HW 9 Stats

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Feb 20, 2024

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You must use software to solve these problems. Copy/paste the output to include with your answer. Answers without the results will receive zero credit. On the other hand, copy/pasted output without any interpretation will receive zero credit. Ch 9 homework problems (5 points each) 9.2 Physical education requirements. In addition to what is in the problem statement graph the data in a side-by-side bar plot and carry out the analyses through the chi square test. State Ho and Ha. Carry out the test. Interpret your findings. 9.2 Physical education requirements. In Exercise 8.41 (page 482), you analyzed data from a study that included 354 higher education institutions: 225 private and 129 public. Among the private institutions, 60 required a physical education course, while among the public institutions, 101 required a course. Your analysis in that exercise focused on the comparison of two proportions. Use these data to construct a two-way table for analysis and find the joint distribution, the marginal distributions, and the conditional distributions . Use these distributions to give a brief summary of the relationship between the type of institution and whether a physical education course is required Ho: Physical education requirements are not associated with the type of school Ha: Physical educations requirements are associated with the type of school. Private Public Total Required 60 101 161 Not required 165 28 193 Total 225 129 354 X-squared = 86.068, df = 1, p-value < 2.2e-16 Marginal distribution of school type: 225 private and 129 public Marginal distribution of physical education status: 161 required and 193 not required Conditional distribution: Probability that a private school requires physical ed : 60/225= .266 Probability that a private school does not require physical ed: 165/225= .733 Probability that a public school requires physical ed: 101/129= .783 Probability that a public school does not require physical ed: 28/129= .217 Probability that a school that requires physical ed is public: 101/161= .627 Probability that a school that requires physical ed is private: 60/161= .372 Probability that a school that does not require physical ed is public: 28/193= .145 Probability that a school that does not require physical ed is private: 165/193=.855 Joint distribution: 60 private and require, 101 public and required, 165 private and not required, 28 public and not required The p-value is approximately zero which means we reject the null. Reject Ho; there is evidence that physical education requirements are associated with the type of school.

Summary: Private schools tend to be more associated with not requiring physical education programs while public schools tend to be more associated with requiring physical education 9.26 Is there a random distribution of trees a-c 9.26 Is there a random distribution of trees? In Example 6.1 (page 329), we examined data concerning the longleaf pine trees in the Wade Tract and concluded that the distribution of trees in the tract was not random . Here is another way to examine the same question. First, we divide the tract into four equal parts, or quadrants, in the east–west direction. Call the four parts Q1 through Q4. Then we take a random sample of 100 trees and count the number of trees in each quadrant. Here are the data: Quadrant Q1 Q2 Q3 Q4 Count 18 22 39 21 a. If the trees are randomly distributed, we expect to find 25 trees in each quadrant. Why? Explain your answer. a. The trees are randomly distributed meaning it is equally likely for a tree to be in each quadrant. This means out of 100 trees, we must have approximately equal number of tress per quadrant equally splitting it up to 25 per quadrant. b. We do not really expect to get exactly 25 trees in each quadrant. Why? Explain your answer.

a. This is because of variation caused by chance. Chance causes variation and are small but the random variations do not significantly affect data. So it is not accurate to expect exactly 25 trees per quadrant but about. c. Perform the goodness-of-fit test for these data to determine if these trees are randomly scattered. Write a short report giving the details of your analysis and your conclusion. a. The p value is greater than 0.01 which means there is not enough statistical evidence to reject the null. Fail to reject Ho; not enough evidence to determine that the trees are not randomly scattered. Chi-squared test for given probabilities X-squared = 10.8, df = 3, p-value = 0.01286 9.33 Are Mexican Americans less likely to be selected as jurors? This is a follow up from HW8. 9.33 Are Mexican Americans less likely to be selected as jurors? Refer to Exercise 8.74 (page 485) concerning Castaneda v. Partida, the case where the Supreme Court review used the phrase “two or three standard deviations” as a criterion for statistical significance. Recall that there were 181,535 persons eligible for jury duty, of whom 143,611 were Mexican Americans. Of the 870 people selected for jury duty, 339 were Mexican Americans. We are interested in finding out if there is an association between being Mexican American and being selected as a juror. Formulate this problem using a two-way table of counts. Construct the 2×2 table using the variables Mexican American or not and juror or not. Find the X2 statistic and its P-value. Square the z statistic that you obtained in Exercise 8.74 and verify that the result is equal to the X2 statistic. > results$observed Juror Nonjuror Mexican American 339 143272 Not Mexican American 531 37393 > results$expected Juror Nonjuror Mexican American 688.2506 142922.75 Not Mexican American 181.7494 37742.25 8.74 a. X-squared= 848.35 -> test stat= sqrt(848.35)= 29.12645 b. P-value= 2.2e-16 X2= 849.99, Df=1, p-value~0 I see that the X2 value and test stat for chapter 8 are the same also with the same p-value. We reject Ho; there is evidence that Mexican Americans are less likely to be chosen as jurors.

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