Name:____________________________________________
Chi Square Practice Problems
Solve all problems using a chi square analysis. You must use statistics to support your answers.
1. A zookeeper hypothesizes that changing the intensity of the light in the primate exhibits will
reduce the amount of aggression between the baboons. In exhibit A, with a lower light intensity,
he observes 36 incidences of aggression over a one month period. In exhibit B, with normal
lights, he observes 42 incidences of aggression. Should he support or reject his hypothesis?

Identify Null Hypothesis: There will be no differences between aggressive
behavior in Exhibit A with lower light intensity than in Exhibit B with higher light
intensity.

Identify Alternative Hypothesis Hypothesis : The animals in exhibit A with lower
light intensity with have a different number of incidence of aggression than an
animal in higher light intensity in exhibit B.

Calculate expected values:
Total incidence: 36 + 42 = 78
Expected= Total / # of categories
Expected = 78/2= 39

Make a table
Category
Exhibit A
Low Light
Intensity
Exhibit B
nirmal light
Totals
Observed
Freq.(O)
36
Expected
Freq.(E)
39
O-E
O-E Squared
O-E squared/E
36-39 = 3
9
9/39= 0.23
42
39
42-39=4
16
16/39 = 0.41
78
0.64

Calculate Chi Square value 0.64.

Calculate Degrees of freedom # categories -1 = 1

Compare your calculated value to critical value, and accept of reject null
hypothesis: Critical value = 3.841 & Calculated Value is 0.64
.
Accept the Null Hypothesis because the critical value is larger than our calculated
value.

Discuss the significance of rejecting or accepting null hypothesis.
The Chi square test suggests that there is no statistically significant difference between the
number of incidences of aggression that occurred between animals in low light (exhibit A)
and normal light (exhibit B). The results of the chi square test suggests that the small
difference that were observed between animals in low light (exhibit A) and normal light
(exhibit B) is probably just due to chance.
2. A high school, students can choose to enter one of three doors. Custodians noticed that door
#3 was always getting broken and suggested that more students use that door because it has a
hands-free opener. Science minded students counted the number of students entering each door
to see if the custodians were right.
Door #1 had 60 students enter | Door #2 had 66 students enter | Door #3 had 80 students enter.
Were the custodians right?

Identify Null Hypothesis: There will be no differences between the number of
students that used the 3 doors.

Identify Alternative Hypothesis Hypothesis : There will be more students that used
door # 3 than door # 1 and door # 2.

Calculate expected values:
Total door uses: 60 + 66 +80 = 206
Expected= Total / # of categories
Expected = 206/3= 69

Make a table
Category
Observed
Freq.(O)
Expected
Freq.(E)
O-E
O-E Squared
O-E squared/E
Door 1
60
69
9
81
81/69 = 1.17
Door 2
66
69
3
9
9/69=0.13
Door 3
80
69
11
121
1.75
Total
206
3.05

Calculate Chi Square value 3.05.

Calculate Degrees of freedom # categories -1 = 2

Compare your calculated value to critical value, and accept of reject null
hypothesis: Critical value = 5.991 & Calculated Value is 3.04

Discuss the significance of rejecting or accepting null hypothesis.
.
The Chi square test suggests that there is no statistically significant difference
between the number of door uses that occurred between doors1,2, 3 . The results of
the chi square test suggests that the small difference that were observed doors
1,2,3,is probably just due to chance.