Answer to Question #210424 in Statistics and Probability for Nathan

Question

An experimental study was conducted by a researcher to determine if a new time slot has an effecton the performance of pupils in mathematics. Fifteen randomly selected learners participated in thestudy. Toward the end of the investigation, a standardized assessment was conducted. The samplemean was X = 85 and the standard deviation s = 3. In the standardization of the test, the mean was75 and the standard deviation was 10. Based on the evidence at hand, is the new time slot effective?Use a = 0.05.​

Answer 1

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 $\sigma_X^2+\sigma_Y^2.$

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

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

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, using 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 $t_c=2.144787.$

The rejection region for this two-tailed test is $R=\{t:|t|>2.144787\}$

The t-statistic is computed as follows:

$$t=\dfrac{\mu_{X-Y}-\mu}{s_{X-Y}/\sqrt{n}}=\dfrac{10-0}{\sqrt{109}/\sqrt{15}}\approx3.709645$$

Since it is observed that $|t|=3.709645>2.144787=t_c,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 0, at the $\alpha=0.05$ significance level.

Using the P-value approach: The p-value for two-tailed $\alpha=0.05, df=14, t=3.709645$ is $p=0.002332,$ and since $p=0.002332<0.05=\alpha,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 0, at the $\alpha=0.05$ significance level.

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