1. 12.2 The Chi-Square Distribution
Advanced Math Topics
12.2 The Chi-Square Distribution

2. 3.0-4.0 G.P.A. 2.0-2.99 G.P.A. Below 2.0 G.P.A. Yes 28 36 11 No 22 44
A teacher surveyed his students to see if they would take a class from him again.
He organized the results by G.P.A. in the table below:
G.P.A.
G.P.A.
Below 2.0 G.P.A.
Yes 28 36 11 No 22 44 9
Is there a significant difference between the answers for each G.P.A.?
Use 5% level of significance.
We can use the Chi-square statistic to answer the question.

3. (O – E)2 Χ2 = Σ E Steps to find the Chi-Square Statistic:
To get the same answers as the back of
book, round all answers to the nearest thousandth.
sum of the top row
1) Find the total proportion p
p =
total sample size
2) Find the expected frequency for each box in the table
E = p(sample size of each column)
3) Find the chi-square statistic
(O – E)2
Χ2 = Σ E
O is the observed frequency
4) Look up the critical value in Table VIII on page A13 using
df = one less than the number of columns
5) If the test statistic is lower than the critical value then accept the null hypothesis
that the proportions from each column are different merely by chance. If the test
statistic is greater than the critical value, then reject the null hypothesis.
The difference in each column is significant.

4. G.P.A.
G.P.A.
Below 2.0 G.P.A.
Yes 28 36 11 No 22 44 9
E = 0.5(50) = (25)
E = 0.5(80) = (40)
E = 0.5(20) = (10)
The two E values in each column
must add to the total
E = 25
E = 40
E = 10
Total: 50 80 20
sum of the top row 75
The null hypothesis is that the proportions
of yes/no responses is not different between
the columns.
p = = =
0.5
total sample size
150
(O – E)2
(28 – 25)2
Χ2 = Σ = =
0.36 E 25
Do this for all six boxes and sum them up.
Χ2 =
0.40 +
0.10 +
0.36 +
0.40 +
0.10 =
1.72
Find the critical value using d.f. = 3-1 = 2 and look under the 0.05 column
Χ2 = 5.991
Since our test statistic (1.72) is less than the critical value (5.991), we accept
the null hypothesis. There is not a significant difference between the responses
of students with different GPA’s.

5. From the HW: P. 606
2) Environmentalists examined 400 dead fish to pinpoint the cause of death. They
found traces of 3 different pesticides. Using a 1% level of significance, test the null
hypothesis that there is no significant difference between the proportion of dead fish
with each of the different pesticides in their bodies.
Pesticide A
Pesticide B
Pesticide C
Yes 51 63 35 No 85 88 78
Answer: The test statistic, 3.199, is less than the critical
value, 9.210, we accept the null hypothesis that there is not
a significant difference between the proportion of dead fish
with each pesticide found in their bodies.

6. P. 606 #2-6, skip #4