* * * For any hypothesis testingproblems in this set, please do only the following: (1) set up the hypotheses in plain English, (2) set up the hypotheses in statistical terms, (3) specify type I and type II errorsfor the hypothesis testing context for the problem, and (4) provide a brief description of the cost implications of each of the two errors for the problem context * * *

1. In 1993, women took an average of 8.5 weeks unpaid leave from their jobs after the birth of a child (U.S. News & World Report, 12/27/1993). Assume that 8.5 weeks is the population meanand 2.2 weeks is the population standard deviation. What is the probabilitythat a simple random sampleof 50 women provides a sample mean unpaid leave of 7.5 to 9.5 weeks after the birth of a child?

2. A sample of 532 BusinessWeek subscribers showed that the mean time a subscriber spends using the Internet and online services is 6.7 hours per week (Business Week 1996 Worldwide Subscriber Study). If the sample standard deviation is 5.8 hours, what is the 95% confidence intervalfor the mean time that the Business Week subscriber population spends on the Internet and online services?

3. Suppose that scores on an aptitude test used for determining admission to graduate study in business are distributed with a mean of 500 and a population standard deviation of 100. If a random sample of 64 applicants from Stephan College has a sample mean of 537, is there any evidence that their mean score is higher than the mean expected of all applicants? (Use = .01)

4. To help your restaurant marketing campaign target the right age levels, you want to find out if there is a statistically significant difference, on the average, between the age of your customers and the age of the general population in town, 43.1 years. A random sample of 50 customers shows an average age of 33.6 years with a standard deviation of 16.2 years. What can you conclude? (Use = .05)

5. External organizational development (OD) professionals provide consulting services in such areas as human resources, training, planning, skills education, industrial psychology, and organizational behavior. The Training and Development Journal (Feb. 1984) conducted a surveyof OD professionals. A random sample of 440 external OD consultants yielded the following summary statisticson daily fees charged:

Sample mean = $720 S = $275

Suppose it is known that other management consultants charge, on average, $800 per day. Do the dataprovide sufficient evidence to indicate that the mean daily fee charged by external OD consultants is less than $800? Test using = .10.

In 1993, women took an average of 8.5 weeks unpaid leave from their jobs after the birth of a child (U.S. News & World Report, 12/27/1993). Assume that 8.5 weeks is the population mean and 2.2 weeks is the population standard deviation. What is the probability that a simple random sample of 50 women provides a sample mean unpaid leave of 7.5 to 9.5 weeks after the birth of a child?

Since z=(xbar-mean)/[sd/sqrt(n)]=(xbar-8.5)/[2.2/sqrt(50)],

P(7.5<xbar<9.5)=P((7.5-8.5)/[2.2/sqrt(50]<Z<(9.5-8.5)/[2.2/sqrt(50])=P(-3.21<Z<3.21)=0.9987 from standard normal table.

A sample of 532 Business Week subscribers showed that the mean time a subscriber spends using the Internet and online services is 6.7 hours per week (Business Week 1996 Worldwide Subscriber Study). If the sample standard deviation is 5.8 hours, what is the 95% confidence interval for the mean time that the Business Week subscriber population spends on the Internet and online services?

From t table, at 95% level with df=532-1=531, the critical value is 1.96.

Hence, a 95% confidence interval for mean time is [6.7-1.96*5.8/sqrt(532), 6.7+1.96*5.8/sqrt(532)]=[6.21, 7.19]

Suppose that scores on an aptitude test used for determining admission to graduate study in business are distributed with a mean of 500 and a population standard deviation of 100. If a random sample of 64 …

The solution gives detailed steps on calculating probability, building confidence interval and performing hypothesis testing under either t or z distribution.