Question

18. Suppose the National Highway Safety Traffic Administration (NHSIA 1 Highway Safety Traffic Administration (NHSTA) wishes to study the average dom sample of 1372 cars, weight of all American-made compact care (Chrysler. Ford, GM). From a random sample or they acquired the sample mean to be 2982 pounds with a standard deviation of 16 pounds. 88% confidence interval for the appropriate parameter (approximate the t-score using approximate the f-score using the Z-score, and assume all conditions are met). a) (2973.269, 2990.731) b) (2975.096, 2988.905) c) (2970.525, 2993.475) d) (2965.264, 2999.105) e) (2972.653, 2973.223) 19. As a confidence interval's confidence level decreases, its margin of error a) Does not change b) Increases c) Decreases 20. For Z N (0,1), which of the following is a false statement? a) P(-2/2 < 2 < </2) = a b) P(Z > 2/2) = a/2 c) P(Z <-- a/2) = 0/2 d) P(12) > ža/2) = 0 21. In a simple linear regression model, it is assumed that the errors C1, ..., en a) All have mean 0 b) Are random and independent c) Are normally distributed d) All have the same variance 02 e) All of the above 22. Recently many companies have been experimenting with telecommuting, allowing employees to work at home on their computers. Among other things, telecommuting is supposed to reduce the number of sick days taken. Suppose that at one firm, it is known that over the past few years employees have taken a mean of 5.4 sick days (use the last two sentences to construct your alternative hypothesis). This year, the firm introduces telecommuting. Management chooses a simple random sample of 80 employees to follow in detail, and, at the end of the year, these employees average 4.7 sick days with a standard deviation of 2.5 days. Let it represent the mean number of sick days for all employees of the firm. Find the p-value for this hypothesis test. a) 0.0022 b) 0.0032 c) 0.0042 d) 0.0052 e) 0.0062

Answer 1

- 18) Given Date.. ; n = 1372, i = 2982 s=165 88% confidance interval fer population mean (c + 2x12x slon) (2982 + 20.064 165/51372 ) (2982 ± 1.5548 x 4-4546) ( 2982 + 6.9260) (2975.074, 2988-9 267 ł (2975.0969 2988.905) 19) Decreases. P(-2x12 C2L 2x12) - x 1 All of the above M 5.4 n = 80 x = 407, S-2.5 Ho: M 25.4vIS HIML5.4. under Ho, Test statistic w ł-4 4.7-5.4 - sion - 2.51550 = 2.50 P- verluer Pty LL) - P(Zrg (-2.5) - P(Z <-2 .5 ) " 0.00 62