Evaluation of working memory training

$t={\displaystyle \frac{\overline{D}-\mu }{s_D/\sqrt{N}}}$

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

$t=\underset{\text{instead of}\text{a paired}t\text{-test}}{\underbrace{{\displaystyle \frac{\overline{\text{Pre}}-\overline{\text{Post}}}{\sqrt{s_{\text{pool}}^2\cdot \frac{1}{2N}}}}}}$, with $s_{\text{pool}}^2={\displaystyle \frac{\left(N-1\right)\cdot s_{\text{Pre}}^2+\left(N-1\right)\cdot s_{\text{Post}}^2}{N+N-2}}$ 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 corresponds to the expression for the $t$-ratio for independent samples with $\frac{1}{2N}\text{.}$ Please keep in mind that $\frac{1}{N}+\frac{1}{N}\ne \frac{1}{2N}\text{.}$

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

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

$t={\displaystyle \frac{\overline{D}-\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Post}}}}/\sqrt{N}}}$

• The result corresponds to the $t$-ratio with the standard deviation for the posttest $s_{\text{Post}}$ 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.

$t={\displaystyle \frac{\overline{D}-\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Pre}}}}/\sqrt{N}}}$

• The result corresponds to the $t$-ratio with the standard deviation for the pretest $s_{\text{Pre}}$ 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.

$t=\overline{D}-{\displaystyle \frac{\mu }{s_D}}/\sqrt{N}$

• Correctly recognised the problem as a $t$-test for paired samples.
• Correctly identified the $t$-ratio for paired samples.
• The result corresponds to the fraction without parentheses around the numerator and the denominator, ${\displaystyle \frac{\left(\overline{D}-\mu \right)}{\left(s_D/\sqrt{N}\right)}}\text{.}$ Please do not forget the parentheses around the numerator and the denominator of a fraction.

$t=\overline{D}-{\displaystyle \frac{\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Pre}}}}}}/\sqrt{N}$

• The result corresponds to the $t$-ratio with the standard deviation for the pretest $s_{\text{Pre}}$ 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 corresponds to the fraction without parentheses around the numerator and the denominator, ${\displaystyle \frac{\left(\overline{D}-\mu \right)}{\left(s_D/\sqrt{N}\right)}}\text{.}$ Please do not forget the parentheses around the numerator and the denominator of a fraction.

$t=\overline{D}-{\displaystyle \frac{\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Post}}}}}}/\sqrt{N}$

• The result corresponds to the $t$-ratio with the standard deviation for the posttest $s_{\text{Post}}$ 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 corresponds to the fraction without parentheses around the numerator and the denominator, ${\displaystyle \frac{\left(\overline{D}-\mu \right)}{\left(s_D/\sqrt{N}\right)}}\text{.}$ Please do not forget the parentheses around the numerator and the denominator of a fraction.

$t=\overline{D}-{\displaystyle \frac{\mu }{s_D}}/N^{\cancel{1/2}}$

• Correctly recognised the problem as a $t$-test for paired samples.
• Correctly identified the $t$-ratio for paired samples.
• The result corresponds to the fraction without parentheses around the numerator and the denominator, ${\displaystyle \frac{\left(\overline{D}-\mu \right)}{\left(s_D/\sqrt{N}\right)}}\text{.}$ Please do not forget the parentheses around the numerator and the denominator of a fraction.
• The result corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.

$t=\overline{D}-{\displaystyle \frac{\mu }{s_D}}/N^{\cancel{1/2}}$

• Correctly recognised the problem as a $t$-test for paired samples.
• Correctly identified the $t$-ratio for paired samples.
• The result corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.
• The result corresponds to the fraction without parentheses around the numerator and the denominator, ${\displaystyle \frac{\left(\overline{D}-\mu \right)}{\left(s_D/N^{1/2}\right)}}\text{.}$ Please do not forget the parentheses around the numerator and the denominator of a fraction.

