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)