Homework SourseIn an effort to promote healthy lifestyles health screenings
| In an effort to promote healthy lifestyles, health screenings are given to employees of a large corporation. In running a promotional trial, 84 out of the 150 people who work in one office for the corporation participate in the health screening. | ||||||||||||||
| a. | Is the above information sufficient for you to be completely certain that more than 50% of all employees of the corporation will participate in the health screening? Why or why not? | |||||||||||||
| b. | In establishing a statistical hypothesis testing of this situation, give the required null and alternative hypotheses for such a test, if it is desired that more than 50% of the employees participate in the health screening. | |||||||||||||
| H0: | p=0.5 | |||||||||||||
| H1: | p>0.5 | |||||||||||||
| c. | Based on your answer in part b, should you use a right-tailed, a left-tailed, or a two-tailed test? Briefly explain how one determines which of the three possibilities is to be used. | |||||||||||||
| d. | Describe the possible Type I error for this situation--make sure to state the error in terms of the percent of employees in the corporation who will participate in the health screenings. | |||||||||||||
| e. | Describe the possible Type II error for this situation--make sure to state the error in terms of the percent of employees in the corporation who will participate in the health screenings. | |||||||||||||
| f. | Determine the appropriate critical value(s) for this situation given a 0.025 significance level. | |||||||||||||
| g. | Determine/calculate the value of the sample\'s test statistic. | |||||||||||||
| h. | Detemine the P-value. | |||||||||||||
| i. | Based upon your work above, is there statistically sufficient evidence in this sample to support that more than 50% of employees will participate in the health screening? Briefly explain your reasoning. | |||||||||||||
| In an effort to promote healthy lifestyles, health screenings are given to employees of a large corporation. In running a promotional trial, 84 out of the 150 people who work in one office for the corporation participate in the health screening. | ||||||||||||||
| a. | Is the above information sufficient for you to be completely certain that more than 50% of all employees of the corporation will participate in the health screening? Why or why not? | |||||||||||||
| b. | In establishing a statistical hypothesis testing of this situation, give the required null and alternative hypotheses for such a test, if it is desired that more than 50% of the employees participate in the health screening. | |||||||||||||
| H0: | p=0.5 | |||||||||||||
| H1: | p>0.5 | |||||||||||||
| c. | Based on your answer in part b, should you use a right-tailed, a left-tailed, or a two-tailed test? Briefly explain how one determines which of the three possibilities is to be used. | |||||||||||||
| d. | Describe the possible Type I error for this situation--make sure to state the error in terms of the percent of employees in the corporation who will participate in the health screenings. | |||||||||||||
| e. | Describe the possible Type II error for this situation--make sure to state the error in terms of the percent of employees in the corporation who will participate in the health screenings. | |||||||||||||
| f. | Determine the appropriate critical value(s) for this situation given a 0.025 significance level. | |||||||||||||
| g. | Determine/calculate the value of the sample\'s test statistic. | |||||||||||||
| h. | Detemine the P-value. | |||||||||||||
| i. | Based upon your work above, is there statistically sufficient evidence in this sample to support that more than 50% of employees will participate in the health screening? Briefly explain your reasoning. | |||||||||||||
Solution
a)
No. We have to test first if this could have just happened by chance.
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b)
Ho: p = 0.5
Ha: p > 0.5
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c)
This is a right tailed test, as Ha used >. One has to just look at the inequality sign Ha uses.
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d)
A type I error is incorrectly rejecting a true null hypothesis.
Thus, it is concluding that the true proportion is greater than 50%, when in fact, it is not.
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