Essay Type (show your work) (10 points each) Chapter 8 1.The weight of a product is measured in pounds. A sample of 64 units is taken from a batch. The sample yielded the following results: = 70 lbs....

1

True/ False
1. False
2. True
3. True
4. False
5. False
6. False
7. False
8. True
9. True
10. False
11. True
12. True
13. False
14. False
15. True
16. False
17. False
18. True
19. False
20. True
21. False
22. False
23. True
24. False
25. True
Multiple Choices
1. C. Narrower for 90% confidence than 95% confidence
2. E. None of the above
2

3. C. 2.769 to 3.231
4. A. .1608 to .2392
5. B. 1549
6. E. 166
7. A. Always
8. C. Less than 
9. B. Decreases
10. E. Reject H0 if t > 1.721
11. B. 1.679
12. A. Always
13. D. The alternative hypothesis states that the two classifications are statistically
independent
14. C. Reject H0 if 
2
> 5.99147
15. D. 16
16. C. Failed to reject H0 and conclude that the quality of the product does not significantly
differ among the three processes
17. D. Error terms depend on the explanatory variable
18. A. MSE
19. E. Sum of Squared differences between actual and predicted Y values
20. D. 0.802
21. C. Positive autocorrelation of error terms
22. E. 4, 17
23. D. 72.63%
24. B. F statistic
25. E. 16
Essay Type
Question 1
Sample
̅
3
Value of t statistics for 99% = 2.575
Standard error

√
√

Confidence interval ̅ ( )
Upper limit of 99% ( )
Lower limit of 99% ( )
99% ( )
Question 2
The given data can be summarized below:
̅
The hypotheses are highlighted below:
The claim of the manager is incorrect.
The claim of the manager is correct.
It is apparent that the valid test in this case is t test because the sample size is lower than 30
(as per central limit theorem) and the standard deviation of the population is unknown.
Hence, t statistics would be used.
Standard error

√

Calculated t statistics



 When significance level is 5%
4

Degree of freedom
The p value for 15 degree of freedom and one tailed test is 0.011321.
It can be seen that the p value is lower than significance level and hence, null hypothesis
would be rejected and alternative hypothesis would be accepted. Therefore, it can be
concluded that claim of the manager is correct i.e. the use of new cashier system will
decrease the average customer waiting time in the store.
 When significance level is 1%
Degree of freedom
The p value for 15 degree of freedom and one tailed test is 0.011321.
It can be seen that the p value is higher than significance level and hence, null hypothesis
would not be rejected. Therefore, it can be concluded that the claim of the manager is
incorrect i.e. the use of new cashier system will decrease the average customer waiting time
in the store.
Question 3
The given data can be summarized below:




( )
The hypotheses are highlighted below:
There is no grade inflation.
There is grade inflation.
In this case, as per central limit theorem, z statistics would be used because the sample size is
significantly higher than 30 and hence, the distribution is assumed to be normal distribution.
Standard error √( ) ( ) ( )
5





 When significance level is 5% ...