# Qnt561 One and Two Samples Sets Week 5

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One- and Two-Sample Tests of Hypothesis, Variance, and Chi-squared Analysis Problem Sets University of Phoenix
QNT 561
August 5, 2011

Chapter 10

Exercise Question 31: A new weight-watching company, Weight Reducers International, advertises that those who join will lose, on the average, 10 pounds the first two weeks with a standard deviation of 2.8 pounds. A random sample of 50 people who joined the new weight reduction program revealed the mean loss to be 9 pounds. At the .05 level of significance, can we conclude that those joining Weight Reducers on average will lose less than 10 pounds? Determine the p-value

To calculate the test statistics:
Z=(9-10)/(2.8/sqrt(50))= -2.525

From the z-table, we find P9z<-2.525)=0.0058

So we reject the null hypothesis. There is strong evidence to suggest that the average weight loss at Weight Reducers is less than 10 pounds.

Exercise Question 32: Dole Pineapple, Inc. is concerned that the 16-ounce can of sliced pineapple is being overfilled. Assume the standard deviation of the process is .03 ounces. The quality control department took a random sample of 50 cans and found that the arithmetic mean weight was 16.05 ounces. At the 5 percent level of significance, can we conclude that the mean weight is greater than 16 ounces? Determine the p-value. Ho: μ ≤ 16| | | | |

Ha: μ > 16| | | | |
α = 0.05| | | | |
critical value z = 1.645| | | |
test statistic = (16.05-16)/(0.03/sqrt(50)) = 11.785|
p-value=0.0000
We reject the null hypothesis since the test statistic is greater than the critical value. We have sufficient evidence to conclude that the mean weight is greater than 16 ounces.

Exercise Question 38: A recent article in the Wall Street Journal reported that the 30-year mortgage rate is now less than 6 percent. A sample of eight small banks in the Midwest revealed the following 30-year rate (in percent): 4.8 5.3 6.5 4.8 6.1 5.8 6.2 5.6 At the .01 significance level, can we conclude that the 30-year mortgage rate for small banks is less than 6 percent? Estimate the p-value.

The p-value is 7.5% which is greater than significance level of 1%. We fail to reject the null hypothesis but we can’t conclude that the rates are less than 6%.

Chapter 11

Exercise Question 27: A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. The information is summarized below:

StatisticMen Women
Sample Mean24.5122.69
Population standard 4.483.86
Deviation
Sample Size3540

At the .01 significance level, is there a difference in the mean number of time mean and women order take-out dinners in a month? What is the p-value? The test statistic: (24.51-22.69)/sqrt(4.48^2/35 + 3.86^2/40) = 1.871