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10
Chi-Square Tests and
the F-Distribution
Elementary Statistics
Larson
Farber
Section 10.1
Goodness of Fit
Chi-Square Distributions
Several important statistical tests use a
probability distribution known as chi square,
denoted .
for 1 or 2 d.f.
for 3 or more d.f.
0
0
is a family of distributions. The graph of the distribution
depends on the number of degrees of freedom (number of
free choices) in a statistical experiment.
The distributions are skewed right and are not
symmetric. The value of is greater than or equal to 0.
Multinomial Experiments
A multinomial experiment is a probability
experiment in which there are a fixed number
of independent trials and there are more than
two possible outcomes for each trial.
•The probability for each outcome is fixed.
•The sum of the probabilities of all possible
outcomes is one.
A chi-square goodness-of-fit test is used to test
whether a frequency distribution fits a specific
distribution.
Chi-Square Test for Goodness-of-Fit
Example: A social service organization claims 50% of all
marriages are the first marriage for both bride and
groom, 12% are first for the bride only, 14% for the
groom only and 24% a remarriage for both.
First Marriage
Bride and Groom
Bride only
Groom only
Neither
%
50
12
14
24
H0: The distribution of first-time marriages is 50% for
both bride and groom, 12% for the bride only, 14% for
the groom only. 24% are remarriages for both.
H1: The distribution of first-time marriages differs from
the claimed distribution.
Goodness-of-Fit Test
Observed frequency, O, is the frequency of the
category found in the sample.
Expected frequency, E, is the calculated
frequency for the category using the specified
distribution. Ei = npi
In a survey of 103 married couples, find the E =
expected number in each category.
First Marriage
%
E = np
Bride and Groom
Bride only
Groom only
Neither
50
12
14
24
103(.50) = 51.50
103(.12) = 12.36
103(.14) = 14.42
103(.24) = 24.72
Chi-Square Test
If the observed frequencies are obtained from a
random sample and each expected frequency is at
least 5, the sampling distribution for the goodnessof-fit test is a chi-square distribution with k – 1
degrees of freedom (where k = the number of
categories).
The test statistic is:
O = observed frequency in each category
E = expected frequency in each category
A social service organization claims 50% of all marriages are the
first marriage for both bride and groom, 12% are first for the bride
only, 14% for the groom only, and 24% a remarriage for both. The
results of a study of 103 randomly selected married couples are
listed in the table. Test the distribution claimed by the agency.
Use
.
First Marriage
Bride and Groom
Bride only
Groom only
Neither
f
55
12
12
24
1. Write the null and alternative hypothesis.
H0: The distribution of first-time marriages is 50% for both bride and
groom, 12% for the bride only, 14% for the groom only. 24% are
remarriages for both.
Ha: The distribution of first-time marriages differs from the claimed
distribution.
2. State the level of significance.
3. Determine the sampling distribution.
A chi-square distribution with 4 – 1 = 3 d.f.
4. Find the critical value.
5. Find the rejection region.
0
2
11.34
6. Find the test statistic.
Bride and groom
Bride only
Groom only
Neither
Total
%
50
12
14
24
100
O
55
12
12
24
103
E
51.5_
12.36
14.42
24.72
103.__
(O – E)2 (O – E) 2/E
12.25__
0.2379
0.1296
0.0105
5.8564
0.4061
0.5184
0.0210
0.6755
= 0.6755
0
11.34
7. Make your decision.
The test statistic 0.6755 does not fall in the rejection region,
so fail to reject H0.
8. Interpret your decision.
The distribution fits the specified distribution for first-time marriages.
Section 10.2
Independence
Test for Independence
A chi-square test may be used to determine whether
two variables (i.e., gender and job performance) are
independent. Two variables are independent if the
occurrence of one of the variables does not affect the
occurrence of the other.
The following contingency table reflects the gender
and job performance evaluation of 220 accountants.
Low
Average
Superior
Total
Male
22
81
9
112
Female
14
75
19
108
Total
36
156
28
220
Expected Values
Assuming the variables are independent, then the expected
value of each cell is:
E1,1 = (112)(36)/220 = 18.33
E1,2 = (112)(156)/220 = 79.42
All other expected values can be found by subtracting
from the total of the row or the column.
Low
Average
Superior
Total
Male
18.33
79.42
14.25
112
Female
17.67
76.58
13.75
108
Total
36
28
220
156
Sampling Distribution
The sampling distribution is a distribution
with degrees of freedom equal to:
(Number of rows – 1) (Number of columns – 1)
Example: Find the sampling distribution for a test of
independence that has a contingency table of 4
rows and 3 columns.
The sampling distribution is a
( 4 – 1) (3 – 1) = 3•2 = 6 d.f.
distribution with
Application
The following table reflects the gender and job performance
evaluation of 220 accountants. Test the claim that gender and
job performance are independent. Use
.
Low
Average
Superior
Total
Male
22
81
9
112
Female
14
75
19
108
Total
36
156
28
220
1. Write the null and alternative hypothesis.
H0: Gender and job performance are independent.
Ha: Gender and job performance are not independent.
2. State the level of significance.
3. Determine the sampling distribution.
Since there are 2 rows and 3 columns, the sampling
distribution is a chi-square distribution with (2 – 1)•(3 – 1) = 2 d.f.
4. Find the critical value.
0
5. Find the rejection region.
5.99
6. Find the test statistic.
Chi-Square Test
O
E
22
81
9
14
75
19
220
18.33
79.42
14.25
17.67
76.58
13.75
220.00
(O – E)2
13.49
2.50
27.61
13.49
2.50
27.61
= 5.51
(O – E)2/E
0.74
0.03
1.94
0.76
0.03
2.01
5.51
0
5.99
7. Make your decision.
The test statistic, 5.51, does not fall in
the rejection region, so fail to reject H0.
