One- and Two-Sample Tests of Hypothesis, Variance, and Chi-Squared Analysis Problem Sets
One- and Two-Sample Tests of Hypothesis, Variance, and Chi-squared Analysis Problem Sets
• 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 Significance 0.05
Population Standard Deviation 2.8
Sample Size 50
Sample Mean 9 Intermediate Calculations
Standard Error of the Mean 0.395979797
Z Test Statistic -2.525381361 Lower-Tail Test
Lower Critical Value -1.644853627 p-Value 0.00577864
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.
• 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 Significance 0.05
Population Standard Deviation 2.8
Sample Size 50
Sample Mean 9 Intermediate Calculations
Standard Error of the Mean 0.395979797
Z Test Statistic -2.525381361 Lower-Tail Test
Lower Critical Value -1.644853627 p-Value 0.00577864
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.