Answer to Question #345508 in Statistics and Probability for Mike

Question

The head of the mathematics department announced that the mean score of Grade 11 students in the second periodic test in statistics was 89, And a standard deviation was 12. One student believed that the main score was less than this, So the student randomly selected 34 students and computer do you mean score, And obtained I mean score of 85. At 0.01 level of significance, Construct the critical regions.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=89$

$H_a:\mu<89$

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 (using t-table) $z_c = -2.3263.$

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

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{85-89}{12/\sqrt{34}}\approx-1.94$$

Since it is observed that $z = -1.94> -2.3263=z_c,$ it is then 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.