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- Apr 14, 2013

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A teacher wants to find out if the order of the exam tasks has an impact on the performance of the students. Therefore, he creates two versions ($ X $ and $ Y $) of an exam in which the exam tasks are arranged differently. The versions are randomly distributed so that $ n $ students receive version $ X $, and $ m = n $ receive version $ Y $ from them. We call the expected score at $ X $ with $ \mu_X $, and the expected score at $ Y $ with $ \mu_Y $. The variances are denoted $ \sigma_X^2 $ and $ \sigma_Y^2 $; it is assumed normal distribution.

(a) Formulate a suitable null hypothesis for the question of the teacher.

(b) Consider that $n = 30, \overline{X} = 79, \overline{Y}= 74, S_X' = 14, S_Y' = 20$. Check the null hypothesis of (a) with significance level $\alpha=5\%$.

(c) Consider that $\overline{X} = 79, \overline{Y}= 80, S_X' = 14, S_Y' = 20$. For which sample size $n$ can we reject the null hypothesis with significance level $\alpha=1\%$ ?

I have done the following:

(a) The null hypothesis is $H_0: \mu_X=\mu_Y$, right?

(b) Since we don't know if we have the same or different variances, we have to test if we have the same $\sigma$ with a F-test.

- If $\sigma_x=\sigma_y$ then we apply a two-samples t-test.
- If $\sigma_x<\sigma_y$ then we apply a Welch-Test

The test is the following:

The null hypothesis and the alternative hypothesis is $H_0:\sigma_Y^2=\sigma_X^2$ and $H_1:\sigma_Y^2>\sigma_X^2$, respectively.

The test statistic is \begin{equation*}F=\frac{{S_Y'}^2}{{S_X'}^2}=\frac{20^2}{14^2}=\frac{400}{196}\approx 2.0408\end{equation*}

$F$ is F-distributed with degres of freedom $\nu_Y=n_Y-1=30-1=29$, $\nu_X=n_X-130-1=29$.

We have that $1-\alpha=95\%$.

The null hypothesis will be rejected if $F>F_{1-\alpha}(\nu_Y, \nu_X)=F_{0.95}(29, 29)$.

It holds that $F_{0.95}(29, 29)=1,86$.

Since $F=2.0408>1.86=F_{0.95}(29, 29)$, we reject the null hypothesis.

So, we apply a Welch-Test.

The zero-hypothesis is $H_0: \mu_X-\mu_Y=0$ and the alternative hypothesis is $H_1:\mu_X-\mu_Y\neq 0$.

The test statistic $T$ for the t-Test with unknown variances \begin{equation*}T=\frac{\overline{X}-\overline{Y}-0}{\sqrt{\frac{{S_X'}^2}{n_X}+\frac{{S_Y'}^2}{n_Y}}}=\frac{79-74}{\sqrt{\frac{14^2}{30}+\frac{20^2}{30}}}=\frac{5}{\sqrt{\frac{196}{30}+\frac{400}{30}}}=\frac{5}{\sqrt{\frac{298}{15}}}\approx 1.1218\end{equation*}

The null hypothesis will be rejected if $|T|>t_{k;1-\alpha/2}$.

The number od degrees of freedom is\begin{align*}k&=\frac{\left (\frac{{S_X'}^2}{n_X}+\frac{{S_Y'}^2}{n_Y}\right )^2}{\frac{1}{n_X-1}\left (\frac{{S_X'}^2}{n_X}\right )^2+\frac{1}{n_Y-1}\left (\frac{{S_Y'}^2}{n_Y}\right )^2}=\frac{\left (\frac{14^2}{30}+\frac{20^2}{30}\right )^2}{\frac{1}{30-1}\left (\frac{14^2}{30}\right )^2+\frac{1}{30-1}\left (\frac{20^2}{30}\right )^2}=\frac{\left (\frac{196}{30}+\frac{400}{30}\right )^2}{\frac{1}{29}\left (\frac{196}{30}\right )^2+\frac{1}{29}\left (\frac{400}{30}\right )^2} \\ & =\frac{\left (\frac{596}{30}\right )^2}{\frac{1}{29}\left (\frac{38416}{900}+\frac{160000}{900}\right )}=\frac{\frac{355216}{900}}{\frac{1}{29}\cdot \frac{198416}{900}}=\frac{355216\cdot 29}{ 198416}=\frac{10301264}{ 198416}\approx 51.9175\end{align*} so $k=52$.

So we get the critical value $t_{k;1-\alpha/2}=t_{52;0.975}=1.67$.

Since $|T|=1.1218<1.67=t_{52;0.975}$ we don't reject the null hypothesis.

Is everything correct?

(c) Do we have to do the same as in (b) just with unknown n?