Question 11 of 20
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A manufacturing company knows that their machines produce parts that are defective on occasion. They have 4 machines producing parts, and want to test
if defective parts are dependent on the machine that produced it. They take a random sample of 321 parts and find the following results. Test at the 0.05
level of significance.
Machine 1
Machine 2
Machine 3
Machine 4

Defective
10
15
16
9
Non-Defective
72
75
66
58
Enter the test statistic - round to 4 decimal places.
1.9943
Answer
Key: 1.9943
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.
Machine 1
Machine 2
Machine 3
Machine 4
Sum
Defective
10
15
16
9
50
Non-Defective
72
75
66
58
271
Sum
82
90
82
67
321
Machine 1
Machine 2
Machine 3
Machine 4
Defective
=82*(50/321)
=90*(50/321)
=82*(50/321)
=67*(50/321)
Non-Defective
=82*(271/321)
=90*(271/321)
=82*(271/321)
=67*(271/321)
Now that we calculated the Expected Counts we need to find the Test Statistic.
Test Stat =
You will need to all 8 Count values but I am only showing you 3 because there isn't room to write out the entire equation.
Question 12 of 20
1.0/ 1.0 Points
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A restaurant 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 13 of 20
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A university changed to a new learning management system during the past school year. The school wants to find out how it’s working for the different
departments – the results in preference found from a survey are below. Run a test for independence at α=0.05.