Click to see additional instructions A local gym is looking in to purchasing

# Click to see additional instructions a local gym is

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Question 11 of 201.0/ 1.0 PointsClick to see additional instructionsA 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 WeightsWeightMachinesEnduranceMachinesAerobicsEquipment0-10 Uses1217251311-30 Uses20189931+ Uses2612119Enter the P-Value - round to 4 decimal places. Make sure you put a 0 in front of the decimal. 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 Aerobics Equipment 0-10 Uses 12 17 25 13 11-30 Uses 20 18 9 9 31+ Uses 26 12 11 9 Sum 58 47 45 31 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 Count we can use Excel to find the p-value. Use =CHISQ.TEST(highlight actual counts, highlight expected counts) = 0.0144
Question 12 of 201.0/ 1.0 PointsA high school offers math placement exams for incoming freshmen to place students into the appropriate math class during their freshman year. Three different middle schools were sampled and the following pass/fail results were found. Run a test for independence at the 0.10 level of significance.School ASchool BSchool CPass403350Fail594567After running an independence test, can it be concluded that pass/fail rates are dependent on school? A.No, it cannot be concludedthat pass/fail rates are dependent on school because the p-value =0.9373.B. No, it cannot be concluded that pass/fail rates are dependent on school because the p-value = 0.0627.C. Yes, it can be concluded that pass/fail rates are dependent on school because the p-value = 0.9373.D. Yes, it can be concluded that pass/fail rates are dependent on school because the p-value = 0.0627.Answer Key:A 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. We need to calculate the Expected Counts. Then sum up the rows and column. You need to find the probability of the row and then multiple it by the column total. School A School B School C Sum Pass 40 33 50 123 Fail 59 45 67 171 Sum 99 78 117 294 School A School B School C Pass =99*(123/294)=78*(123/294) =117*(123/294)
Fail=99*(171/294)=78*(171/294) =117*(171/294)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) = 0.93730.9373 > .10, Do Not Reject Ho. No, it cannot be concluded that pass/fail rates are dependent onschool.
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