Answer to Question #350616 in Statistics and Probability for "Ning"

Question

Professor Balenciaga has reported her students' grades for several semesters and the average for all the grades ofthese students is 83. Her new class of 28 students seem to be higher than the average of ability and she wants to demonstrate that the current class is superior to the previous classes according to their average. Is there sufficient evidence for the class average of 86.2 and the standard deviation of 12 present to support her argument that the current class is superior? Using the 0.05 significance level.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le83$

$H_1:\mu>83$

This corresponds to a right-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 $\alpha = 0.05,$ $df=n-1=29$ and the critical value for a right-tailed test is $t_c =1.699127.$

The rejection region for this right-tailed test is $R = \{t:t>1.699127\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{86.2-83}{12/\sqrt{30}}=1.4606$$

Since it is observed that $t=1.4606<1.699127=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for right-tailed, $df=29$ degrees of freedom, $t=1.4606$ is $p=0.077438,$ and since $p=0.077438>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is greater than 83, at the $\alpha = 0.05$ significance level.