Answer to Question #137447 in Statistics and Probability for soong

Question

Q3 60 smokers were questioned about the number of hours they sleep each day. We want to test the hypothesis that the smokers need less sleep than the general public which needs an average of 8 hours of sleep. If the sample mean is 7.5 and known population standard deviation is 0.2, what can you conclude using significance level of 0.01.

Answer 1

The provided sample mean is $\bar{X}=7.5$ and the known population standard deviation is $\sigma=0.2,$ and the sample size is $n=60.$

The following null and alternative hypotheses need to be tested:

$H_0: \mu\geq8$

$H_1:\mu<8$

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 $z_c=-2.33.$

The rejection region for this left-tailed test is $R=\{z:z<-2.33\}.$

The z-statistic is computed as follows:

$$Z=\dfrac{\bar{X}-\mu}{\sigma/\sqrt{n}}=\dfrac{7.5-8}{0.2/\sqrt{60}}\approx-19.365$$

Since it is observed that $z=-19.365<-2.33=z_c,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is less than 8, at the 0.01 significance level.

Using the P-value approach: The p-value is $p<0.00001,$and since $p\approx0<0.01,$ p it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is less than 8, at the 0.01 significance level.

We can conclude that the smokers need less sleep than the general public which needs an average of 8 hours of sleep.