Answer to Question #202537 in Statistics and Probability for Delmundo

Question

The mean score of the first periodic exam in Statistics is 89 and the standard deviation is 12. One student believed that the mean was less than this, she then randomly selected 34 students and computed their mean score. She obtained that the mean score of 85 at 99% confidence. Test the student belief. Use both method to evaluate

Answer 1

Hypothesized Population Mean $\mu=89$

Standard Deviation $\sigma=12$

Sample Size $n=34$

Sample Mean $\bar{x}=85$

Significance Level $\alpha=0.01$

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

$H_0: \mu\geq89$

$H_1: \mu<89$

This corresponds to 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 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:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{85-89}{12/\sqrt{34}}\approx-1.94365$$

Since it is observed that $z=-1.94365>-2.3263=z_c,$ 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.

Using the P-value approach: The p-value for left-tailed, the significance level$\alpha=0.01, z=-1.94365,$ is $p=P(Z<-1.94365)=0.02597,$ and since $p=0.02597>0.01=\alpha,$ 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.