1. in 1995, 10.8% of All U.S. Families Had Incomes Below the Poverty Level, As Reported

Homework 9

1. In 1995, 10.8% of all U.S. families had incomes below the poverty level, as reported by the Census Bureau in Current Population Reports. During that same year, of 368 randomly selected families whose householder had a Bachelor's degree or more, 9 had incomes below the poverty level. At the 1% significance level, do the data provide sufficient evidence to conclude that in 1995, families whose householder had a Bachelor's degree or more had a lower percentage earning incomes below the poverty level than the national percentage of 10.8%? (Use rejection region approach.)

2. The Chicago Title Insurance Company publishes statistics on recent home buyers in the The Guarantor. According to that publication, 83.1% of home buyers in 1995 purchased single-family houses. Out of 2544 randomly selected home buyers for this year, 2081 purchased single-family houses. Do the data provide sufficient evidence to conclude that this year's percentage of home buyers purchasing single-family houses is different from the 1995 figure of 83.1%? Use = 0.024. (Use p-value approach.)

3. Problem 5.44

4. An office manager has implemented an incentive plan that she thinks will reduce the mean time required to handle a customer complaint. The mean time for handling a complaint was 30 minutes prior to implementing the incentive plan. After the plan was in place for several months, a random sample of the records of 38 customers who had complaints revealed a mean time of 28.7 minutes with a standard deviation of 3.8 minutes. Is there sufficient evidence that the incentive plan has reduced the mean time to handle a complaint? Us 0.01 level of significance. (Note that one has to be very cautious about using a 2 sided confidence interval to draw conclusion about a one tailed test. That is why 5.78 d did not mention a significance level.) To avoid careless mistakes, perform the test instead of drawing the conclusion from a confidence interval when dealing with 1-sided test. Or, you have to compute the 1-sided confidence bound in order to use it to draw conclusion about the test). For this problem, add that we want to perform the test at a level of significance of 0.01.

5. Use Minitab to solve following: A study was conducted of 90 adult male patients following a new treatment for congestive heart failure. One of the variables measured on the patients was the increase in exercise capacity (in minutes) over a 4-week treatment perios. The previous treatment regime had produced an average increase of µ = 2 minutes. The researchers wanted to evaluate whether the new treatment had increased the value of µ in comparison to the previous treatment. The data yielded a sample mean of 2.17 and S = 1.05 Calculate the power of the test for ua = 2.1 and ua = 2.5 for both an alpha of 0.05 and 0.01. Comment on the effect these changes have on the power of the test. Also, find the minimum sample size needed to have a power of 0.80 for an alpha of 0.05 to detect an effect size of 0.2.

Minitab Steps:

Select STAT > Power and Sample Size > One Sample t

Enter 90 50 for sample size, 0.1 0.5 for differences, and 1.05 for Standard Deviation

Click Options and select Greater Than for alternative and first enter 0.05 for significance level and click OK twice

Go back and change significance level from 0.05 to 0.01

For sample size calculations return and clear sample sizes, enter 0.2 for difference, 0.8 power and change alpha back to 0.05