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: | ||||||||||||||
| H1: | ||||||||||||||
| 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: | ||||||||||||||
| H1: | ||||||||||||||
| 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
Set Up Hypothesis
Null, H0:P=0.5
Alternate, H1: P>0.5
Test Statistic
No. Of Success chances Observed (x)=84
Number of objects in a sample provided(n)=150
No. Of Success Rate ( P )= x/n = 0.56
Success Probability ( Po )=0.5
Failure Probability ( Qo) = 0.5
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.56-0.5/(Sqrt(0.25)/150)
Zo =1.4697
| Zo | =1.4697
Critical Value
The Value of |Z | at LOS 0.01% is 2.33
We got |Zo| =1.47 & | Z | =2.33
Make Decision
Hence Value of |Zo | < | Z | and Here we Do not Reject Ho
P-Value: Right Tail - Ha : ( P > 1.46969 ) = 0.07082
Hence Value of P0.01 < 0.07082,Here We Do not Reject Ho
We don\'t have indicate that more than 50% of all employees of the corporation will participate in the health screening
