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

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:

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.

0.0/ 1.0 Points