An expirement study was conducted by a researcher to determine if a new time slot has an effect on the performance of pupils in Mathematics. Forty randomly selected learners participated in the study. Toward the end of the investigations, a standardized assessment was conducted. The sample mean was 78 and the standard devation of 13. In the standardization of the test, the mean was 63 and the standard deviation was 6. Use 5% level of significance.

Difference of two independent normal variables

Let $X$ have a normal distribution with mean $\mu_X$ and variance $\sigma_X^2.$

Let $Y$ have a normal distribution with mean $\mu_Y$ and variance $\sigma_Y^2.$

If $X$ and $Y$are independent, then $X-Y$will follow a normal distribution with mean $\mu_X-\mu_Y$ and

variance $\sigma_X^2+\sigma_Y^2.$

$\mu_{X-Y}=78-63=15$$s_{X-Y}=\sqrt{(13)^2+(6)^2}=\sqrt{205}$

The following null and alternative hypotheses need to be tested:

$H_0:\mu=0$

$H_1:\mu\not=0$

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, the sample standard deviation will be used.

Based on the information provided, the significance level is $\alpha=0.05,$

$df=n-1=40-1=39$ degrees of freedom, and the critical value for a two-tailed test is $t_{0.025,39}= 2.022691$

The rejection region for this two-tailed test is $R=\{t:|t|\gt 2.022691\}$

The t-statistic is computed as follows:

$t=\dfrac{\mu_{X-Y}-\mu}{s_{X-Y}/\sqrt{n}}=\dfrac{15-0}{\sqrt{205}\over\sqrt{40}}\approx6.6259$

Since it is observed that $|t|=6.6259\gt 2.022691=t_{0.025,39},$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the new time slot is effective at the $\alpha=0.05$ significance level.