Question

A large number of companies are trying to reduce the cost of prescription drug benefits by requiring employees to purchase drugs through a mandatory mail-order program. Suppose a large national retail company is considering requiring a mandated mail order program. The company is concerned about employee attitude and satisfaction levels and wishes to determine employee attitude regarding such a mail order prescription program. The company will move to the mail order system if more than 50% of the surveyed employees provide positive feedback on the program. The company disseminates information on the program and randomly surveys 600 employees. The results showed 310 employees favored the plan while the remainder of the surveyed employees responded unfavorably.  Answer the following questions:

A. Perform the appropriate test to determine if more than 50% of the company's employees favored the mail-order prescription program. Perform the test using the P-VALUE method. Use an alpha of .05 and be sure to show all required steps.

B. Define a Type I Error for this problem. Be specific - generic definitions will not be accepted.

C. Define a Type II Error for this problem. Be specific - generic definitions will not be accepted.

Answer 1

A. The null hypothesis would be that and the alternate hypothesis would be . The test statistic would be or or . The P-value can be obtained online or by a simple R-command as below.

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> pnorm(0.8186)
[1] 0.7934927

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Hence, the p-value of this z-statistic would be 0.7934927. At an alpha of 0.05, rejecting the null would require that the p-value of the statistic is greater than 0.95. But, as the p value is less than 0.95, we fail to reject the null. This means that the sample proportion of 0.5167 is not significantly greater than 0.5.

B. The type-I error would be when we reject the null, but it is true, and its probability is alpha. In this case, the type-I error would be to reject the hypothesis that the proportion is less than or equal to 0.5, when in truth, the proportion is indeed less than or equal to 0.5. The probability of this error is 0.05 or 5%.

C. The type-II error would be when we accept the null, but it is false, and its probability is 1-alpha. In this case, the type-II error would to fail to reject (meaning accept) the hypothesis that the proportion is less than or equal to 0.5, when in truth, the proportion is is greater than 0.5.