Chapter 10
Chi-Square Tests and the FDistribution
Larson/Farber 4th ed
1
Chapter Outline
•
•
•
•
10.1 Goodness of Fit
10.2 Independence
10.3 Comparing Two Variances
10.4 Analysis of Variance
Larson/Farber 4th ed
2
Section 10.1
Goodness of Fit
Larson/Farber 4th ed
3
Section 10.1 Objectives
• Use the chi-square distribution to test whether a
frequency distribution fits a claimed distribution
Larson/Farber 4th ed
4
Multinomial Experiments
Multinomial experiment
• A probability experiment consisting of a fixed
number of trials in which there are more than two
possible outcomes for each independent trial.
• A binomial experiment had only two possible
outcomes.
• The probability for each outcome is fixed and each
outcome is classified into categories.
Larson/Farber 4th ed
5
Multinomial Experiments
Example:
• A radio station claims that the distribution of music
preferences for listeners in the broadcast region is as
shown below.
Distribution of music Preferences
Classical
4% Oldies
2%
Country
36% Pop
18%
Gospel
11% Rock
29%
Each outcome is
classified into
categories.
Larson/Farber 4th ed
The probability for
each possible
outcome is fixed.
6
Chi-Square Goodness-of-Fit Test
Chi-Square Goodness-of-Fit Test
• Used to test whether a frequency distribution fits an
expected distribution.
• The null hypothesis states that the frequency
distribution fits the specified distribution.
• The alternative hypothesis states that the frequency
distribution does not fit the specified distribution.
Larson/Farber 4th ed
7
Chi-Square Goodness-of-Fit Test
Example:
• To test the radio station’s claim, the executive can
perform a chi-square goodness-of-fit test using the
following hypotheses.
H0: The distribution of music preferences in the
broadcast region is 4% classical, 36% country,
11% gospel, 2% oldies, 18% pop, and 29% rock.
(claim)
Ha: The distribution of music preferences differs from
the claimed or expected distribution.
Larson/Farber 4th ed
8
Chi-Square Goodness-of-Fit Test
• To calculate the test statistic for the chi-square
goodness-of-fit test, the observed frequencies and the
expected frequencies are used.
• The observed frequency O of a category is the
frequency for the category observed in the sample
data.
Larson/Farber 4th ed
9
Chi-Square Goodness-of-Fit Test
• The expected frequency E of a category is the
calculated frequency for the category.
 Expected frequencies are obtained assuming the
specified (or hypothesized) distribution. The
expected frequency for the ith category is
Ei = npi
where n is the number of trials (the sample size)
and pi is the assumed probability of the ith
category.
Larson/Farber 4th ed
10
Example: Finding Observed and
Expected Frequencies
A marketing executive randomly
selects 500 radio music listeners
from the broadcast region and asks
each whether he or she prefers
classical, country, gospel, oldies,
pop, or rock music. The results are
shown at the right. Find the
observed frequencies and the
expected frequencies for each type
of music.
Larson/Farber 4th ed
Survey results
(n = 500)
Classical
8
Country
210
Gospel
72
Oldies
10
Pop
75
Rock
125
11
Solution: Finding Observed and
Expected Frequencies
Observed frequency: The number of radio music
listeners naming a particular type of music
Survey results
(n = 500)
Classical
8
Country
210
Gospel
72
Oldies
10
Pop
75
Rock
125
Larson/Farber 4th ed
observed frequency
12
Solution: Finding Observed and
Expected Frequencies
Expected Frequency: Ei = npi
Type of
music
Classical
Country
Gospel
Oldies
Pop
Rock
Larson/Farber 4th ed
% of
listeners
4%
36%
11%
2%
18%
29%
Observed
frequency
8
210
72
10
75
125
n = 500
Expected
frequency
500(0.04) = 20
500(0.36) = 180
500(0.11) = 55
500(0.02) = 10
500(0.18) = 90
500(0.29) = 145
13
Chi-Square Goodness-of-Fit Test
For the chi-square goodness-of-fit test to be used, the
following must be true.
1. The observed frequencies must be obtained by using
a random sample.
2. Each expected frequency must be greater than or
equal to 5.
Larson/Farber 4th ed
14
Chi-Square Goodness-of-Fit Test
• If these conditions are satisfied, then the sampling
distribution for the goodness-of-fit test is approximated
by a chi-square distribution with k – 1 degrees of
freedom, where k is the number of categories.
• The test statistic for the chi-square goodness-of-fit test is
2
(
O

