Answer to Question #338811 in Statistics and Probability for Sarah

Question

An educational researcher found that the average entrance core of the incoming freshmen in a university is 84. A random sample of 24 students from a public school was then selected and found out that their average score in the entrance exam is 88 with a standard deviation of 16. Is there any evidence to show that the sample students from public school performed better than the rest in the entrance examination using the 0.01 level of significance?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le84$

$H_a:\mu>84$

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 $\alpha = 0.01,$ $df=n-1=23$ degrees of freedom, and the critical value for a right-tailed test is $t_c = 2.499867.$

The rejection region for this right-tailed test is $R = \{t: t >2.499867\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{88-84}{16/\sqrt{24}}\approx1.224745$$

Since it is observed that $t = 1.224745 \le 2.499867=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for right-tailed test, $df=23$ degrees of freedom, $t=1.224745$ is $p = 0.116534,$ and since $p =0.116534 \ge 0.01=\alpha,$ 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 84, at the $\alpha = 0.01$ significance level.