In a 2011 study, medical researchers from Rutgers University randomly assigned men to take selenium, vitamin E, both selenium and vitamin E, or placebo. The researchers then followed up to see if the men developed prostate cancer. In this table, the observed count appears above the expected count in each cell. Observed 8177 8117 8147 8167 Expected 8180.3 8166.3 8133.5 8127.9 32608 Observed 575 620 555 529 Prostate Cancer Expected 571.7 570.7 568.5 568.1 2279 Reference: Klein, E.A., |.M. Thompson, C.M. Tangen, et al. (21 co-authors). 2011. Vitamin E and the risk of prostate cancer: the selenium and vitamin E cancer prevention trial (SELECT). Journal of the American Medical Association 306: 1549-1556. Researchers conducted a chi-square test of homogeneity to determine if there are statistically significant differences among the proportions taking selenium, vitamin E, both selenium and vitamin E, or placebo. Chi-square test results: chi-square test statistic = 7.7832, P-value = 0.051. If researchers include an exploratory data analysis with the test of homogeneity, they will calculate some percentages. If they use the count of men taking selenium who develop prostate cancer (575), what other number will they use to calculate a percentage relevant to their analysis? 0 8,752 0 2,279 0 34,887 Suppose that we want to investigate whether curfews correlate with differences in grades for students in middle school. We select a random sample of 81 middle school students. The variables are curfew (yeso) and grade (a letter grade that represents the average grade across courses). Is there an association between grade and curfew? Or are these two variables independent? We use the data to conduct a chi-square test of independence at the 5% level. In this results table, the observed count appears above the expected count in each cell. What can we conclude? Curfew A B C D Total Yes 10 28 15 1 54 8.667 30 14 1.333 No 3 17 6 1 27 4.333 15 7 0.6667 Total 13 45 21 2 81 Chi-Square Test Statistic DF Value P-value Chi-square 3 1.4796703 0.687 Q There is a statistically significant association between curfew and grades. 0 Nothing, because the conditions for use of the chi-square test are not met. 0 There is not a statistically significant association between curfew and grades. 0 A curfew helps students to perform better academically. Workers at a city owned swimming pool are investigating if the pool users accurately represent the neighborhood population. For three weeks, pool workers survey users about their neighborhood as the users sign-in. The table below provides the distribution of residents in the neighborhoods and the counts observed by pool workers. You can treat these 240 people as a random sample from the population of pool users. Pool workers conducted a chi-square goodness-offit test at the 5% significance level. The chi-square value is 1.3892 and the P-value is 0.7081. What can the pool workers conclude? O The proportion of pool users from Longfellow is significantly greater than 0.30. O The proportion of pool users from Piedmont is significantly less than 0.20. The distribution of pool users is significantly different from the distribution of residents in the O neighborhood. The distribution of pool users is not significantly different from the distribution of residents in the neighborhood. How patriotic are you? Would you say extremely patriotic, very patriotic, somewhat patriotic, or not especially patriotic? Below is the data from Gallup polls that asked this question of a random sample of US. adults in 1999 and a second independent random sample in 2010. We conducted a chi-square test of homogeneity to determine if there are statistically significant differences in the distribution of responses for these two years. In this results table, the observed count appears above the expected count in each cell. extremely patriotic very patriotic somewhat patriotic not especially patriotic Total 1999 193 466 284 51 994 257.2 443.8 237.3 55.72 2010 324 426 193 61 1004 259.8 448.2 239.7 56.28 Total 51 7 892 477 1 12 1998 Statistic DF Value P-value Chi-square 3 53.19187