<p>AMS572.01 Quiz 9 Fall 2011 (*Take home team project, due Friday, November 18, before class) </p><p>Team Number:______Name of Group Leader:______</p><p>Instruction: This is a take home exam. You should first do it individually, and then discuss among your entire team (as divided for the team project.) ONE solution ONLY is expected from each team. Please provide complete solutions for full credit. </p><p>The Hutterite Brethren is a religious group that is essentially a closed population with almost all marriages within the group. The following table shows the distribution of sons in families with 7 children whose mothers were born between 1879 and 1936.</p><p>Sons 0 1 2 3 4 5 6 7</p><p>Count 0 6 14 25 21 22 9 1</p><p>(a) At the significance level of 0.05, please provide two approaches to test the hypothesis that the number of sons follows a binomial distribution with p = 0.5, where p is the probability that a child is male. (b) Discuss whether the two test approaches proposed in part (a) are equivalent or not. Please provide thorough discussions and justifications for full credit. </p><p>Answer: </p><p>For the first two approaches (1 & 2) proposed, the alternative hypothesis is that the distribution of the number of sons born may not necessarily be a binomial distribution. Therefore, the general model for the distribution of the number of sons born in a m-children family is a multinomial distribution with the probability of observing 0, 1, 2, …, m sons to be P1, </p><p>P2, .., Pm, and the only restriction is that . </p><p>Approach 1 -- .</p><p>* You will notice that Approach 1 is the Chi-square goodness-of-fit test we have learned in class.</p><p>1 (a) Using , the results below could be obtained.</p><p>Sons 0 0 0.766 - 1 6 5.359 0.003 2 14 16.078 0.269 3 25 26.797 0.120 4 21 26.797 1.254</p><p>5 22 16.078 2.181 6 9 5.359 2.452 7 1 0.766 - Total 98 6.278 Note that cell 0 was combined with cell 1, and cell 7 was combined with cell 6, to satisfy the requirement that no cell can have and no more than 1/5th of the can be distribution with p=0.5 is a plausible distribution.</p><p>, so we couldn’t reject .</p><p>Approach 2 -- .</p><p>* Approach 2 is the likelihood ratio test with large sample approximation.</p><p>(a) After combining cell 0 with 1, and cell 1 with 7, define .</p><p>Then under , , , . </p><p>.</p><p>Under , we only have , then , </p><p>.</p><p>, and .</p><p>Thus, , then , so we couldn’t reject .</p><p>(Under , there are 5 unknown parameters, since , thus the df of chi-square distribution is 5)</p><p>, this is where comes from.</p><p>(b) , and they both use as threshold, so 2 approaches are equivalent.</p><p>3 In the following two approaches (3 & 4), the alternative hypothesis is that the distribution is still binomial but p, the probability of giving birth to a son at each single birth, does not necessarily have to be 0.5.</p><p>Approach 3 -- the distribution is still binomial but p, the probability of giving birth to a son at each single birth, does not necessarily have to be 0.5</p><p>* Approach 3 is the likelihood ratio test for one population proportion with large sample approximation.</p><p>,</p><p>,</p><p>,</p><p>.</p><p>Thus, 0.276, then , , so we couldn’t reject .</p><p>Approach 4 -- the distribution is still binomial but p, the probability of giving birth to a son at each single birth, does not necessarily have to be 0.5</p><p>* Approach 4 is the large sample approximate Z-test for one population proportion.</p><p>(a) , .</p><p>, so we couldn’t reject .</p><p>(b) Approaches 3 and 4 are equivalent, </p><p> since .</p><p>However they are not equivalent to Approaches 1 & 2 because they have different alternative hypotheses! 5</p>