MATH 302 STATISTICS Week 8 Test Prep Knowledge Check Homework Practice Questions and Answers

Week 8 Knowledge Check Homework Practice Questions Part 1 of 4 - Chi-Square Test for Homogeneity 8.0/ 8.0 Points Question 1 of 20 1.0/ 1.0 Points Click to see additional instructions Students at a high school are asked to evaluate their experience in the class at the end of each school year. The courses are evaluated on a 1-4 scale – with 4 being the best experience possible. In the History Department, the courses typically are evaluated at 10% 1’s, 15% 2’s, 34% 3’s, and 41% 4’s. Mr. Goodman sets a goal to outscore these numbers. At the end of the year he takes a random sample of his evaluations and finds 10 1’s, 13 2’s, 48 3’s, and 52 4’s. At the 0.05 level of significance, can Mr. Goodman claim that his evaluations are significantly different than the History Department’s? Enter the test statistic - round to 4 decimal places. Test statistic= 3.0021 Answer Key: 3.0021 Feedback: 1's 2's 3's 4's Observed Counts 10 13 48 52 Expected Counts 123 *.10 = 12.3 123*.15 = 18.45 123*.34 = 41.82 123*.41 = 50.43 Test Stat = Question 2 of 20 1.0/ 1.0 Points Click to see additional instructions A company manager believes that a person’s ability to be a leader is directly correlated to their zodiac sign. He never selects someone to chair a committee without first evaluating their zodiac sign. An irate employee sets out to prove her manager wrong. She claims that if zodiac sign truly makes a difference in leadership, then a random sample of 210 CEO’s in our country would reveal a difference in zodiac sign distribution. She finds the following zodiac signs for her random sample of 210 CEO’s: Births Signs 25 Aries 13 Taurus 17 Gemini 21 Cancer 16 Leo 18 Virgo 15 Libra 16 Scorpio 20 Sagittarius 11 Capricorn 23 Aquarius 15 Pisces Can she conclude that zodiac sign makes a difference in whether or not a person makes a good leader? Enter the p-value - round to 4 decimal places. Make sure you put the 0 in front of the decimal. p-value = 0.4798 Answer Key: 0.4798 Feedback: The Expected Count is all the same value. 210*(1/12) = 17.5 Births Signs Expected Count 25 Aries 17.5 13 Taurus 17.5 17 Gemini 17.5 21 Cancer 17.5 16 Leo 17.5 18 Virgo 17.5 15 Libra 17.5 16 Scorpio 17.5 20 Sagittarius 17.5 11 Capricorn 17.5 23 Aquarius 17.5 15 Pisces 17.5 Use Excel to find the p-value =CHISQ.TEST(Highlight Observed, Highlight Expected) Question 3 of 20 1.0/ 1.0 Points Click to see additional instructions An urban economist is curious if the distribution in where Oregon residents live is different today than it was in 1990. She observes that today there are approximately 3,109 thousand residents in NW Oregon, 902 thousand residents in SW Oregon, 244 thousand in Central Oregon, and 102 thousand in Eastern Oregon. She knows that in 1990 the breakdown was as follows: 72.7% NW Oregon, 20.7% SW Oregon, 4.8% Central Oregon, and 2.8% Eastern Oregon. Can she conclude that the distribution in residence is different today at a 0.05 level of significance? Enter the test statistic - round to 4 decimal places. Test statistic= 10.1714 Answer Key: 10.1714 Feedback: NW Oregon SW Oregon Central Oregon Eastern Oregon Observed Counts 3109 902 244 102 Expected Counts 4357*.727 = 3167.539 4357*.207= 901.899 4357*.048= 209.136 4357*.028= 121.996 Test Stat = Question 4 of 20 1.0/ 1.0 Points Click to see additional instructions A large department store is curious about what sections of the store make the most sales. The manager has data from ten years prior that show 30% of sales come from Clothing, 25% Home Appliances, 18% Housewares, 13% Cosmetics, 12% Jewelry, and 2% Other. In a random sample of 550 current sales, 188 came from Clothing, 153 Home Appliances, 83 Housewares, 54 Cosmetics, 61 Jewelry, and 11 Other. At α=0.10, can the manager conclude that the distribution of sales among the departments has changed? Enter the test statistic - round to 4 decimal places. Test statistic = 12.2012 Answer Key: 12.2012 Feedback: Clothing Home App. Housewares Cosmetics Jewelry Other Observed Counts 188 153 83 54 61 11 Expected Counts 550*.30 = 165 550*.25 = 137.5 550*.18 = 99 550*.13 = 71.5 550*.12 = 66 550*.02= 11