$t=\overline{D}-{\displaystyle \frac{\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Pre}}}}}}/N^{\cancel{1/2}}$

• The result corresponds to the $t$-ratio with the standard deviation for the pretest $s_{\text{Pre}}$ 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 corresponds to the fraction without parentheses around the numerator and the denominator, ${\displaystyle \frac{\left(\overline{D}-\mu \right)}{\left(s_D/\sqrt{N}\right)}}\text{.}$ Please do not forget the parentheses around the numerator and the denominator of a fraction.
• The result corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.

$t=\overline{D}-{\displaystyle \frac{\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Pre}}}}}}/N^{\cancel{1/2}}$

• The result corresponds to the $t$-ratio with the standard deviation for the pretest $s_{\text{Pre}}$ 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 corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.
• The result corresponds to the fraction without parentheses around the numerator and the denominator, ${\displaystyle \frac{\left(\overline{D}-\mu \right)}{\left(s_D/N^{1/2}\right)}}\text{.}$ Please do not forget the parentheses around the numerator and the denominator of a fraction.

$t=\overline{D}-{\displaystyle \frac{\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Post}}}}}}/N^{\cancel{1/2}}$

• The result corresponds to the $t$-ratio with the standard deviation for the posttest $s_{\text{Post}}$ 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 corresponds to the fraction without parentheses around the numerator and the denominator, ${\displaystyle \frac{\left(\overline{D}-\mu \right)}{\left(s_D/\sqrt{N}\right)}}\text{.}$ Please do not forget the parentheses around the numerator and the denominator of a fraction.
• The result corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.

$t=\overline{D}-{\displaystyle \frac{\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Post}}}}}}/N^{\cancel{1/2}}$

• The result corresponds to the $t$-ratio with the standard deviation for the posttest $s_{\text{Post}}$ 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 corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.
• The result corresponds to the fraction without parentheses around the numerator and the denominator, ${\displaystyle \frac{\left(\overline{D}-\mu \right)}{\left(s_D/N^{1/2}\right)}}\text{.}$ Please do not forget the parentheses around the numerator and the denominator of a fraction.

$t={\displaystyle \frac{\overline{D}\cancel{-\mu }}{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Post}}}}/\sqrt{N}}}$

• The result corresponds to the $t$-ratio with the standard deviation for the posttest $s_{\text{Post}}$ 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 corresponds to the $t$-ratio, when the null hypothesis $\mu$ has been omitted. Please do not forget $\mu$ in the $t$-ratio.

$t={\displaystyle \frac{\overline{D}\cancel{-\mu }}{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Pre}}}}/\sqrt{N}}}$

• The result corresponds to the $t$-ratio with the standard deviation for the pretest $s_{\text{Pre}}$ 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 corresponds to the $t$-ratio, when the null hypothesis $\mu$ has been omitted. Please do not forget $\mu$ in the $t$-ratio.

$t={\displaystyle \frac{\overline{D}-\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Post}}}}/N^{\cancel{1/2}}}}$

• The result corresponds to the $t$-ratio with the standard deviation for the posttest $s_{\text{Post}}$ 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 corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.

$t={\displaystyle \frac{\overline{D}\cancel{-\mu }}{s_D/\sqrt{N}}}$

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

$t={\displaystyle \frac{\overline{D}-\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Pre}}}}/N^{\cancel{1/2}}}}$

• The result corresponds to the $t$-ratio with the standard deviation for the pretest $s_{\text{Pre}}$ 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 corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.

$t={\displaystyle \frac{\overline{D}-\mu }{s_D/N^{\cancel{1/2}}}}$

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

$t={\displaystyle \frac{\overline{D}\cancel{-\mu }}{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Post}}}}/N^{\cancel{1/2}}}}$

• The result corresponds to the $t$-ratio with the standard deviation for the posttest $s_{\text{Post}}$ 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 corresponds to the $t$-ratio, when the null hypothesis $\mu$ has been omitted. Please do not forget $\mu$ in the $t$-ratio.
• The result corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.

