Question #349674

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.


1
Expert's answer
2022-06-12T15:45:53-0400

The following null and alternative hypotheses need to be tested:

H0:μ=80H_0:\mu=80

H1:μ80H_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 α=0.02,\alpha = 0.02, df=n1=28df=n-1=28 and the critical value for two-tailed test is tc=2.46714.t_c =2.46714.

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

The t-statistic is computed as follows:


t=xˉμs/n=658010/29=8.0777t=\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=tc,|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=28df=28 degrees of freedom, t=8.0777t=-8.0777 is p=0,p=0, and since p=0<0.02=α,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 μ\mu

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


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