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For a statistics class project, a college student randomly samples 75 men who exercise at a gym regularly and 68 women who exercise at a gym regularly. The college student believes that on average men spend more time at the gym each week. The college student records the number of minutes each person exercises in a given week. The college student conducts a hypothesis test at the 5% significance level. Use the summary statistics below to conduct a hypothesis test using your calculator or calculator

User Yclkvnc
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Answer:

Explanation:

The record of the statistics and the summary statistics which are the missing files in the question are attached below.

From the given information:

The null hypothesis and the alternative hypothesis can be represented by :


\mathbf{H_o:} There is no difference between the average time spend by men and women at gym each week


\mathbf{H_i:} The average time spend by men is greater than the average time spend by women at the gym each week

From the summary statistics in the attached file below:

The p-value = 0.3253

Level of significance = 5% = 0.05

Therefore; it is obvious that the p-value is greater than the level of significance i.e (0.3253 > 0.05)

Hence; there is no enough evidence to reject the null hypothesis

CONCLUSION: We conclude that the mean number of minutes exercised per week is larger for men than women at the this gym.

For a statistics class project, a college student randomly samples 75 men who exercise-example-1
For a statistics class project, a college student randomly samples 75 men who exercise-example-2
User Tawanda
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