Answer to Question #348910 in Statistics and Probability for Nicole

Question

II. Determine the given of the problems below and formulate the null and alternative hypothesis both in words and symbols. Write your answer in your notebook. Please follow the format in the examples.

1. A health specialist wants to determine the average number of hours a person exercise in a day during the quarantine period. She found out that the mean number of hours a person exercise 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

$H_0:$ the average number of hours a person exercise in a day is 80 minutes.

$H_a:$ the average number of hours a person exercise in a day is not 80 minutes.

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 a 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.