8. Interpret your decision.
Gender and job evaluation are independent
variables. Do not hire accountants based on
their gender, since gender does not influence
job performance levels.
Section 10.3
Comparing Two
Variances
Two-Sample Test for Variances
To compare population variances,
and
, use the F-distribution.
Let s12 and s22 represent the sample variances of two different
populations. If both populations are normal and the population
variances,
and
, are equal, then the sampling distribution
is called an F-distribution. s12 always represents the larger of
the two variances.
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
d.f.N = 8
d.f.D = 20
0
1
2
3
4
5
F-Test for Variances
To test whether variances of two normally distributed
populations are equal, randomly select a sample from
each population.
Let s12 and s22 represent the sample variances where
The test statistic is:
The sampling distribution is an F distribution with
numerator d.f. = n1 – 1 and denominator d.f. = n2 – 1.
In F-tests for equal variances, only use the right tail
critical value. For a right-tailed test, use the critical value
corresponding to the one in the table for the given .
For a two-tail test, use the right-hand critical value
corresponding to
.
Application
An engineer wants to perform a t-test to see if the mean gas
consumption of Car A is lower than that for Car B. A random
sample of gas consumption of 16 Car A’s has a standard
deviation of 4.5. A random sample of the gas consumption of 22
Car B’s has a standard deviation of 4.2. Should the engineer
use the t-test with equal variances or the one for unequal
variances? Use
.
1. Write the null and alternative hypothesis.
2. State the level of significance.
Since the sample variance
for Car A is larger than that
for Car B, use s12 to
represent the sample
variance for car A.
3. Determine the sampling distribution
An F distribution with d.f.N = 15, d.f.D = 21
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
4. Find the critical value.
5. Find the rejection region.
0.025
0
1
2
2.53
6. Find the test statistic.
3
4
5
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0.025
0
1
2
3
4
5
7. Make your decision.
Since F = 1.148 does not fall in the rejection region, fail to
reject the null hypothesis.
8. Interpret your decision.
There is not enough evidence to reject the claim that the
variances are equal. In performing a t-test for the means of
the two populations, use the test for equal variances.
Section 10.4
Analysis of Variance
ANOVA
One-way analysis of variance (ANOVA) is a hypothesis testing
technique that is used to compare means from three or more
populations.
H0:
(All population means are equal.)
Ha: At least one of the means is different from the others.
The variance is calculated in two
different ways and the ratio of the
two values is formed.
1. MSB, Mean Square Between, the variance between
samples, measures the differences related to the treatment
given to each sample.
2. MSW , Mean Square Within, the variance within samples,
measures the differences related to entries within the same
sample. The variance within samples is due to sampling error.
Mean Square Between
Each group is given a different “treatment.” The variation from
the grand mean (mean of all values in all groups) is
measured. The treatment (or factor) is the variable that
distinguishes members of one sample from another.
First calculate SSB and then divide by k – 1, the
degrees of freedom. (k = the number of treatments
or factors.)
Mean Square Within
Calculate SSW and divide by N – k, the degrees of freedom.
•If MSB is close in value to MSW, the variation is not
attributed to different effects the different treatments have
on the variable. The ratio of the two measures (F-ratio) is
close to 1.
•If MSB is significantly greater than MSW, the variation is
probably due to differences in the treatments or factors, and
the F-ratio will differ significantly from 1.
Analysis of Variance
The table shows the annual amount spent on reading (in $) for a
random sample of American consumers from four regions. At
, can you conclude that the mean annual amounts spent
are different?
Northeast
Midwest
South
308
246
103
58
169
143
246
141
164
109
158
119
220
167
99
144
76
214
316
108
1. Write the null and alternative hypothesis.
West
223
184
221
269
199
171
204
H0:
(All population means are equal.)
Ha: At least one of the means is different from the others.
2. State the level of significance.
3. Determine the sampling distribution.
An F distribution with d.f.N = 3, d.f.D = 23
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
4. Find the critical value.
5. Find the rejection region.
0.10
0
1
2
3
2.34
4
5
6. Find the test statistic.
Northeast
Midwest
South
West
308
58
141
109
220
144
316
246
169
246
158
167
76
103
143
164
119
99
214
108
223
184
221
269
199
171
204
Calculate the mean and variance for each sample.
210.14
177.00
135.71
1020.80
4050.05
1741.39
Calculate
the mean of all values.
Mean Square Between
1
2
3
4
mean
n
185.14
177.00
135.71
210.14
7
6
7
7
66.26
0.00
1704.86
1098.26
463.8
0.0
11934.0
7687.8
n
1
2
3
4
7
6
7
7
s2
9838.66
4050.05
1741.39
1020.80
59031.9
20250.2
10448.4
6124.8
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0.10
0
1
2
2.53
3
4
5
7. Make your decision.
Since F = 1.669 does not fall in the rejection region, fail to
reject the null hypothesis.
8. Interpret your decision.
There is not enough evidence to support the claim that
the means are not all equal. Expenses for reading are
the same for all four regions.
Minitab Output
One-way Analysis of Variance
Source
Factor
Error
Total
Analysis of Variance
DF
SS
MS
F
3 20085 6695 1.61
23 95857 4168
26 15942
P
0.215
Using the P-value method, fail to reject the null
hypothesis, since 0.215 > 0.10. There is not enough
evidence to support that the amount spent on reading is
different in different regions.