$t={\displaystyle \frac{\overline{D}\cancel{-\mu }}{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Pre}}}}/N^{\cancel{1/2}}}}$

• The result corresponds to the $t$-ratio with the standard deviation for the pretest $s_{\text{Pre}}$ 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 corresponds to the $t$-ratio, when the null hypothesis $\mu$ has been omitted. Please do not forget $\mu$ in the $t$-ratio.
• The result corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.

$t={\displaystyle \frac{\overline{D}\cancel{-\mu }}{s_D/N^{\cancel{1/2}}}}$

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

$t=\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Pre}}}}-{\displaystyle \frac{\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Pre}}}}}}/\sqrt{N}$

• The result corresponds to the $t$-ratio with the standard deviation for the pretest $s_{\text{Pre}}$ 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.
• The result corresponds to the $t$-ratio with the mean $\overline{\text{Pre}}$ instead of the mean change score $\overline{D}$. Please insert the mean change score $\overline{D}$ 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 corresponds to the fraction without parentheses around the numerator and the denominator, ${\displaystyle \frac{\left(\overline{D}-\mu \right)}{\left(s_D/\sqrt{N}\right)}}\text{.}$ Please do not forget the parentheses around the numerator and the denominator of a fraction.

$t=\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Pre}}}}-{\displaystyle \frac{\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Pre}}}}}}/\sqrt{N}$

• The result corresponds to the $t$-ratio with the mean $\overline{\text{Pre}}$ instead of the mean change score $\overline{D}$. Please insert the mean change score $\overline{D}$ into the $t$-ratio.
• The result corresponds to the $t$-ratio with the standard deviation for the pretest $s_{\text{Pre}}$ 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 corresponds to the fraction without parentheses around the numerator and the denominator, ${\displaystyle \frac{\left(\overline{D}-\mu \right)}{\left(s_D/\sqrt{N}\right)}}\text{.}$ Please do not forget the parentheses around the numerator and the denominator of a fraction.

$t=\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Pre}}}}-{\displaystyle \frac{\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Pre}}}}}}/N^{\cancel{1/2}}$

• The result corresponds to the $t$-ratio with the standard deviation for the pretest $s_{\text{Pre}}$ 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.
• The result corresponds to the $t$-ratio with the mean $\overline{\text{Pre}}$ instead of the mean change score $\overline{D}$. Please insert the mean change score $\overline{D}$ 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 corresponds to the fraction without parentheses around the numerator and the denominator, ${\displaystyle \frac{\left(\overline{D}-\mu \right)}{\left(s_D/\sqrt{N}\right)}}\text{.}$ Please do not forget the parentheses around the numerator and the denominator of a fraction.
• The result corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.

$t=\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Pre}}}}-{\displaystyle \frac{\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Pre}}}}}}/N^{\cancel{1/2}}$

• The result corresponds to the $t$-ratio with the standard deviation for the pretest $s_{\text{Pre}}$ 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.
• The result corresponds to the $t$-ratio with the mean $\overline{\text{Pre}}$ instead of the mean change score $\overline{D}$. Please insert the mean change score $\overline{D}$ 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 corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.
• The result corresponds to the fraction without parentheses around the numerator and the denominator, ${\displaystyle \frac{\left(\overline{D}-\mu \right)}{\left(s_D/N^{1/2}\right)}}\text{.}$ Please do not forget the parentheses around the numerator and the denominator of a fraction.

$t=\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Pre}}}}-{\displaystyle \frac{\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Pre}}}}}}/N^{\cancel{1/2}}$

• The result corresponds to the $t$-ratio with the mean $\overline{\text{Pre}}$ instead of the mean change score $\overline{D}$. Please insert the mean change score $\overline{D}$ into the $t$-ratio.
• The result corresponds to the $t$-ratio with the standard deviation for the pretest $s_{\text{Pre}}$ 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 corresponds to the fraction without parentheses around the numerator and the denominator, ${\displaystyle \frac{\left(\overline{D}-\mu \right)}{\left(s_D/\sqrt{N}\right)}}\text{.}$ Please do not forget the parentheses around the numerator and the denominator of a fraction.
• The result corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.

