Test Stat You will need to use all 16 Count values but I am only showing you

# Test stat you will need to use all 16 count values

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Test Stat = You will need to use all 16 Count values but I am only showing you the first and last value because there
isn't room to write out the entire equation. Question 12 of 20 1.0/ 1.0 Points Click to see additional instructions A high school runs a survey asking students if they participate in sports. The results are found below. Run an independence test for the data at α=0.01 . Freshmen Sophomores Juniors Seniors Yes 75 88 55 42 No 30 28 38 40 Enter the test statistic - round to 4 decimal places. 16.2406 Answer Key:16.2406 Feedback: We are running a Chi-Square Test for Independence. Copy and Paste the table into Excel. You are given the Observed Counts in the table. Next you need to sum the rows and columns. Once you have those you need to calculate the Expected Counts. You need to find the probability of the row and then multiple it by the column total. Freshmen Sophomores Juniors Yes 75 88 55 No 30 28 38 Sum 105 116 Freshmen Sophomores Juniors Yes =105*(260/396) =116*(260/396) =93*(260/396) No =105*(136/396) =116*(136/396) =93*(136/396) Now that we calculated the Expected Counts we need to find the Test Statistic. Test Stat = You will need to all 8 Count values but I am only showing you 3 because there isn't room to write out the entire equation. Question 13 of 20 1.0/ 1.0 Points Click to see additional instructions A local gym is looking in to purchasing more exercise equipment and runs a survey to find out the preference in exercise equipment amongst their members. They categorize the members based on how frequently they use the gym each month – the results are
below. Run an independence test at the 0.01 level of significance. Free Weights Weight Machines Endurance Machines Aerobics Equipment 0-10 Uses 12 17 25 13 11-30 Uses 20 18 9 9 31+ Uses 26 12 11 9 Enter the P -Value - round to 4 decimal places. 0.0144 Answer Key:0.0144 Feedback: We are running a Chi-Square Test for Independence. Copy and Paste the table into Excel. You are given the Observed Counts in the table. Next you need to sum the rows and columns. Once you have those you need to calculate the Expected Counts. You need to find the probability of the row and then multiple it by the column total. Free Weights Weight Machines Endurance Machines 0-10 Uses 12 17 25 11-30 Uses 20 18 9 31+ Uses 26 12 11 Sum 58 47 Free Weights Weight Machines Endurance Machines 0-10 Uses =58*(67/181) =47*(67/181) =45*(67/181) 11-30 Uses =58*(56/181) =47*(56/181) =45*(56/181) 31+ Uses =58*(58/181) =47*(58/181) =45*(58/181) Now that we calculated the Expected Count we can use Excel to find the p-value. Use =CHISQ.TEST(highlight actual counts, highlight expected counts) = 0.0144 Question 14 of 20 1.0/ 1.0 Points A public opinion poll surveyed a simple random sample of 550 voters in Oregon. The respondents were asked which political party they identified with most and were categorized by residence. Results are shown below. Decide if voting preference is independent from location of residence. Let α=0.05 .
Republican Democrat Independent NW Oregon 85 103 22 SW Oregon 45 66 10 Central Oregon 46 53 9 Eastern Oregon 67 33 11 Can it be concluded that voting preference is dependent on location of residence?
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