McClass

*Table 1.* Observed HDRS scores at T0, EOT, and $D = T0 - EOT$
HDRS	T0	EOT	$D$
Average	$27.4$	$23.6$	$3.8$
SD	$6.7$	$8.0$	$3.2$

Question

Does rfCBT lead to a relevant reduction (i.e., more than $μ = 2.9$ units) in mean HDRS scores between baseline (T0) and End of Treatment (EOT)? Please report the $t$ -ratio.

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

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.

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

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

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

The result matches the $t$ -ratio with the standard deviation for EOT $s_{EOT}$ 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.

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

The result matches the $t$ -ratio with the standard deviation for EOT $s_{EOT}$ 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.

$t = | \frac{\bar{D} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{T0}}} / \sqrt{N}} |$

The result matches the $t$ -ratio with the standard deviation for T0 $s_{T0}$ 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.

$t = \frac{\bar{D} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{T0}}} / \sqrt{N}}$

The result matches the $t$ -ratio with the standard deviation for T0 $s_{T0}$ 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.

$t = \underset{instead of a paired t ‑ test}{\underset{⏟}{\frac{\bar{T0} - \bar{EOT}}{{[s_{pool}^{2} \cdot (1 / N + 1 / N)]}^{1 / 2}}}}$ , with $s_{pool}^{2} = \frac{(N - 1) \cdot s_{T0}^{2} + (N - 1) \cdot s_{EOT}^{2}}{N + N - 2}$ 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 $t$ -ratio without the square root in the denominator. Please do not forget to take the square root of the error variance $\sqrt{s_{pool}^{2} \cdot (1 / N + 1 / N)}$ .

$t = \underset{instead of a paired t ‑ test}{\underset{⏟}{\frac{\bar{T0} - \bar{EOT}}{\sqrt{s_{pool}^{2} \cdot (1 / N + 1 / N)}}}}$ , with $s_{pool}^{2} = \frac{(N - 1) \cdot s_{T0}^{2} + (N - 1) \cdot s_{EOT}^{2}}{N + N - 2}$ denoting the pooled variance

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

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

The result matches the $t$ -ratio with the standard deviation for EOT $s_{EOT}$ 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 with the null hypothesis $μ$ omitted. Please do not forget $μ$ in the the $t$ -ratio.

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

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.

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

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.

$t = \frac{\bar{D} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{T0}}} / \sqrt{N}}$

The result matches the $t$ -ratio with the standard deviation for T0 $s_{T0}$ 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 with the null hypothesis $μ$ omitted. Please do not forget $μ$ in the the $t$ -ratio.

$t = | \bar{D} - \frac{μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{T0}}}} / \sqrt{N} |$

The result matches the $t$ -ratio with the standard deviation for T0 $s_{T0}$ 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.

$t = \bar{D} - \frac{μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{T0}}}} / \sqrt{N}$

The result matches the $t$ -ratio with the standard deviation for T0 $s_{T0}$ 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.

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

The result matches the $t$ -ratio with the standard deviation for EOT $s_{EOT}$ 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.

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

The result matches the $t$ -ratio with the standard deviation for EOT $s_{EOT}$ 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.

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

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 $N$ .

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

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 $N$ .

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

The result matches the $t$ -ratio with the standard deviation for EOT $s_{EOT}$ 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 $N$ .

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

The result matches the $t$ -ratio with the standard deviation for EOT $s_{EOT}$ 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 $N$ .

$t = | \bar{D} - \frac{μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{T0}}}} / N^{1 / 2} |$

The result matches the $t$ -ratio with the standard deviation for T0 $s_{T0}$ 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 $N$ .

$t = \bar{D} - \frac{μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{T0}}}} / N^{1 / 2}$

The result matches the $t$ -ratio with the standard deviation for T0 $s_{T0}$ 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 $N$ .

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

The result matches the $t$ -ratio with the standard deviation for EOT $s_{EOT}$ 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 $N$ .

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

The result matches the $t$ -ratio with the standard deviation for EOT $s_{EOT}$ 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 $N$ .

