Answer to Question #178389 in Statistics and Probability for Jii

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. One student who 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 student’s belief.

Answer 1

We have that:

$\mu=89$

$\sigma = 12$

$n=34$

$\bar x = 85$

$\alpha = 0.01$

$H_0:\mu = 89$

$H_a:\mu<89$

The hypothesis test is left-tailed.

Since the population standard deviation is known and the sample size is large (>30) we use the z-test.

The critical value for $\alpha = 0.01$ is Z_0.01 = –2.33

The critical region is z < –2.33

Test statistic:

$$Z_{test}=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt n}}=\frac{85-89}{\frac{12}{\sqrt 34}}=-1.94$$

Since –1.94 > –2.33 thus the Z_test does not fall in the rejection region we fail to reject the null hypothesis.

There is no sufficient evidence to support the student’s belief. We are 99% confident to conclude that the mean score of Grade 9 students in the first periodic examination in Mathematics is 89.