$t=\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Pre}}}}-{\displaystyle \frac{\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Pre}}}}}}/N^{\cancel{1/2}}$

• The result corresponds to the $t$-ratio with the mean $\overline{\text{Pre}}$ instead of the mean change score $\overline{D}$. Please insert the mean change score $\overline{D}$ into the $t$-ratio.
• The result corresponds to the $t$-ratio with the standard deviation for the pretest $s_{\text{Pre}}$ 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 corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.
• The result corresponds to the fraction without parentheses around the numerator and the denominator, ${\displaystyle \frac{\left(\overline{D}-\mu \right)}{\left(s_D/N^{1/2}\right)}}\text{.}$ Please do not forget the parentheses around the numerator and the denominator of a fraction.

$t=\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Post}}}}-{\displaystyle \frac{\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Post}}}}}}/\sqrt{N}$

• The result corresponds to the $t$-ratio with the posttest mean $\overline{\text{Post}}$ instead of the Mean change score $\overline{D}$. Please insert the mean change score $\overline{D}$ into the $t$-ratio.
• The result corresponds to the $t$-ratio with the standard deviation for the posttest $s_{\text{Post}}$ 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 corresponds to the fraction without parentheses around the numerator and the denominator, ${\displaystyle \frac{\left(\overline{D}-\mu \right)}{\left(s_D/\sqrt{N}\right)}}\text{.}$ Please do not forget the parentheses around the numerator and the denominator of a fraction.

$t=\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Post}}}}-{\displaystyle \frac{\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Post}}}}}}/\sqrt{N}$

• The result corresponds to the $t$-ratio with the standard deviation for the posttest $s_{\text{Post}}$ 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.
• The result corresponds to the $t$-ratio with the posttest mean $\overline{\text{Post}}$ instead of the Mean change score $\overline{D}$. Please insert the mean change score $\overline{D}$ 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 corresponds to the fraction without parentheses around the numerator and the denominator, ${\displaystyle \frac{\left(\overline{D}-\mu \right)}{\left(s_D/\sqrt{N}\right)}}\text{.}$ Please do not forget the parentheses around the numerator and the denominator of a fraction.

$t=\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Post}}}}-{\displaystyle \frac{\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Post}}}}}}/N^{\cancel{1/2}}$

• The result corresponds to the $t$-ratio with the posttest mean $\overline{\text{Post}}$ instead of the Mean change score $\overline{D}$. Please insert the mean change score $\overline{D}$ into the $t$-ratio.
• The result corresponds to the $t$-ratio with the standard deviation for the posttest $s_{\text{Post}}$ 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 corresponds to the fraction without parentheses around the numerator and the denominator, ${\displaystyle \frac{\left(\overline{D}-\mu \right)}{\left(s_D/\sqrt{N}\right)}}\text{.}$ Please do not forget the parentheses around the numerator and the denominator of a fraction.
• The result corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.

$t=\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Post}}}}-{\displaystyle \frac{\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Post}}}}}}/N^{\cancel{1/2}}$

• The result corresponds to the $t$-ratio with the posttest mean $\overline{\text{Post}}$ instead of the Mean change score $\overline{D}$. Please insert the mean change score $\overline{D}$ into the $t$-ratio.
• The result corresponds to the $t$-ratio with the standard deviation for the posttest $s_{\text{Post}}$ 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 corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.
• The result corresponds to the fraction without parentheses around the numerator and the denominator, ${\displaystyle \frac{\left(\overline{D}-\mu \right)}{\left(s_D/N^{1/2}\right)}}\text{.}$ Please do not forget the parentheses around the numerator and the denominator of a fraction.

