Solutions to PSTAT 120C Practice Final: Hypothesis Testing and Chi-Square Goodness of Fit, Exams of Asian literature

Download Solutions to PSTAT 120C Practice Final: Hypothesis Testing and Chi-Square Goodness of Fit and more Exams Asian literature in PDF only on Docsity! PSTAT 120C: Solutions to Practice Final Questions June 5, 2009 1. The outcomes were Lost Won Lost Won Lost Lost Won Lost Won Lost Lost Lost Won Lost Won Lost (a) Calculate a P value for testing the null hypothesis that the outcome of each successive game is independent of the others. The number of Runs is R = 13 out of a sample of 6 Wins and 10 Losses. From Table 871, the P -value is P{R ≥ 13} = 1− P{R ≤ 12} = 1− .99 = .01. Therefore, P = 0.02. (b) We reject the null hypothesis that these outcomes are independent. There is a significant negative association between successive outcomes. A win increases the probability that the next outcome will be a loss. 2. (a) The expected value for each cell in this table comes out to be A B C F Total Juniors 14.1 33.0 23.0 2.9 73 Seniors 9.9 23.0 16.0 2.1 51 Total 24 56 39 5 124 The expected number of F’s for both Juniors and Seniors is less than 4 so the normal approximation is not appropriate. The best solution for this table is to aggregate the number of C’s and F’s into one column. The new table is A B C or F Juniors 16 32 25 Seniors 8 24 19 (b) The expected values for this table are A B C or F Juniors 14.1 33.0 25.9 73 Seniors 9.9 23.0 18.1 52 Total 24 56 44 124 Thus the test statistic is X2 = (16− 14.1)2 14.1 + (32− 33)2 33 + (25− 25.9)2 25.9 + + (8− 9.9)2 9.9 + (24− 23)2 23 + (19− 18.1)2 18.1 = 0.748 (c) There are 2 rows and 3 columns which leads to df = (2− 1)(3− 1) = 2. Thus, the critical value for a χ2 distribution for α = 0.01 is 9.21 (from Table 6). 1 (d) We conclude that there is not significant difference between the grades that Juniors are getting and the grades that the Seniors are getting. 3. The mean of the data is x̄ = 7.81 and thus the maximum likelihood estimator of λ̂ = 1/7.81 = 0.128 We can calculate the expected value for each cell in the table by using the formula for expo- nential probabilities P{X > t} = e−λt Thus P{X > 3} P{X > 6} P{X > 9} P{X > 14} 0.681 0.464 0.316 0.167 From which we can calculate(by subtraction) 0 < X ≤ 3 3 < X ≤ 6 6 < X ≤ 9 9 < X ≤ 14 X > 14 P 0.319 0.217 0.148 0.149 0.167 E 9.57 6.52 4.44 4.48 5.00 The test statistic is therefore X2 = (2− 9.57)2 9.57 + (13− 6.52)2 6.52 + (4.44− 3)2 4.44 + (9− 4.48)2 4.48 + (3− 5)2 5 = 18.26 This statistic has approximately a χ2 distribution with df = m − 2 = 3 degrees of freedom (remember we lose one degree of freedom because we estimated λ.) The critical value for α = 0.05 is 7.814 so there is a significant departure from the exponential model in this data. 4. If we assume that P(A) = P(B), then the expected number of observations in those two cells should be the same. Likewise for B and C. We could estimate P(A) = 12 ( 34+30 100 ) . The expected values under the null hypothesis are Event A B C D Observed 34 30 12 24 Expected 32 32 18 18 The chi-squared statistic is χ2 = 4 32 + 4 32 + 36 18 + 36 18 = 4.25 This χ2 has two degrees of freedom because we estimated two means. The value 4.15 is not significant when compare to the critical value χ2.95,2 = 5.991. Therefore, we accept the hypothesis that the probabilities are equal. 5. The table with its marginals is 2 Men No Show Deferred Served Total White 225 133 20 378 Black 47 52 7 106 Hispanic 87 52 2 141 Women No Show Deferred Served Total White 213 117 19 349 Black 56 68 5 129 Hispanic 100 63 5 168 (a) The expected number of Hispanic women that served on a jury is 168(58) 1271 = 7.6664 The expected number of Hispanic men that served on a jury is 141(58) 1271 = 6.4343 (b) The normal approximation is appropriate because the smallest marginal values lead to the expectation 106(58) 1271 = 4.837 which is greater than 4 which is our rule of thumb. (c) If the test statistic is X2 = 28.75, then the critical value is χ210,0.01 = 23.2093 and we conclude that there still is a significant relationship. 10. (a) For 38 observations with ȳ = 11.129, µ̂ = E(µ | ȳ) = ȳ ( nτ2 nτ2 + σ2 ) + η ( σ2 nτ2 + σ2 ) = 11.129 ( 38(5) 38(5) + 1 ) + 10 ( 1 191 ) = 11.1231 (b) The conditional variance is Var(µ | ȳ) = 1(5) 191 The 95% credible interval 11.1231± 1.96 √ 5 191 11.1231− 1.96 √ 5 191 = 10.806 11.1231 + 1.96 √ 5 191 = 11.44 5 (c) To generate an estimate on the original scale we need eµ. The Bayes estimator is E eµ. From the MGF we get E eX = exp [ µ+ 1 2 σ2 ] For the posterior distribution, E (eµ | ȳ) = exp [ 11.1231 + 5 191(2) ] = $68, 610 6