$t = \underset{instead of a paired t ‑ test}{\underset{⏟}{\frac{\bar{T0} - \bar{EOT}}{\sqrt{s_{pool}^{2} \cdot \frac{1}{2 N}}}}}$ , with $s_{pool}^{2} = \frac{(N - 1) \cdot s_{T0}^{2} + (N - 1) \cdot s_{EOT}^{2}}{N + N - 2}$ 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 $\frac{1}{2 N}$ under the square root. Please keep in mind that $\frac{1}{N} + \frac{1}{N} \neq \frac{1}{2 N}$ .

$t = | \frac{\bar{D} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{T0}}} / N^{1 / 2}} |$

The result matches the $t$ -ratio with the standard deviation for T0 $s_{T0}$ 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 $N$ .

$t = \frac{\bar{D} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{T0}}} / N^{1 / 2}}$

The result matches the $t$ -ratio with the standard deviation for T0 $s_{T0}$ 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 $N$ .

$t = \underset{instead of a paired t ‑ test}{\underset{⏟}{\frac{\bar{T0} - \bar{EOT}}{{(s_{pool}^{2} \cdot \frac{1}{2 N})}^{1 / 2}}}}$ , with $s_{pool}^{2} = \frac{(N - 1) \cdot s_{T0}^{2} + (N - 1) \cdot s_{EOT}^{2}}{N + N - 2}$ 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 $t$ -ratio without the square root in the denominator. Please do not forget to take the square root of the error variance $\sqrt{s_{pool}^{2} \cdot (1 / N + 1 / N)}$ .
The result matches the expression for the $t$ -ratio for independent samples with $\frac{1}{2 N}$ under the square root. 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 with the null hypothesis $μ$ omitted. Please do not forget $μ$ in the the $t$ -ratio.

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

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 $N$ .

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

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 $N$ .

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

The result matches the $t$ -ratio with the EOT average $\bar{EOT}$ 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 EOT $s_{EOT}$ 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.

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

The result matches the $t$ -ratio with the EOT average $\bar{EOT}$ 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 EOT $s_{EOT}$ 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.

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

The result matches the $t$ -ratio with the EOT average $\bar{EOT}$ 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 EOT $s_{EOT}$ 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 with the null hypothesis $μ$ omitted. Please do not forget $μ$ in the the $t$ -ratio.

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

The result matches the $t$ -ratio with the standard deviation for EOT $s_{EOT}$ 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 with the null hypothesis $μ$ omitted. Please do not forget $μ$ in the the $t$ -ratio.
The result matches the $t$ -ratio without the square root around $N$ . Please do not forget the square root around $N$ .

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

The result matches the $t$ -ratio with the EOT average $\bar{EOT}$ 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 EOT $s_{EOT}$ 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.

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

The result matches the $t$ -ratio with the EOT average $\bar{EOT}$ 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 EOT $s_{EOT}$ 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.

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

The result matches the $t$ -ratio with the EOT average $\bar{EOT}$ 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 EOT $s_{EOT}$ 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 $N$ .

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

The result matches the $t$ -ratio with the EOT average $\bar{EOT}$ 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 EOT $s_{EOT}$ 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 $N$ .

$t = \frac{\bar{D} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{T0}}} / N^{1 / 2}}$

The result matches the $t$ -ratio with the standard deviation for T0 $s_{T0}$ 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 with the null hypothesis $μ$ omitted. Please do not forget $μ$ in the the $t$ -ratio.
The result matches the $t$ -ratio without the square root around $N$ . Please do not forget the square root around $N$ .

$t = | \frac{\underset{instead of \bar{D}}{\underset{⏟}{\bar{T0}}} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{T0}}} / \sqrt{N}} |$

The result matches the $t$ -ratio with the T0 average $\bar{T0}$ 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 T0 $s_{T0}$ 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.

$t = \frac{\underset{instead of \bar{D}}{\underset{⏟}{\bar{T0}}} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{T0}}} / \sqrt{N}}$

The result matches the $t$ -ratio with the T0 average $\bar{T0}$ 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 T0 $s_{T0}$ 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.

$t = \frac{\underset{instead of \bar{D}}{\underset{⏟}{\bar{T0}}} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{T0}}} / \sqrt{N}}$

The result matches the $t$ -ratio with the T0 average $\bar{T0}$ 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 T0 $s_{T0}$ 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 with the null hypothesis $μ$ omitted. Please do not forget $μ$ in the the $t$ -ratio.

