Question

An experimental study was conducted by a researcher to determine if a new time slot has an effect on the performance of pupils in mathematics. Fifteen randomly selected learners participated in the study. Toward the end of the investigation, a standardized assessment was conducted. The sample mean was 𝑿̅ = 𝟖𝟓 and the standard deviation s = 3. in the standardization of the test, the mean was 75 and the standard deviation was 10. based on the evidence at hand, is the new timeslot effective?

Answer 1

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 $Z=X-Y$will follow a normal distribution with mean $\mu_X-\mu_Y$ and

variance $\sigma_X^2+\sigma_Y^2.$

$\mu_{X-Y}=85-75=10$

$s_{X-Y}=\sqrt{(3)^2+(10)^2}=\sqrt{109}$

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 $\alpha=0.05,$

$df=n-1=15-1=14$ degrees of freedom, and the critical value for a two-tailed test is $t_{0.025,14}= 2.145$

The rejection region for this two-tailed test is $R=\{t:|t|\gt 2.145\}$

The t-statistic is computed as follows:

$$t=\dfrac{\mu_{X-Y}-\mu}{s_{X-Y}/\sqrt{n}}=\dfrac{10-0}{\sqrt{109}\over\sqrt{15}}\approx3.71$$

Now, $t=3.71\gt t_{0.025,14}=2.145$ thus, we reject the null hypothesis and conclude that the data provides sufficient evidence to show that the new time slot is effective at 5% significance level.