$t=\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Post}}}}-{\displaystyle \frac{\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Post}}}}}}/N^{\cancel{1/2}}$

• The result corresponds to the $t$-ratio with the standard deviation for the posttest $s_{\text{Post}}$ 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.
• The result corresponds to the $t$-ratio with the posttest mean $\overline{\text{Post}}$ instead of the Mean change score $\overline{D}$. Please insert the mean change score $\overline{D}$ 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 corresponds to the fraction without parentheses around the numerator and the denominator, ${\displaystyle \frac{\left(\overline{D}-\mu \right)}{\left(s_D/\sqrt{N}\right)}}\text{.}$ Please do not forget the parentheses around the numerator and the denominator of a fraction.
• The result corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.

$t=\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Post}}}}-{\displaystyle \frac{\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Post}}}}}}/N^{\cancel{1/2}}$

• The result corresponds to the $t$-ratio with the standard deviation for the posttest $s_{\text{Post}}$ 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.
• The result corresponds to the $t$-ratio with the posttest mean $\overline{\text{Post}}$ instead of the Mean change score $\overline{D}$. Please insert the mean change score $\overline{D}$ 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 corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.
• The result corresponds to the fraction without parentheses around the numerator and the denominator, ${\displaystyle \frac{\left(\overline{D}-\mu \right)}{\left(s_D/N^{1/2}\right)}}\text{.}$ Please do not forget the parentheses around the numerator and the denominator of a fraction.

$t={\displaystyle \frac{\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Post}}}}-\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Post}}}}/\sqrt{N}}}$

• The result corresponds to the $t$-ratio with the posttest mean $\overline{\text{Post}}$ instead of the Mean change score $\overline{D}$. Please insert the mean change score $\overline{D}$ into the $t$-ratio.
• The result corresponds to the $t$-ratio with the standard deviation for the posttest $s_{\text{Post}}$ 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.

$t={\displaystyle \frac{\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Post}}}}-\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Post}}}}/\sqrt{N}}}$

• The result corresponds to the $t$-ratio with the standard deviation for the posttest $s_{\text{Post}}$ 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.
• The result corresponds to the $t$-ratio with the posttest mean $\overline{\text{Post}}$ instead of the Mean change score $\overline{D}$. Please insert the mean change score $\overline{D}$ into the $t$-ratio.
• Correctly recognised the problem as a $t$-test for paired samples.
• Correctly identified the $t$-ratio for paired samples.

$t={\displaystyle \frac{\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Post}}}}\cancel{-\mu }}{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Post}}}}/\sqrt{N}}}$

• The result corresponds to the $t$-ratio with the posttest mean $\overline{\text{Post}}$ instead of the Mean change score $\overline{D}$. Please insert the mean change score $\overline{D}$ into the $t$-ratio.
• The result corresponds to the $t$-ratio with the standard deviation for the posttest $s_{\text{Post}}$ 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 corresponds to the $t$-ratio, when the null hypothesis $\mu$ has been omitted. Please do not forget $\mu$ in the $t$-ratio.

$t={\displaystyle \frac{\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Post}}}}\cancel{-\mu }}{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Post}}}}/\sqrt{N}}}$

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

$t={\displaystyle \frac{\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Pre}}}}-\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Pre}}}}/\sqrt{N}}}$

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

$t={\displaystyle \frac{\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Pre}}}}-\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Pre}}}}/\sqrt{N}}}$

• The result corresponds to the $t$-ratio with the mean $\overline{\text{Pre}}$ instead of the mean change score $\overline{D}$. Please insert the mean change score $\overline{D}$ into the $t$-ratio.
• The result corresponds to the $t$-ratio with the standard deviation for the pretest $s_{\text{Pre}}$ 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.

$t={\displaystyle \frac{\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Pre}}}}\cancel{-\mu }}{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Pre}}}}/\sqrt{N}}}$

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

$t={\displaystyle \frac{\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Pre}}}}\cancel{-\mu }}{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Pre}}}}/\sqrt{N}}}$

• The result corresponds to the $t$-ratio with the mean $\overline{\text{Pre}}$ instead of the mean change score $\overline{D}$. Please insert the mean change score $\overline{D}$ into the $t$-ratio.
• The result corresponds to the $t$-ratio with the standard deviation for the pretest $s_{\text{Pre}}$ 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 corresponds to the $t$-ratio, when the null hypothesis $\mu$ has been omitted. Please do not forget $\mu$ in the $t$-ratio.