$t = | \underset{instead of \bar{D}}{\underset{⏟}{\bar{T0}}} - \frac{μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{T0}}}} / \sqrt{N} |$

The result matches the $t$ -ratio with the T0 average $\bar{T0}$ 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 T0 $s_{T0}$ 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.

$t = \underset{instead of \bar{D}}{\underset{⏟}{\bar{T0}}} - \frac{μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{T0}}}} / \sqrt{N}$

The result matches the $t$ -ratio with the T0 average $\bar{T0}$ 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 T0 $s_{T0}$ 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.

$t = | \underset{instead of \bar{D}}{\underset{⏟}{\bar{T0}}} - \frac{μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{T0}}}} / N^{1 / 2} |$

The result matches the $t$ -ratio with the T0 average $\bar{T0}$ 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 T0 $s_{T0}$ 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 $N$ .

$t = \underset{instead of \bar{D}}{\underset{⏟}{\bar{T0}}} - \frac{μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{T0}}}} / N^{1 / 2}$

The result matches the $t$ -ratio with the T0 average $\bar{T0}$ 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 T0 $s_{T0}$ 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 $N$ .

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

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

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

The result matches the $t$ -ratio with the EOT average $\bar{EOT}$ 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 EOT $s_{EOT}$ 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 $N$ .

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

The result matches the $t$ -ratio with the EOT average $\bar{EOT}$ 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 EOT $s_{EOT}$ 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 $N$ .

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

The result matches the $t$ -ratio with the EOT average $\bar{EOT}$ 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 EOT $s_{EOT}$ 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 with the null hypothesis $μ$ omitted. Please do not forget $μ$ in the the $t$ -ratio.
The result matches the $t$ -ratio without the square root around $N$ . Please do not forget the square root around $N$ .

$t = | \frac{\underset{instead of \bar{D}}{\underset{⏟}{\bar{T0}}} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{T0}}} / N^{1 / 2}} |$

The result matches the $t$ -ratio with the T0 average $\bar{T0}$ 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 T0 $s_{T0}$ 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 $N$ .

$t = \frac{\underset{instead of \bar{D}}{\underset{⏟}{\bar{T0}}} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{T0}}} / N^{1 / 2}}$

The result matches the $t$ -ratio with the T0 average $\bar{T0}$ 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 T0 $s_{T0}$ 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 $N$ .

$t = \frac{\underset{instead of \bar{D}}{\underset{⏟}{\bar{T0}}} - μ}{\underset{instead of s_{D}}{\underset{⏟}{s_{T0}}} / N^{1 / 2}}$

The result matches the $t$ -ratio with the T0 average $\bar{T0}$ 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 T0 $s_{T0}$ 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 with the null hypothesis $μ$ omitted. Please do not forget $μ$ in the 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 EOT average $\bar{EOT}$ into the $t$ -ratio. Use the change scores instead.
Do not insert the T0 average $\bar{T0}$ into the $t$ -ratio. Use the change scores instead.
Do not insert the standard deviation for EOT $s_{EOT}$ into the $t$ -ratio. Use the change scores instead.
Do not insert the standard deviation for T0 $s_{T0}$ into the $t$ -ratio. Use the change scores instead.
Do not omit the null hypothesis $μ$ in the $t$ -ratio.

Question

Does rfCBT lead to a relevant reduction (i.e., more than $μ = 2.9$ units) in mean HDRS scores between baseline (T0) and End of Treatment (EOT)? Please report the two-tailed $p$ -value.

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

Feedback

Waiting for response...

Solution(s)

The system accepts the following correct response(s)

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

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

Incorrect responses

The system detects the following incorrect responses

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

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

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

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.

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

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

Traps

Avoid these traps

Do not report the density of the $t$ -distribution.
Do not report the lower one-tailed $p$ -value.
Do not report the upper one-tailed $p$ -value.

Question

Determine the confidence interval for the average change in the patients’ HDRS scores.

This is a $t$ -test for paired samples.
The formula to calculate the lower and upper bound of the confidence interval is $\bar{D} \pm T_{1 - α / 2} (N - 1 df) \cdot s_{D} / \sqrt{N}$ .
Make sure to use the $1 - α / 2$ -quantile of the $t$ -distribution.

Feedback

Waiting for response...

