Answer to Question #349674 in Statistics and Probability for veronica

Question

A health specialist wants to determine the average number of hours a person

exercises in a day during the quarantine period. She found out that the mean number

of hours a person exercises in a day during the quarantine period is 80 minutes. A

random sample of 29 persons were surveyed and found that their mean is 65 minutes

and a standard deviation of 10 minutes. Test the hypothesis at 2% level of significance

and assume that the population is normally distributed.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=80$

$H_1:\mu\not=80$

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.02,$ $df=n-1=28$ and the critical value for two-tailed test is $t_c =2.46714.$

The rejection region for this two-tailed test is $R = \{t:|t|>2.46714\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{65-80}{10/\sqrt{29}}=-8.0777$$

Since it is observed that $|t|=8.0777>2.46714=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=28$ degrees of freedom, $t=-8.0777$ is $p=0,$ and since $p=0<0.02=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is different than 80, at the $\alpha = 0.02$ significance level.