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.