Solution(s)

The system accepts the following correct response(s)

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

Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the formula for the upper and lower bound of the confidence interval for a mean value.
Correctly used the $1 - α / 2$ -quantile of the $t$ -distribution.

Incorrect responses

The system detects the following incorrect responses

$\bar{D} \pm \underset{instead of T_{1 - α / 2} (N - 1 df)}{\underset{⏟}{t}} \cdot s_{D} / \sqrt{N}$ , 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 and lower bound of the confidence interval for a mean value.
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 $t$ -ratio without the square root around $N$ .

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

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 $1 - α / 2$ -quantile of the $t$ -distribution.

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

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 square root around $N$ . Please do not forget the square root around $N$ .
Correctly used the $1 - α / 2$ -quantile of the $t$ -distribution.

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

Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the formula for the upper and lower bound of the confidence interval for a mean value.
The result matches the confidence interval based on the standard Normal distribution. Please insert the quantile of the $t$ -distribution into the formula for the confidence interval.

$\bar{D} \pm \underset{instead of T_{1 - α / 2} (N - 1 df)}{\underset{⏟}{t}} \cdot s_{D} / N^{1 / 2}$ , 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 and lower bound of the confidence interval for a mean value.
The result matches the confidence interval without 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 $t$ -ratio without the square root around $N$ .

$\bar{D} \pm \underset{instead of T_{1 - α / 2} (N - 1 df)}{\underset{⏟}{t}} \cdot s_{D} / \sqrt{N}$ , 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 and lower bound of the confidence interval for a mean value.
The result matches the confidence interval based on the observed $t$ -statistic. Please use the quantile of the $t$ -distribution instead.

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

Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the formula for the upper and lower bound of the confidence interval for a mean value.
The result matches the confidence interval without square root around $N$ . Please do not forget the square root around $N$ .
Correctly used the $1 - α / 2$ -quantile of the $t$ -distribution.

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

Correctly recognised the problem as a $t$ -test for paired samples.
Correctly identified the formula for the upper and lower bound of the confidence interval for a mean value.
The result matches the confidence interval without 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 into the formula for the confidence interval.

$\bar{D} \pm \underset{instead of T_{1 - α / 2} (N - 1 df)}{\underset{⏟}{t}} \cdot s_{D} / N^{1 / 2}$ , 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 and lower bound of the confidence interval for a mean value.
The result matches the confidence interval without 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.

Traps

Avoid these traps

Determine the confidence interval based on the quantiles of the $t$ -distribution.
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.
If you calculate the confindence interval with SPSS, keep in mind that SPSS subtracts $μ$ from the two bounds of the CI (which must be undone).

Phase II clinical study

Question

Result 1

Result 2

1.81...1.88

1.81...1.88

0.73...0.75

0.73...0.75

0.87...0.90

0.87...0.90

1.52...1.56

2.39...2.44

3.13...3.17

3.65...3.67

3.65...3.67

3.71...3.77

3.72...3.74

3.72...3.74

3.73...3.75

3.73...3.75

3.76...3.78

3.76...3.78

3.78...3.80

3.78...3.80

3.78...3.80

3.78...3.80

4.68...5

4.68...5

4.78...4.90

5.56...6

5.56...6

6.09...6.27

7.73...7.92

11.12...12.86

11.12...12.86

17.13...17.29

17.13...17.29

19.52...19.71

19.94...21.12

23.57...23.59

23.57...23.59

23.62...23.64

23.62...23.64

23.68...25.34

24.04...24.31

24.04...24.31

26.89...27.17

27.36...27.38

27.36...27.38

27.42...27.44

27.42...27.44

47.37...54.29

109.10...115.23

109.10...115.23

124.36...131.34

153.31...163.67

153.31...163.67

171.43...182.94

Question

0.068...0.076

0.034...0.038

0.070...0.078

0.96...0.97

Question

2.82...2.83to4.76...4.77

-2.41...-1.55to9.14...10.00

-0.08...-0.07to1.86...1.87

0.74...0.75to1.04...1.05

2.84...2.86to4.73...4.75

2.85...2.99to4.60...4.74

2.89...2.93to4.66...4.70

3.64...3.65to3.94...3.95

3.65...3.66to3.93...3.94

3.65...3.67to3.92...3.94