Gold Silver Bronze Sum Sum 11 42 42 Germany 16 9 6 31 31 Canada 13

# Gold silver bronze sum sum 11 42 42 germany 16 9 6 31

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Gold Silver Bronze Sum Sum Norway 17 14 11 42 42 Germany 16 9 6 31 31 Canada 13 9 11 33 33 United States 12 10 7 29 29 Sum Sum 58 58 42 42 35 35 135 135 Gold Silver Bronze Norway =58*(42/135) =42*(42/135) =35*(42/135) Germany =58*(31/135) =42*(31/135) =35*(31/135) Canada =58*(33/135) =42*(33/135) =35*(33/135) United States =58*(33/135) =42*(29/135) =35*(29/135)
1.0/ 1.0 Points Now that we calculated the Expected Counts we need to find the Test Statistic. Test Stat = You will need to all 12 Count values but I am only showing you 3 because there isn't room to write out the entire equation. Question 13 of 20 Click to see additional instructions An electronics store has 4 branches in a large city. They are curious if sales in any particular department are different depending on location. They take a random sample of 4 purchases throughout the 4 branches – the results are recorded below. Run an independence test for the data below at the 0.05 level of significance. Appliances TV Computers Cell Phones Branch 1 56 28 63 24 Branch 2 44 22 55 27 Branch 3 53 17 49 33 Branch 4 51 31 66 29 Enter the P -Value - round to 4 decimal places. Make sure you put a 0 in front of the decimal. ! 0.6099 Answer Key: Answer Key: 0.6099 Feedback: 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. Appliances TV Computers Cell Phones Sum Branch 1 56 28 63 24 171 Branch 2 44 22 55 27 148 Branch 3 53 17 49 33 152 Branch 4 51 31 66 29 177 Sum 204 98 233 113 648 Appliances TV Computers Cell Phones Branch 1 =204*(171/648) =98*(171/648) =233*(171/648) =113*(171/648)
0.0/ 1.0 Points Branch 2 =204*(148/648) =98*(148/648) =233*(148/648) =113*(148/648) Branch 3 =204*(152/648) =98*(152/648) =233*(152/648) =113*(152/648) Branch 4 =204*(177/648) =98*(177/648) =233*(177/648) =113*(177/648) 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) = 0.6099 Question 14 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 test statistic - round to 4 decimal places. % 15.8713 Answer Key: Answer Key: 15.8803 Feedback: 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 Aerobics Equipment Sum 0-10 Uses 12 17 25 13 67 11-30 Uses 20 18 9 9 56 31+ Uses 26 12 11 9 58 Sum 58 47 45 31 181
1.0/ 1.0 Points Free Weights Weight Machines Endurance Machines Aerobics Equipment 0-10 Uses =58*(67/181) =47*(67/181) =45*(67/181) =31*(67/181) 11-30 Uses =58*(56/181) =47*(56/181) =45*(56/181) =31*(56/181) 31+ Uses =58*(58/181) =47*(58/181) =45*(58/181) =31*(58/181) Now that we calculated the Expected Counts we need to find the Test Statistic.