E
)
2  
E
The test is always
a right-tailed test.
where O represents the observed frequency of each
category and E represents the expected frequency of each
category.
Larson/Farber 4th ed
15
Chi-Square Goodness-of-Fit Test
In Words
1. Identify the claim. State the
null and alternative
hypotheses.
In Symbols
State H0 and Ha.
2. Specify the level of
significance.
Identify .
3. Identify the degrees of
freedom.
d.f. = k – 1
4. Determine the critical
value.
Use Table 6 in
Appendix B.
Larson/Farber 4th ed
16
Chi-Square Goodness-of-Fit Test
In Words
In Symbols
5. Determine the rejection region.
6. Calculate the test statistic.
7. Make a decision to reject or fail
to reject the null hypothesis.
(O  E)2
 
E
2
If χ2 is in the
rejection region,
reject H0.
Otherwise, fail to
reject H0.
8. Interpret the decision in the
context of the original claim.
Larson/Farber 4th ed
17
Example: Performing a Goodness of Fit
Test
Use the music preference data to perform a chi-square
goodness-of-fit test to test whether the distributions are
different. Use α = 0.01.
Distribution of
music preferences
Classical
4%
Country
36%
Gospel
11%
Oldies
2%
Pop
18%
Rock
29%
Larson/Farber 4th ed
Survey results
(n = 500)
Classical
8
Country
210
Gospel
72
Oldies
10
Pop
75
Rock
125
18
Solution: Performing a Goodness of Fit
Test
• H0: music preference is 4% classical, 36% country,
11% gospel, 2% oldies, 18% pop, and 29% rock
• Ha: music preference differs from the claimed or
expected distribution
• Test Statistic:
• α = 0.01
• d.f. = 6 – 1 = 5
• Rejection Region
• Decision:
0.01
0
Larson/Farber 4th ed
15.086
• Conclusion:
χ2
19
Solution: Performing a Goodness of Fit
Test
Type of
music
Classical
Country
Gospel
Oldies
Pop
Rock
Observed
frequency
8
210
72
10
75
125
Expected
frequency
20
180
55
10
90
145
2
(
O

E
)
2  
E
(8  20)2 (210  180)2 (72  55)2 (10  10)2 (75  90)2 (125  145)2






20
180
55
10
90
145
 22.713
Larson/Farber 4th ed
20
Solution: Performing a Goodness of Fit
Test
• H0: music preference is 4% classical, 36% country,
11% gospel, 2% oldies, 18% pop, and 29% rock
• Ha: music preference differs from the claimed or
expected distribution
• Test Statistic:
• α = 0.01
χ2 = 22.713
• d.f. = 6 – 1 = 5
• Rejection Region
• Decision: Reject H0
0.01
0
Larson/Farber 4th ed
χ2
15.086
22.713
There is enough evidence to
conclude that the distribution
of music preferences differs
from the claimed distribution.
21
Example: Performing a Goodness of Fit
Test
The manufacturer of M&M’s candies claims that the
number of different-colored candies in bags of dark
chocolate M&M’s is uniformly distributed. To test this
claim, you randomly select a bag that contains 500 dark
chocolate M&M’s. The results are shown in the table on
the next slide. Using α = 0.10, perform a chi-square
goodness-of-fit test to test the claimed or expected
distribution. What can you conclude? (Adapted from
Mars Incorporated)
Larson/Farber 4th ed
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Example: Performing a Goodness of Fit
Test
Color
Brown
Yellow
Red
Blue
Orange
Green
Frequency
80
95
88
83
76
78
n = 500
Larson/Farber 4th ed
Solution:
• The claim is that the
distribution is uniform, so
the expected frequencies of
the colors are equal.
• To find each expected
frequency, divide the sample
size by the number of colors.
• E = 500/6 ≈ 83.3
23
Solution: Performing a Goodness of Fit
Test
• H0: Distribution of different-colored candies in bags
of dark chocolate M&Ms is uniform
• Ha: Distribution of different-colored candies in bags
of dark chocolate M&Ms is not uniform
• Test Statistic:
• α = 0.10
• d.f. = 6 – 1 = 5
• Rejection Region
• Decision:
0.10
0
Larson/Farber 4th ed
9.236
• Conclusion:
χ2
24
Solution: Performing a Goodness of Fit
Test
2
(
O