$t={\displaystyle \frac{\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Post}}}}-\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Post}}}}/N^{\cancel{1/2}}}}$

• The result corresponds to the $t$-ratio with the posttest mean $\overline{\text{Post}}$ instead of the Mean change score $\overline{D}$. Please insert the mean change score $\overline{D}$ into the $t$-ratio.
• The result corresponds to the $t$-ratio with the standard deviation for the posttest $s_{\text{Post}}$ 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 corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.

$t={\displaystyle \frac{\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Post}}}}-\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Post}}}}/N^{\cancel{1/2}}}}$

• The result corresponds to the $t$-ratio with the standard deviation for the posttest $s_{\text{Post}}$ 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.
• The result corresponds to the $t$-ratio with the posttest mean $\overline{\text{Post}}$ instead of the Mean change score $\overline{D}$. Please insert the mean change score $\overline{D}$ 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 corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.

$t={\displaystyle \frac{\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Post}}}}\cancel{-\mu }}{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Post}}}}/N^{\cancel{1/2}}}}$

• The result corresponds to the $t$-ratio with the posttest mean $\overline{\text{Post}}$ instead of the Mean change score $\overline{D}$. Please insert the mean change score $\overline{D}$ into the $t$-ratio.
• The result corresponds to the $t$-ratio with the standard deviation for the posttest $s_{\text{Post}}$ 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 corresponds to the $t$-ratio, when the null hypothesis $\mu$ has been omitted. Please do not forget $\mu$ in the $t$-ratio.
• The result corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.

$t={\displaystyle \frac{\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Post}}}}\cancel{-\mu }}{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Post}}}}/N^{\cancel{1/2}}}}$

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

$t={\displaystyle \frac{\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Pre}}}}-\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Pre}}}}/N^{\cancel{1/2}}}}$

• The result corresponds to the $t$-ratio with the standard deviation for the pretest $s_{\text{Pre}}$ 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.
• The result corresponds to the $t$-ratio with the mean $\overline{\text{Pre}}$ instead of the mean change score $\overline{D}$. Please insert the mean change score $\overline{D}$ 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 corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.

$t={\displaystyle \frac{\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Pre}}}}-\mu }{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Pre}}}}/N^{\cancel{1/2}}}}$

• The result corresponds to the $t$-ratio with the mean $\overline{\text{Pre}}$ instead of the mean change score $\overline{D}$. Please insert the mean change score $\overline{D}$ into the $t$-ratio.
• The result corresponds to the $t$-ratio with the standard deviation for the pretest $s_{\text{Pre}}$ 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 corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.

$t={\displaystyle \frac{\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Pre}}}}\cancel{-\mu }}{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Pre}}}}/N^{\cancel{1/2}}}}$

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

$t={\displaystyle \frac{\underset{\text{instead of}\overline{D}}{\underbrace{\overline{\text{Pre}}}}\cancel{-\mu }}{\underset{\text{instead of}{s}_D}{\underbrace{s_{\text{Pre}}}}/N^{\cancel{1/2}}}}$

• The result corresponds to the $t$-ratio with the mean $\overline{\text{Pre}}$ instead of the mean change score $\overline{D}$. Please insert the mean change score $\overline{D}$ into the $t$-ratio.
• The result corresponds to the $t$-ratio with the standard deviation for the pretest $s_{\text{Pre}}$ 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 corresponds to the $t$-ratio, when the null hypothesis $\mu$ has been omitted. Please do not forget $\mu$ in the $t$-ratio.
• The result corresponds to the $t$-ratio without the square root around $N$. Please do not forget the square root around $N$.