps

Do not flip the numerator and denominator of the test statistic.
Do not forget to add parentheses around $(\frac{1}{n_{VR}} + \frac{1}{n_{Box}})$ .
Remember the parentheses around the different elements in the denominator.
Please do not forget school math, $\frac{1}{n_{VR}} + \frac{1}{n_{Box}} \neq \frac{1}{n_{VR} + n_{Box}} .$
Do not forget to raise the $z$ -value to the square to obtain the $χ^{2}$ -statistic.
Do not forget to raise the $z$ -value to the square to obtain the $χ^{2}$ -statistic.
The numerator of the test statistic includes the difference between the success proportions $p_{VR}$ and $p_{Box}$ .

Training of surgical skills

Question

0.20...0.81

0...0.02

0.06...0.13

0.06...0.13

0.08...0.32

0.29...0.57

0.29...0.57

0.30...1.45

0.45...0.90

0.45...0.90

0.55...1.21

0.55...1.21

0.69...3.24

0.83...1.80

0.83...1.80

1.11...2.21

1.11...2.21

1.23...4.85

1.33...1.41

1.33...1.41

1.77...1.99

1.77...3.41

1.77...3.41

3.16...11.57

5.64...6.26

5.64...6.26

7.88...14.40

7.88...14.40

8.72...10.01

8.72...10.01

15.99...20.01

15.99...20.01

31.88...39.07

62.23...207.36

76.16...100.01

255.99...400.01