E
)
2  
E
Color
Brown
Yellow
Red
Blue
Orange
Green
Observed
frequency
80
95
88
83
76
78
Expected
frequency
83.3
83.3
83.3
83.3
83.3
83.3
(80  83.3) 2 (95  83.3) 2 (88  83.3) 2 (83  83.3) 2 (76  83.3) 2 (78  83.3)2






83.3
83.3
83.3
83.3
83.3
83.3
 3.016
Larson/Farber 4th ed
25
Solution: Performing a Goodness of Fit
Test
• H0: Distribution of different-colored candies in bags
of dark chocolate M&Ms is uniform
• Ha: Distribution of different-colored candies in bags
of dark chocolate M&Ms is not uniform
• Test Statistic:
• α = 0.01
χ2 = 3.016
• d.f. = 6 – 1 = 5
• Rejection Region
• Decision: Fail to Reject H0
0.10
0
3.016
Larson/Farber 4th ed
9.236
χ2
There is not enough evidence
to dispute the claim that the
distribution is uniform.
26
Section 10.1 Summary
• Used the chi-square distribution to test whether a
frequency distribution fits a claimed distribution
Larson/Farber 4th ed
27
Section 10.2
Independence
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Section 10.2 Objectives
• Use a contingency table to find expected frequencies
• Use a chi-square distribution to test whether two
variables are independent
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Contingency Tables
r  c contingency table
• Shows the observed frequencies for two variables.
• The observed frequencies are arranged in r rows and
c columns.
• The intersection of a row and a column is called a
cell.
Larson/Farber 4th ed
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Contingency Tables
Example:
• The contingency table shows the results of a random
sample of 550 company CEOs classified by age and
size of company.(Adapted from Grant Thornton LLP, The
Segal Company)
Age
Company 39 and
70 and
40 - 49 50 - 59 60 - 69
size
under
over
Small /
Midsize
42
69
108
60
21
Large
5
18
85
120
22
Larson/Farber 4th ed
31
Finding the Expected Frequency
• Assuming the two variables are independent, you can
use the contingency table to find the expected
frequency for each cell.
• The expected frequency for a cell Er,c in a
contingency table is
Expected frequency Er,c 
Larson/Farber 4th ed
(Sum of row r)  (Sum of column c)
Sample size
32
Example: Finding Expected Frequencies
Find the expected frequency for each cell in the
contingency table. Assume that the variables, age and
company size, are independent.
Age
Company
size
Small /
Midsize
Large
Total
39 and
70 and
40 - 49 50 - 59 60 - 69
under
over
Total
42
69
108
60
21
300
5
18
85
120
22
250
47
87
193
180
43
550
marginal totals
Larson/Farber 4th ed
33
Solution: Finding Expected Frequencies
Er,c 
(Sum of row r)  (Sum of column c)
Sample size
Age
Company
size
39 and
70 and
40 - 49 50 - 59 60 - 69
under
over
Total
Small /
Midsize
42
69
108
60
21
300
Large
Total
5
47
18
87
85
193
120
180
22
43
250
550
300  47
E1,1 
 25.64
550
Larson/Farber 4th ed
34
Solution: Finding Expected Frequencies
Age
39 and
under
40 - 49
50 - 59
60 - 69
70 and
over
Total
Small /
Midsize
42
69
108
60
21
300
Large
5
18
85
120
22
250
Total
47
87
193
180
43
550
Company size
E1,2 
E1,4
300  87
 47.45
550
300  180

