Answer to Question #344526 in Statistics and Probability for Mary rose Burlas

Question

the subject group head of shs mathematics department announced that the mean percentage score of grade 11 learning in the 3rd quarter grade in statistic and probability was it in 9 in the standard deviation was 12 one learned who believed that the mean percentage score was less than this randomly selected 34 learner in computed their mean scores obtained a mean percentage score of 85 at 0.01 level of significance test the learners belief

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge89$

$H_1:\mu<89$

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 $R = \{z:z<-2.3263\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{85-89}{12/\sqrt{34}}\approx-1.9437$$

Since it is observed that $z|=-1.9437>-2.3263=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=P(z<-1.9437)= 0.025966,$ and since $p= 0.025966>0.01=\alpha,$ 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.