Answer to Question #349353 in Statistics and Probability for Cecel

Question

The head of the math department announced that the mean score of grade 11 students in the first periodic examination isn mathematics was 89 and the standard deviation was 12. One student believed that the mean score was less than this, randomly selected 34 students and computed their mean score. She obtained a mean score of 85. At 0.01 level of significance, test the students belief.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=89$

$H_1:\mu\not=89$

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha = 0.01,$ and the critical value for a two-tailed test is $z_c = 2.5758.$

The rejection region for this two-tailed test is $R = \{z:|z|>2.5758\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{85-89}{12/\sqrt{34}}=-1.94365$$

Since it is observed that $|z|=1.94365<2.5758=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=2P(z<-1.94365)=0.051938,$ and since $p=0.051938>0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is different than 89, at the $\alpha = 0.01$ significance level.