 98.18
550
Larson/Farber 4th ed
E1,3 
E1,5
300  193
 105.27
550
300  43

 23.45
550
35
Solution: Finding Expected Frequencies
Age
39 and
under
40 - 49
50 - 59
60 - 69
70 and
over
Total
Small /
Midsize
42
69
108
60
21
300
Large
5
18
85
120
22
250
Total
47
87
193
180
43
550
Company size
E2,1 
250  47
 21.36
550
E2,4 
Larson/Farber 4th ed
E2,2 
250  180
 81.82
550
250  87
250  193
 39.55 E2,3 
 87.73
550
550
E2,5 
250  43
 19.55
550
36
Chi-Square Independence Test
Chi-square independence test
• Used to test the independence of two variables.
• Can determine whether the occurrence of one variable
affects the probability of the occurrence of the other
variable.
Larson/Farber 4th ed
37
Chi-Square Independence Test
For the chi-square independence test to be used, the
following must be true.
1. The observed frequencies must be obtained by using
a random sample.
2. Each expected frequency must be greater than or
equal to 5.
Larson/Farber 4th ed
38
Chi-Square Independence Test
• If these conditions are satisfied, then the sampling
distribution for the chi-square independence test is
approximated by a chi-square distribution with
(r – 1)(c – 1) degrees of freedom, where r and c are the
number of rows and columns, respectively, of a
contingency table.
• The test statistic for the chi-square independence test is
2
(
O

E
)
2  
E
The test is always a
right-tailed test.
where O represents the observed frequencies and E
represents the expected frequencies.
Larson/Farber 4th ed
39
Chi-Square Independence Test
In Words
1. Identify the claim. State the
null and alternative
hypotheses.
2. Specify the level of
significance.
3. Identify the degrees of
freedom.
4. Determine the critical value.
Larson/Farber 4th ed
In Symbols
State H0 and Ha.
Identify .
d.f. = (r – 1)(c – 1)
Use Table 6 in
Appendix B.
40
Chi-Square Independence Test
In Words
In Symbols
5. Determine the rejection
region.
6. Calculate the test statistic.
7. Make a decision to reject or
fail to reject the null
hypothesis.
8. Interpret the decision in the
context of the original claim.
Larson/Farber 4th ed
(O  E)2
 
E
2
If χ2 is in the
rejection region,
reject H0.
Otherwise, fail to
reject H0.
41
Example: Performing a χ2 Independence
Test
Using the age/company size contingency table, can you
conclude that the CEOs ages are related to company
size? Use α = 0.01. Expected frequencies are shown in
parentheses.
Age
Company size
39 and
under
40 - 49
60 - 69
70 and
over
Small /
Midsize
42
(25.64)
69
108
60
(47.45) (105.27) (98.18)
21
(23.45)
300
Large
5
(21.36)
18
(39.55)
85
(87.73)
120
(81.82)
22
(19.55)
250
Total
47
87
193
180
43
550
Larson/Farber 4th ed
50 - 59
Total
42
Solution: Performing a Goodness of Fit
Test
•
•
•
•
•
H0: CEOs’ ages are independent of company size
Ha: CEOs’ ages are dependent on company size
α = 0.01
d.f. = (2 – 1)(5 – 1) = 4
• Test Statistic:
Rejection Region
• Decision:
0.01
0
Larson/Farber 4th ed
13.277
χ2
43
Solution: Performing a Goodness of Fit
Test
2
(
O

E
)
2  
E
(42  25.64) 2 (69  47.45) 2 (108  105.27) 2 (60  98.18) 2 (21  23.45) 2





25.64
47.45
105.27
98.18
23.45
(5  21.36) 2 (18  39.55) 2 (85  87.73) 2 (120  81.82) 2 (22  19.55)2





