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.