Download Quiz Solutions for Proportions and Hypothesis Testing and more Quizzes Probability and Statistics in PDF only on Docsity! Quiz # 7 June 2, 2008 Name: Directions: Make sure to read each problem carefully. Make sure to show all of your work to receive full credit. Problem 1. To run a significance test comparing two proportions, we need to calculate the “pooled proportion” at some point. Supposing our conditions for inference are met, p̂1 = 0.3647, and p̂2 = 0.4015, the pooled proportion is equal to: (a) p̂ = 0.3647 (the smaller of the two proportions) (b) p̂ = 0.4015 (the larger of the two proportions) (c) p̂ = 0.3831 (the mean of the two proportions) (d) p̂ = 0.0368 (the difference of the two proportions) (e) p̂ cannot be determined as we don’t have the sample sizes, or numbers of successes. (f) None of the above. The pooled proportion is the sum of the successes divided by the sum of the sample sizes. If we had either the number of successes or the sample sizes, then we could make this calculation. Problem 2. Suppose that a magician has two weighted coins, and he thinks that coin A will come up heads a larger proportion of the time than coin B will. The magician decides to test this. If pA is the proportion of heads from coin A, and pB is the proportion of heads from coin B, what should his hypotheses be? (a) H0 : pA = pB; Ha : pA 6= pB (b) H0 : pA = pB; Ha : pA < pB (c) H0 : pA = pB; Ha : pA > pB (d) H0 : pA 6= pB; Ha : pA = pB (e) H0 : p̂A = p̂B; Ha : p̂A 6= p̂B (f) H0 : p̂A = p̂B; Ha : p̂A > p̂B (g) None of the above. Problem 3. A scientist on Mars wanted to know about the proportion of Martian children that played the game “Earth Invaders” at least once a week. She collected an SRS of 143 Martian children and found that 13 of them played the game at least once a week. Give a 95% confidence interval for the proportion p of all Martian children that play “Earth Invaders” at least once a week. Solution. Our conditions aren’t met for a large sample confidence interval, so we use a plus four. To find the confidence interval, we first calculate that p̃ = 15 147 ≈ 0.1020 and the z∗ for the 95% confidence level is z∗ = 1.96. Hence our confidence interval is p̃± z∗ √ p̃(1− p̃) n + 4 = 0.1020± 0.0489 = (0.0531, 0.1510)
