Answer to Question #335048 in Statistics and Probability for Kael

Question

The Head of the Math Department announced that the mean score of Grade 11 students in St. James High School in the medterm examination in statistics and probability was 88 and the standard deviation was 10. One student who believed that the mean score was less than this, randomly selected 35 students and computed their mean score. She obtained a mean score of 84. At 0.01 level of significance, test the student’s belief.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge88$

$H_a:\mu<88$

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 $R = \{z: z < -2.3263\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{84-88}{10/\sqrt{35}}\approx-2.3664$$

Since it is observed that $z = -2.3664 < -2.3263=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p = P(Z<-2.3664)=0.008981,$ and since $p = 0.008981 < 0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is less than 88, at the $\alpha = 0.01$ significance level.