21.36
39.55
87.73
81.82
19.55
 77.9
Larson/Farber 4th ed
44
Solution: Performing a Goodness of Fit
Test
•
•
•
•
•
H0: CEOs’ ages are independent of company size
Ha: CEOs’ ages are dependent on company size
α = 0.01
d.f. = (2 – 1)(5 – 1) = 4
• Test Statistic:
χ2 = 77.9
Rejection Region
0.01
0
Larson/Farber 4th ed
χ2
13.277
77.9
• Decision: Reject H0
There is enough evidence to
conclude CEOs’ ages are
dependent on company size.
45
Section 10.2 Summary
• Used a contingency table to find expected frequencies
• Used a chi-square distribution to test whether two
variables are independent
Larson/Farber 4th ed
46
Section 10.3
Comparing Two Variances
Larson/Farber 4th ed
47
Section 10.3 Objectives
• Interpret the F-distribution and use an F-table to find
critical values
• Perform a two-sample F-test to compare two
variances
Larson/Farber 4th ed
48
F-Distribution
• Let s12 and s22 represent the sample variances of two
different populations.
• If both populations are normal and the population
variances σ 12 and σ 22 are equal, then the sampling
distribution of
s12
F 2
s2
is called an F-distribution.
Larson/Farber 4th ed
49
Properties of the F-Distribution
1. The F-distribution is a family of curves each of
which is determined by two types of degrees of
freedom:
 The degrees of freedom corresponding to the
variance in the numerator, denoted d.f.N
 The degrees of freedom corresponding to the
variance in the denominator, denoted d.f.D
2. F-distributions are positively skewed.
3. The total area under each curve of an F-distribution
is equal to 1.
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50
Properties of the F-Distribution
4. F-values are always greater than or equal to 0.
5. For all F-distributions, the mean value of F is
approximately equal to 1.
d.f.N = 1 and d.f.D = 8
d.f.N = 8 and d.f.D = 26
d.f.N = 16 and d.f.D = 7
d.f.N = 3 and d.f.D = 11
F
1
Larson/Farber 4th ed
2
3
4
51
Critical Values for the F-Distribution
1. Specify the level of significance .
2. Determine the degrees of freedom for the numerator,
d.f.N.
3. Determine the degrees of freedom for the denominator,
d.f.D.
4. Use Table 7 in Appendix B to find the critical value. If
the hypothesis test is
a. one-tailed, use the  F-table.
b. two-tailed, use the ½ F-table.
Larson/Farber 4th ed
52
Example: Finding Critical F-Values
Find the critical F-value for a right-tailed test when
α = 0.05, d.f.N = 6 and d.f.D = 29.
Solution:
The critical value is F0 = 2.43.
Larson/Farber 4th ed
53
Example: Finding Critical F-Values
Find the critical F-value for a two-tailed test when
α = 0.05, d.f.N = 4 and d.f.D = 8.
Solution:
• When performing a two-tailed hypothesis test using
the F-distribution, you need only to find the righttailed critical value.
• You must remember to use the ½α table.
1
1
  (0.05)  0.025
2
2
Larson/Farber 4th ed
54
Solution: Finding Critical F-Values
½α = 0.025, d.f.N = 4 and d.f.D = 8
The critical value is F0 = 5.05.
Larson/Farber 4th ed
55
Two-Sample F-Test for Variances
To use the two-sample F-test for comparing two
population variances, the following must be true.
1. The samples must be randomly selected.
2. The samples must be independent.
3. Each population must have a normal distribution.
Larson/Farber 4th ed
56
Two-Sample F-Test for Variances
• Test Statistic
s12
F 2
s2
where s12 and s22 represent the sample variances with
s12  s22.
• The degrees of freedom for the numerator is
d.f.N = n1 – 1 where n1 is the size of the sample
having variance s12.
• The degrees of freedom for the denominator is
d.f.D = n2 – 1, and n2 is the size of the sample having
variance s22.
Larson/Farber 4th ed
57
Two-Sample F-Test for Variances
In Words
1. Identify the claim. State the
null and alternative
hypotheses.
In Symbols
State H0 and Ha.
2. Specify the level of
significance.
Identify .
3. Identify the degrees of
freedom.
d.f.N = n1 – 1
d.f.D = n2 – 1
4. Determine the critical value.
Use Table 7 in
Appendix B.
Larson/Farber 4th ed
58
Two-Sample F-Test for Variances
In Words
