Answer to Question #336799 in Statistics and Probability for Samantha

Question

APPLICATION:

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 x̄= 75 and s=5. In standardization of the test, the mean was

65 and standard deviation was 8. Based on the evidence at

hand, is the new time slot effective? Use α = 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 $\sigma_X^2+\sigma_Y^2.$

$$\mu_{X-Y}=75-65=10$$ $$s_{X-Y}=\sqrt{(5)^2+(8)^2}=\sqrt{89}$$

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:

$$t=\dfrac{\mu_{X-Y}-\mu}{s_{X-Y}/\sqrt{n}}=\dfrac{10-0}{\sqrt{89}/\sqrt{15}}\approx4.105354$$

Since it is observed that $|t|=4.105354>2.144787=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed $\alpha=0.05,$ $df=14, t=4.105354$ is $p=0.001071,$ and since $p=0.001071<0.05=\alpha,$ 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.