•Exercises 19 and 20 (Ch. 17)

Chapter 10

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. Answer

H0: Mean weight lose = 10 pound (µ = 10)

H1: Mean weight lose < 10 pound (µ < 10)

Test Statistic used is Z test . Given that = 9, n= 50, = 2.8 Decision rule: Reject the null hypothesis, if the p value is less than the significance level 0.05. Details

Z Test of Hypothesis for the Mean

Data

Null Hypothesis =10

Level of Significance0.05

Population Standard Deviation2.8

Sample Size50

Sample Mean9

Intermediate Calculations

Standard Error of the Mean0.395979797

Z Test Statistic-2.525381361

Lower-Tail Test

Lower Critical Value-1.644853627

p-Value0.00577864

Reject the null hypothesis

P value = P (Z < -2.5253) = 0.00578

Conclusion: Reject the null hypothesis, since the P-value is less than the significance level. The sample provides enough evidence to conclude that weights lose is less than 10 pounds. 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. Answer

The null hypothesis tested is

H0: Mean weight of the can of sliced pineapple = 16 ounces. (µ = 16) The alternative hypothesis is

H1: Mean weight of the can of sliced pineapple > 16 ounces. (µ > 16) Significance level =0.05

The test statistic used is . Given that = 16.05, n= 50, = 0.03 Rejection criteria: Reject the null hypothesis, if the calculated value of the test statistic is greater than the critical value at the 0.05 significance level. Details

Z Test of Hypothesis for the Mean

Data

Null Hypothesis =16

Level of Significance0.05

Population Standard Deviation0.03

Sample Size50

Sample Mean16.05

Intermediate Calculations

Standard Error of the Mean0.004242641

Z Test Statistic11.78511302

Upper-Tail Test

Upper Critical Value1.644853627

p-Value0

Reject the null hypothesis

Conclusion: Reject the null hypothesis, since the calculated value of test statistic is greater than the critical value. The sample provides enough evidence to support the claim that the mean weight of the can is greater than 16 ounces. P-value = P (Z > 11.78511302) = 0.

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 rates (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. Answer

The null hypothesis tested is

H0: The 30-year mortgage rate for small banks is = 6 % (µ = 6%) The alternative hypothesis is

H1: The 30-year mortgage rate for small banks is < 6 % (µ < 6%) Significance level = 0.01

Test Statistic used is .

Decision rule: Reject the null hypothesis, if the value of test statistic is less than the critical value of t at 7 df. Details

t Test for Hypothesis of the Mean

Data

Null Hypothesis =6

Level of Significance0.01

Sample Size8

Sample Mean5.6375

Sample Standard Deviation...