Test 2 Answer Guide

15 November 2000

Test 2 Answer Guide

Question 1. The four methods (aside from "rigged"!) are random, stratified random, cluster, and systematic. See 17 October class notes and the beginning of Chapter 7.

Question 2. The null hypothesis H₀ is "nothing much going on" -- µ = 1. The alternative hypothesis H_a is µ < 1. A Type I error (false rejection of H₀) would lead to a false accusation of the supermarket. A Type II error (false acceptance of H₀) would result in customers being cheated. See the early part of Chapter 9 and 31 October class notes.

Question 3. Attack: a biased estimator may, if the bias is reasonably small, be superior to an unbiased estimator with very large variance. See Section 8.2 and 24 October class notes.

Question 4. Continuous and bounded both ends: uniform, triangular, beta. Continuous unbounded on both ends: normal, Student's-t. Continuous bounded on exactly one end: lognormal (logarithms can't be negative!), chi-square, F, gamma, Weibull, exponential.

Question 5. Attack: There may be a strong non-linear relationship. See Section 6.5 and 5 October class notes.

Question 6. This question is similar to assigned homework from chapter 6. Using a double integral, c = 3 to keep the probability 1. Similarly, the probability that 0 <= x <= 0.25 and simultaneously 0 <= y <= 0.5 is 0.022. Since the product of the marginal densities in x and y equals the density function, x and y are independent by definition 6.9 page 274.

Question 7.
For group X, 115 has z-score = 3.0. For group Y, 115 has z-score 2.4. So 0.0013 is the fraction of group X qualifying and 0.0082 is the fraction of group Y qualifying. So the expected fraction of the workers from group X is 0.0013/(0.0013+0.0082) or less than one-seventh and the expected fraction from group Y is greater than six-sevenths. Shocked? This problem has much to say to those who would whine "Our test may be culturally or racially biased, but only a little!"

Question 8.
Use the chi-square test from section 8.10. See assigned homework problem 8.60.

Question 9.
Use the t-test for small samples (box page 366). See assigned homework problem 8.28.

Question 10.
Use the new interval construction method presented in class and illustrated with exercise 8.44 on 31 October. You presumably used this method to solve exercise 8.48.

Question 11.
This question is similar to exercise 7.24 page 318, and uses the central limit theorem. The breaking strength of the cable is normally distributed with mean 15,000. From box page 236, the variance of the strength of each wire is 450. The variance of the strength of the cable is 45,000, so the pertinent standard deviation is 212.1. The pertinent z-score is (14,705 - 15000)/212.1 = -1.39. The chance a standard normal z is less than this is about 0.08, which is the probability of breakage.

Question 12.
This question is similar to exercise 9.4 page 429. Use table 3 page 1092, as you did for Test 1 problem 10 or homework exercise 4.70. If l = 4.0, the chance of 7 or more such storms = 1 - 0.8893 = 0.1107, causing an incorrect rejection of H₀ (Type I error). Hence a = 0.1107. If l really is 5.0, the chance of 6 or fewer such storms, causing an incorrect acceptance of H₀ (Type II error), is 0.7622. Hence b = 0.7622.