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.