Answer to Question #344057 in Statistics and Probability for niclene

Question

Yna Celestine believes that the average amount of time spent by her classmates

in studying their self learning module in Math per week is less than 300 minutes

with a standard deviation of 45 minutes. She took a random sample of 35 students

in their class and found out that average time spent for studying was 285 minutes.

Test the claim at the 0.05 level of significance.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge300$

$H_1:\mu<300$

This corresponds to a one-tailed 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 $\alpha = 0.05,$ and the critical value for a left-tailed test is $z_c =- 1.6449.$

The rejection region for this lrft-tailed test is $R = \{z:z<-1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{285-300}{45/\sqrt{35}}\approx-1.9720$$

Since it is observed that $z=-1.9720<-1.6449=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(z<-1.9720)=0.024305,$ and since $p=0.024305<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is less than 300, at the $\alpha = 0.05$ significance level.