Answer Key 180789 Feedback We are running a Chi Square Test for Independence

Answer key 180789 feedback we are running a chi

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Answer Key: 18.0789 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
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NW Location NE Location SE Location Will Go Out 66 40 45 Won’t Go Out 20 25 20 0.0/ 1.0 Points calculate the Expected Counts. You need to find the probability of the row and then multiple it by the column total. Republican Democrat Independent Sum NW Oregon 85 103 22 210 SW Oregon 45 66 10 121 Central Oregon 46 53 9 108 Eastern Oregon 67 33 11 111 Sum 243 255 52 550 Republican Democrat Independent NW Oregon =243*(210/550) =255*(210/550) =52*(210/550) SW Oregon =243*(121/550) =255*(121/550) =52*(121/550) Central Oregon =243*(108/550) =255*(108/550) =52*(108/550) Eastern Oregon =243*(111/550) =255*(111/550) =52*(111/550) Now that we calculated the Expected Counts we need to find the Test Statistic. Test Stat = You will need to use all 12 Count values but I am only showing you 3 because there isn't room to write out the entire equation. Question 10 of 20 Click to see additional instructions A restaurant chain that has 3 locations in Portland is trying to determine which of their 3 locations they should keep open on New Year’s Eve. They survey a random sample of customers at each location and ask each whether or not they plan on going out to eat on New Year’s Eve. The results are below. Run a test for independence to decide if the proportion of customers that will go out to eat on New Year’s Eve is dependent on location. Use α=0.05 . Enter the P -Value - round to 4 decimal places. Make sure you put a 0 in front of the decimal. Answer Key: 0.1294 Feedback:
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0.0/ 1.0 Points 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. NW Location NE Location SE Location Sum Will Go Out 66 40 45 151 Won’t Go Out 20 25 20 65 Sum 86 65 65 216 NW Location NE Location SE Location Will Go Out =86*(151/216) =65*(151/216) =65*(151/216) Won’t Go Out =86*(65/216) =65*(65/216) =65*(65/216) 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) = 0.129 Question 11 of 20 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. Make sure you put a 0 in front of the decimal. 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.
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0.0/ 1.0 Points
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