Answer to Question #345152 in Statistics and Probability for Mariane

Question

The Head of the Math Department announced that the mean score of

Grade 9 students in the first periodic examination in Mathematics was 89 and

the standard deviation was 12. 0ne student, who believed that the mean

score was less than this, randomly selected 25 students and computed their

mean score. She obtained a mean score of 85. At a 0.01 level of significance,

test the student's beliefs.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=89$

$H_1:\mu<89$

This corresponds to a left-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 left-tailed test is $z_c = -2.3263.$

The rejection region for this left-tailed test is $R = \{z<-2.3263\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{85-89}{12/\sqrt{25}}=-1.6667$$

6. Since it is observed that $z=-1.6667>-2.3263=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for left-tailed is $p=P(Z<-1.6667)=0.047787,$ and since $p=0.047787>0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

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

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