Chi Square Practice Problems Name: _____________________________ Class: _____
Directions: 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?
Null Hypoth: There is no statistically significant difference between the observed results and the expected results. (Changing light intensity does not reduce baboon aggression.)
Light intensity
B, normal light
A, lower light
Aggressive acts
42
36
Total - 78
You would expect the values to be equal if there was no light difference, so just divide by 2 to get expected results…
Expected values Aggressive acts
B, normal light 39
A, lower light 39
Total - 78
Normal light
Lower light o
42
36 e
39
39
(o-e)
3
-3
(o-e)
2
9
9
(o-e)
2
/e
9/39 = .23
9/39 = .23
Deg of freed – 1, critical value at .05 = 3.84
X2 = .46
.46 < 3.84, so accept the null – light intensity does not have an effect on aggression.

2. At 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?
Null Hypoth: There is no statistically significant difference between the observed results and the expected results. (No door is chosen in preference to the others.)
Door
#1
#2
#3
Observed use
60 students
66 students
80 students
Expected use
68.67 students
68.67 students
68.67 students
Door
#1
#2
#3 o
60
66
80
206 students 206/3 = 69 expected for each door e
68.67
68.67
68.67
(o-e)
-8.67
-2.67
11.33
(o-e)
2
75.1689
7.1289
128.3689
(o-e)
2
/e
1.0946
.1038
1.8694
Deg of freedom
– 2, critical value at .05 = 5.99
3.068 < 5.99, accept the null – no preferred door choice
X2 = 3.068

3. A scientist predicts that the kittens born with a congenital birth defect will be 25% based on the hypothesis that it is caused be a recessive gene in that breed of cat. After surveying several litters, he found that 44 out of 125 kittens had the defect. Is his hypothesis correct?
Null Hypoth: There is no statistically significant difference between the observed results and the expected results. (The birth defect seen in these kittens is not due to anything other than Mendelian Genetics.)
Type of kitten born Observed
Normal 81
Congenital defect 44
Expected
94
31
Normal kittens o
81
Congenital defect 44
125 total kittens observed e
94
31
(o-e)
-13
13
(o-e)
2
169
169
(o-e)
2
/e
1.798
5.45
X2 = 7.25
Deg of freedom
– 1, critical value at .05 = 3.84
7.25 > 3.84, reject the null – something other than Mendelian Genetics with recessive genes is affecting the birth of kittens with a congenital defect.

4. Suppose you take a random sample of 30 students who are using a new math text and a second sample of 30 students who are using a more traditional text. You compare student achievement on the state test given to all students at the end of the course. Based on state test performance, would you recommend the new math book?
Observed results
New Textbook
Old Textbook
Passed State Test
26
22
Failed State Test
4
8
Null Hypoth: There is no statistically significant difference between the observed results and the expected results.
(Success on the state test was not due to differences in the math textbook.)
Now you need to determine the expected values to get a comparison…this is done with a simple addition of what you have from the observed data…here’s a formula for how to do that!
Expected results
New text
Old text
NT passed
NT failed
OT passed
OT failed o
26
4
22
8
Passed state test Failed state test
30x48/60 = 24 30x12/60 = 6
30x48/60 = 24 30x12/60 = 6 e
24
6
24
6
(o-e)
2
-2
-2
2
(o-e)
2
4
4
4
4
(o-e)
2
/e
.167
.667
.167
.667
X2 = 1.67
Deg of freedom
– 3, critical value at .05 = 7.82
1.67< 7.82 accept the null
– Success on the state test was not due to differences in the math textbook.

5. In a study of the effectiveness of an antipsychotic drug, patients treated with the drug were compared to patients
Observed results
Placebo
Antipsychotic totals receiving a placebo. In terms of the number relapsing, 698 of 1,068 patients relapsed after taking the placebo while
639 out of 2,127 patients relapsed after taking the antipsychotic drug. Test the prediction that the antipsychotic is significantly more effective in preventing relapse than the placebo.
Null Hypoth: There is no statistically significant difference between the observed results and the expected results.
(The antipsychotic drug is no more effective than the placebo.)
(use the tables for formulas given in #4)…
No relapse
370
1488
1858
Relapse
698
639
1337 totals
1068
2127
3195
Expected results
Placebo
Antipsychotic o
No relapse
621.01
1236.92 e
Relapse
447.26
890.74
(o-e) (o-e)
2
(o-e)
2
/e
621 -251 63,001 101.45 Placebo no relapse
370
Placebo relapse 698
Antipsych no relapse
1488
Antipsych relapse 639
447
1237
891
251
251
-252
63,001
63,001
63,504
140.94
50.93
71.27
X2 = 364.59
Deg of freedom – 3, critical value at .05 = 7.82
364.59 > 7.82 Reject the null – The antipsychotic drug is significantly more effective in preventing relapse.

6. A student makes a monohybrid cross with Drosophila (fruit flies). She crosses two heterozygotes for the white eye.
Ww x Ww. She expects to see a 3:1 phenotypic ratio of Red eyes (WW and Ww) to white eyes (ww) – this is her null hypothesis. She rears the next generation through to adult flies and counts the following numbers:
White eyes 210
Wild type (red eyes) 680
Perform a chi square analysis on these results and find out if it is close enough to 3:1 to fail to reject her null hypothesis.
Make sure to show all work and explain your conclusions.
Null Hypoth: There is no statistically significant difference between the observed results and the expected results. (Eye color in fruit flies follows Mendelian Genetics and nothing else is affecting it.)
Eye color
White
Red
Observed
210
680
Expected
223
667
White
Red o
210
680 e
223
667
(o-e)
-13
13
(o-e)
2
169
169
(o-e)
2
/e
.76
.25
Deg of freedom – 1, critical value at .05 = 3.84
X2 = 1.01
1.01 < 3.84, Accept the null – eye color in fruit flies follows Mendelian Genetics.