Inference for Categorical Data 293 in C#
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Calculator Tip: It s a bit of a digression, but if you actually wanted to do the experiment in question 1, you would need to have an efficient way of counting the number of each outcome You certainly don t want to simply scroll through all 500 entries and tally each one Even sorting them first and then counting would be tedious (more so if n were bigger than 4) The easiest way is to draw a histogram of the data and then TRACE to get the totals Once you have your 500 values from randBin in L1, go to STAT PLOTS and set up a histogram for L1 Choose a WINDOW something like [ 05,45,1, 1,300,1,1] Be sure that Xscl is set to 1 You may need to adjust the Ymax from 300 to get a nice picture on your screen Then simply TRACE across the bars of the histogram and read the value of n for each outcome off of the screen The reason for having x go from 05 to 45 is so that the (integer) outcomes will be in the middle of each bar of the histogram A chisquare test for the homogeneity of proportions is conducted on three populations and one categorical variable that has four values Computation of the chisquare statistic yields X 2 = 172 Is this finding significant at the 001 level of significance Which of the following best describes the difference between a test for independence and a test for homogeneity of proportions Discuss the correctness of each answer a There is no difference because they both produce the same value of the chisquare test statistic b A test for independence has one population and two categorical variables, whereas a test for homogeneity of proportions has more than one population and only one categorical variable c A test for homogeneity of proportions has one population and two categorical variables, whereas a test for independence has more than one population and only one categorical variable d A test for independence uses count data when calculating chisquare and a test for homogeneity uses percentages or proportions when calculating chisquare Compute the expected value for the cell that contains the frog You are given the marginal distribution

