1. Using Chi-Square
Both chi-square tests are calculated the same ways and provide the
same information, but answer two very different questions…
Chi-square test for homogeneity tests whether the distribution of a
categorical variable is the same for each of several populations or
treatments.
Chi-square test for association/independence tests whether two
categorical variables are associated in some population of interest.
It is common to see questions asking about association that require a
test for homogeneity and questions asking about differences between
proportions of variables when a test for association is required.
2. Using Chi-Square
Instead of paying attention to how the question is asked…. look at
how the data is produced
If the data came from two or more independent random samples or
treatment groups in a randomized experiment – do a chi-square test for
homogeneity
If the data came from a single random samples, with the individuals
classified according to two categorical variables – do a chi-square test
for association/independence
3. Using Chi-Square
Are men and women equally likely to suffer lingering fear from
watching scary movies as children? Researchers asked a random
sample of 117 college students to write narrative accounts of their
exposure to scary movies before the age of 13. More than one-fourth
of the students said that some of the fright symptoms are still present
when they are awake.
Fright
Symptoms
Male Female Total
Yes 7 29 36
No 31 50 81
Total 35 79 117
Expected Counts are printed below
observed counts & chi-square
contributions are below expected counts
MALE FEMALE Total
Yes
7
11.69
1.883
29
24.31
0.906
36
No
31
26.31
0.837
50
54.69
0.403
81
Total 38 79 117
Chi-Sq = 4.028, DF = 1, P-value = 0.045
4. Using Chi-Square
Explain why a chi-square test for association/independence and not a
chi-square test for homogeneity should be used.
Fright
Symptoms
Male Female Total
Yes 7 29 36
No 31 50 81
Total 35 79 117
Expected Counts are printed below
observed counts & chi-square
contributions are below expected counts
MALE FEMALE Total
Yes
7
11.69
1.883
29
24.31
0.906
36
No
31
26.31
0.837
50
54.69
0.403
81
Total 38 79 117
Chi-Sq = 4.028, DF = 1, P-value = 0.045
5. Using Chi-Square
Researchers decided to use the null hypothesis
H0: gender and ongoing fright symptoms are independent in the
population of interest.
State the correct alternative hypothesis.
Fright
Symptoms
Male Female Total
Yes 7 29 36
No 31 50 81
Total 35 79 117
Expected Counts are printed below
observed counts & chi-square
contributions are below expected counts
MALE FEMALE Total
Yes
7
11.69
1.883
29
24.31
0.906
36
No
31
26.31
0.837
50
54.69
0.403
81
Total 38 79 117
Chi-Sq = 4.028, DF = 1, P-value = 0.045
6. Using Chi-Square
Interpret the P-value in context. What conclusion would you draw at a
α = 0.01
Fright
Symptoms
Male Female Total
Yes 7 29 36
No 31 50 81
Total 35 79 117
Expected Counts are printed below
observed counts & chi-square
contributions are below expected counts
MALE FEMALE Total
Yes
7
11.69
1.883
29
24.31
0.906
36
No
31
26.31
0.837
50
54.69
0.403
81
Total 38 79 117
Chi-Sq = 4.028, DF = 1, P-value = 0.045
7. Using Chi-Square
We can turn quantitative variable in a categorical variable by grouping
intervals of values (quantities) together
(think about histograms)
Tax income brackets
Income City 1 City 2
Under $10,000 70 62
$10,000 to $19,999 52 63
$20,000 to $24,999 69 50
$25,000 to $34,999 22 19
$35,000 or more 28 24
8. Using Chi-Square
Large Sample Size condition
Observed
Counts:
Expected
Counts:
Pg. 721
Living Arrangements 19 20 21 22 Totals
Parents’ Home 324 378 337 318 1357
Another Person’s Home 37 47 40 38 162
Your Own Place 116 279 372 487 1254
Group Quarters 58 60 49 25 192
Other 5 2 3 9 19
Total 540 766 801 877 2984
Living Arrangements 19 20 21 22 Totals
Parents’ Home 245.57 348.35 364.26 398.82 1357
Another Person’s Home 29.32 41.59 43.49 47.61 162
Your Own Place 226.93 321.90 336.61 368.55 1254
Group Quarters 34.75 49.29 54.51 56.43 192
Other 3.44 4.88 5.10 5.58 19
Total 540 766 801 877 2984