One- and Two-Sample Tests of Hypothesis, Variance, and Chi-Squared Analysis Problem Sets
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: = 10 pounds
H1: < 10 pounds
Reject the null hypothesis if Z < -1.65
9.0 – 10.00 Z= ---------------- = -2.53 2.8/sqrt(50)
The null hypothesis is rejected, the average weight loss is less than 10 pounds.
p-value = .5000 - .4943 = .0057
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:
H0: = = 16 ounces
H1: = > 16 pounds
Reject the null hypothesis if Z > 1.65
16.05 – 16.00 Z= ---------------- = 11.79 .03/sqrt(50)
The null hypothesis is rejected, since 11.79 is greater than 1.65. The 16-ounce cans can be overfilled.
p-value ~ 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):
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:
H0: = 6 percent
H1: < 6 percent
Reject the null hypothesis if t < -2.99
5.638 – 6.0 t= ---------------- = -1.61
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: = 10 pounds
H1: < 10 pounds
Reject the null hypothesis if Z < -1.65
9.0 – 10.00 Z= ---------------- = -2.53 2.8/sqrt(50)
The null hypothesis is rejected, the average weight loss is less than 10 pounds.
p-value = .5000 - .4943 = .0057
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:
H0: = = 16 ounces
H1: = > 16 pounds
Reject the null hypothesis if Z > 1.65
16.05 – 16.00 Z= ---------------- = 11.79 .03/sqrt(50)
The null hypothesis is rejected, since 11.79 is greater than 1.65. The 16-ounce cans can be overfilled.
p-value ~ 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):
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:
H0: = 6 percent
H1: < 6 percent
Reject the null hypothesis if t < -2.99
5.638 – 6.0 t= ---------------- = -1.61