Stat 250.3
November 19, 2003
Activity 9: Chi-Square test
Situation 1: Professionals in a workplace were asked if they needed help with statistical analyses. The
following contingency table shows the results by highest educational degree earned (ugrad=Bachelor’s
degree only, grad=Graduate degree).
ugrad
grad
Total
yes
no
Total
392
362.01
121
150.99
513
1013
1042.99
465
435.01
1478
1405
586
1991
Chi-Sq = 11.370
DF = 1, P-Value = 0.001
A. What are the null and alternative hypotheses?
B. Are the conditions for the chi-square test met?
C. What is the test-statistic? ____ = ________
What is the p-value?
D. What can we conclude?
E. What can we conclude for the relationship between the proportions in these two groups? Can
we decide if one of the two proportios is greater than the other?
We could have done a different test in this case. Notice that the response variable was whether or not an
individual needed help with statistical analyses. Also notice that there are two independent groups, those with
bachelor’s degrees and those with graduate degrees. So, in this case, we could have done a two-proportion ztest to determine if there is a difference between the proportions of those with bachelor’s degrees and those
with graduate degrees who need help with statistical analyses…
What if we had done a two-proportion z test, instead? Below is the Minitab output:
Test and CI for Two Proportions
Sample
ugrad
grad
X
392
1013
N
513
1478
Sample p
0.764133
0.685386
Estimate for p(ugrad) – p(grad): 0.0787469
95% CI for p(ugrad) - p(grad): (0.0350424, 0.122451)
Test for p(ugrad) - p(grad) = 0 (vs not = 0): Z = 3.37
F. What are the hypotheses of this test?
P-Value = 0.001
Stat 250.3
November 19, 2003
G. What is the test statistic? _____= _________What is the p-value?
H. Compare the test statistic and the p-value with the ones of the chi-suqre test.
I. What is our conclusion?
J. What can we say about the relationship between these two tests?
Situation 2: Is there a relationship between an individual’s view on capital punishment and his/her
religion? A sample of adults provides the following results:
Rows: cappun
Catholic
238
78
316
Favor
Oppose
All
Columns: religion
Jewish
23
7
30
None
105
33
138
Other Protestant
25
753
12
206
37
959
All
1144
336
1480
Chi-Square = 3.641, DF = 4, P-Value = 0.457
Cell Contents -Count
A. Write the null and alternative hypotheses for this situation.
B. What is the p-value? ___________ What is the test statistic? ___=______.
C. What can you conclude?
A. If the data are NOT summarized:
Minitab 13 Direstions: Select Stat > Tables > Cross Tabulation, enter the two variables of interest in the box
of Classification variables, select the “Chi Square Analysis” option and “Above the Expected Count” option.
Minitab 14 Direstions: Select Stat > Tables > Cross Tabulation and Chi-Square …, enter one of the
variables in the box “For Rows” and the other in the box “For Columns”, click on the “Chi Square” button
and select Chi-Square Analysis and Expected Cell Counts options.
B. If the data are summarized in a 2-way table: You need to type in the table using r rows and c
columns in the Minitab spreadsheet. Then,
Minitab 13 and 14 Direstions: Select Stat > Tables > Chi Square Test (Table in Worksheet) and then enter
the columns that contain the 2-way table in the appropriate box.