Question 9 of 20
1.0/ 1.0 Points
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.
NW Location
NE Location
SE Location
Will Go Out
66
40
45
Won’t Go Out
20
25
20
.

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.
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) = 0.1294
Question 10 of 20
1.0/ 1.0 Points
A 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 A
School B
School C
Pass
40
33
50
Fail
59
45
67
After running an independence test, can it be concluded that pass/fail rates are dependent on
school?

B.
Yes, it can be concluded that pass/fail rates are dependent on school because the p-value
= 0.9373.
C.
Yes, it can be concluded that pass/fail rates are dependent on school because the p-value
= 0.0627.
D.
No, it cannot be concluded that pass/fail rates are dependent on school because the p-
value = 0.9373.
Answer Key: D
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) = 0.9373
0.9373 > .10, Do Not Reject Ho. No, it cannot be concluded that pass/fail rates are dependent on
school.