Goodness of Fit Test for Proportions
of Multinomial Population
•Chi-square distribution
•Hypotheses test/Goodness of fit test
Chi-square distribution
With 2 degrees
of freedom
With 5 degrees
of freedom
With 10 degrees
of freedom
2
0
Chi-Square Distribution
a2
• We will use the notation
to denote the
value for the chi-square distribution that
provides an area of a to the right of the
stated a2 value.
• For example, there is a .95 probability of
obtaining a 2 (chi-square) value such that

2
.975
 
2
2
.025
For 9 d.f. and a = .975
Selected Values from the Chi-Square Distribution Table
Area in Upper Tail
Degrees
of Freedom
5
6
7
8
9
10
.99
0.554
0.872
1.239
1.647
2.088
.975
0.831
1.237
1.690
2.180
2.700
.95
1.145
1.635
2.167
2.733
3.325
.90
1.610
2.204
2.833
3.490
4.168
2.558 3.247
3.940
4.865 15.987 18.307 20.483 23.209
2
Our

.975
.10
9.236
10.645
12.017
13.362
14.684
value
.05
11.070
12.592
14.067
15.507
16.919
.025
12.832
14.449
16.013
17.535
19.023
.01
15.086
16.812
18.475
20.090
21.666
.025
Area in
Upper Tail
= .975
2
0 2.700
For 9 d.f. and a = .025
Selected Values from the Chi-Square Distribution Table
Area in Upper Tail
Degrees
of Freedom
5
6
7
8
9
10
.99
0.554
0.872
1.239
1.647
2.088
.975
0.831
1.237
1.690
2.180
2.700
.95
1.145
1.635
2.167
2.733
3.325
.90
1.610
2.204
2.833
3.490
4.168
2.558 3.247
3.940
4.865 15.987 18.307 20.483 23.209
Our
.10
9.236
10.645
12.017
13.362
14.684
2
value
.025
.05
11.070
12.592
14.067
15.507
16.919
.025
12.832
14.449
16.013
17.535
19.023
.01
15.086
16.812
18.475
20.090
21.666
Area in Upper
Tail = .025
2
0
19.023
For 9 d.f. and a = .10
Selected Values from the Chi-Square Distribution Table
Area in Upper Tail
Degrees
of Freedom
5
6
7
8
9
10
.99
0.554
0.872
1.239
1.647
2.088
.975
0.831
1.237
1.690
2.180
2.700
.95
1.145
1.635
2.167
2.733
3.325
.90
1.610
2.204
2.833
3.490
4.168
2.558 3.247
3.940
4.865 15.987 18.307 20.483 23.209
Our .10 value
2
.10
9.236
10.645
12.017
13.362
14.684
.05
11.070
12.592
14.067
15.507
16.919
.025
12.832
14.449
16.013
17.535
19.023
.01
15.086
16.812
18.475
20.090
21.666
Area in Upper
Tail = .10
0
14.684
2
•
•
•
•
For 9 d.f. and
For 8 d.f. and
For 6 d.f. and
For 10 d.f. and
and .025
=16.919, a = .05
a2 =3.49, a = .90
a2 =16.812, a = .01
a2 =18.9, a = between .05
a2
Hypothesis (Goodness of Fit) Test
for Proportions of a Multinomial Population
This is simply a hypothesis test to see if the hypothesized population
proportions agree with the observed population proportions from
our sample.
1. Set up the null and alternative hypotheses.
H 0 : P1  P10 , P2  P20 , ... , Pk  Pk 0
H a : At least oneof the cell probabilities differs from the hypothesized value
2. Select a random sample and record the observed
frequency, fi , for each of the k categories.
3. Assuming H0 is true, compute the expected
frequency, ei , in each category by multiplying the
category probability by the sample size.
Hypothesis (Goodness of Fit) Test
for Proportions of a Multinomial Population
4. Compute the value of the test statistic.
2
(
f

e
)
2   i i
ei
i 1
k
where:
fi = observed frequency for category i
ei = expected frequency for category i
k = number of categories
Note: The test statistic has a chi-square distribution
with k – 1 df provided that the expected frequencies
are 5 or more for all categories.
Hypothesis (Goodness of Fit) Test
for Proportions of a Multinomial Population
5. Rejection rule:
p-value approach:
Reject H0 if p-value < a
Critical value approach:
Reject H0 if
 2  a2
where a is the significance level and
there are k - 1 degrees of freedom
Multinomial Distribution Goodness of Fit Test
• Example: Finger Lakes Homes (A)
Finger Lakes Homes manufactures
four models of prefabricated homes,
a two-story colonial, a log cabin, a
split-level, and an A-frame. To help
in production planning, management
would like to determine if previous
customer purchases indicate that there
is a preference in the style selected.
Multinomial Distribution Goodness of Fit Test
• Example: Finger Lakes Homes (A)
The number of homes sold of each
model for 100 sales over the past two
years is shown below.
SplitAModel Colonial Log Level Frame
# Sold
30
20
35
15
Multinomial Distribution Goodness of Fit Test

Hypotheses
H0: pC = pL = pS = pA = .25
Ha: The population proportions are not
pC = .25, pL = .25, pS = .25, and pA = .25
where:
pC = population proportion that purchase a colonial
pL = population proportion that purchase a log cabin
pS = population proportion that purchase a split-level
pA = population proportion that purchase an A-frame
Multinomial Distribution Goodness of Fit Test

Rejection Rule
Reject H0 if p-value < .05 or 2 > 7.815.
With a = .05 and
k-1=4-1=3
degrees of freedom
Do Not Reject H0
Reject H0
7.815
2
Multinomial Distribution Goodness of Fit Test
• Expected Frequencies
e1 = .25(100) = 25
e3 = .25(100) = 25
e2 = .25(100) = 25
e4 = .25(100) = 25
• Test Statistic
2
2
2
2
(
30

25
)
(
20

25
)
(
35

25
)
(
15

25
)
2 



25
25
25
25
=1+1+4+4
= 10
Multinomial Distribution Goodness of Fit Test
• Conclusion Using the p-Value Approach
Area in Upper Tail
.10
.05
.025
.01
.005
2 Value (df = 3)
6.251 7.815 9.348 11.345 12.838
Because 2 = 10 is between 9.348 and 11.345, the
area in the upper tail of the distribution is between
.025 and .01.
The p-value < a . We can reject the null hypothesis.
Multinomial Distribution Goodness of Fit Test
• Conclusion Using the Critical Value Approach
2 = 10 > 7.815
We reject, at the .05 level of significance,
the assumption that there is no home style
preference.