5. Determine the rejection
region.
6. Calculate the test statistic.
7. Make a decision to reject or
fail to reject the null
hypothesis.
8. Interpret the decision in the
context of the original
claim.
Larson/Farber 4th ed
In Symbols
s12
F 2
s2
If F is in the
rejection region,
reject H0.
Otherwise, fail to
reject H0.
59
Example: Performing a Two-Sample FTest
A restaurant manager is designing a system that is
intended to decrease the variance of the time customers
wait before their meals are served. Under the old
system, a random sample of 10 customers had a
variance of 400. Under the new system, a random
sample of 21 customers had a variance of 256. At
α = 0.10, is there enough evidence to convince the
manager to switch to the new system? Assume both
populations are normally distributed.
Larson/Farber 4th ed
60
Solution: Performing a Two-Sample FTest
•
•
•
•
•
Because 400 > 256, s12  400 and s22  256
H0: σ12 ≤ σ22
• Test Statistic:
s12 400
Ha: σ12 > σ22
F 2 
 1.56
256
s2
α = 0.10
d.f.N= 9 d.f.D= 20
• Decision: Fail to Reject H0
There is not enough evidence
Rejection Region:
to convince the manager to
switch to the new system.
0.10
0
Larson/Farber 4th ed
1.561.96
F
61
Example: Performing a Two-Sample FTest
You want to purchase stock in a company and are
deciding between two different stocks. Because a
stock’s risk can be associated with the standard
deviation of its daily closing prices, you randomly select
samples of the daily closing prices for each stock to
obtain the results. At α = 0.05, can you conclude that
one of the two stocks is a riskier investment? Assume
the stock closing prices are normally distributed.
Stock A
n2 = 30
s2 = 3.5
Larson/Farber 4th ed
Stock B
n1 = 31
s1 = 5.7
62
Solution: Performing a Two-Sample FTest
•
•
•
•
•
Because 5.72 > 3.52, s12  5.72 and s22  3.52
H0: σ12 = σ22
• Test Statistic:
s12 5.7 2
Ha: σ12 ≠ σ22
F  2  2  2.65
s2
3.5
½α = 0. 025
d.f.N= 30 d.f.D= 29
• Decision: Reject H0
There is enough evidence to
Rejection Region:
support the claim that one of
the two stocks is a riskier
0.025
investment.
0
Larson/Farber 4th ed
2.092.65
F
63
Section 10.3 Summary
• Interpreted the F-distribution and used an F-table to
find critical values
• Performed a two-sample F-test to compare two
variances
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Section 10.4
Analysis of Variance
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Section 10.4 Objectives
• Use one-way analysis of variance to test claims
involving three or more means
• Introduce two-way analysis of variance
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One-Way ANOVA
One-way analysis of variance
• A hypothesis-testing technique that is used to
compare means from three or more populations.
• Analysis of variance is usually abbreviated ANOVA.
• Hypotheses:
 H0: μ1 = μ2 = μ3 =…= μk (all population means are
equal)
 Ha: At least one of the means is different from the
others.
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One-Way ANOVA
In a one-way ANOVA test, the following must be true.
1. Each sample must be randomly selected from a
normal, or approximately normal, population.
2. The samples must be independent of each other.
3. Each population must have the same variance.
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One-Way ANOVA
Variance between samples
Variance within samples
1. The variance between samples MSB measures the
differences related to the treatment given to each
sample and is sometimes called the mean square
between.
Test statistic 
2. The variance within samples MSW measures the
differences related to entries within the same sample.
This variance, sometimes called the mean square
within, is usually due to sampling error.
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One-Way Analysis of Variance Test
• If the conditions for a one-way analysis of variance
are satisfied, then the sampling distribution for the
test is approximated by the F-distribution.
• The test statistic is
MS B
F
MSW
• The degrees of freedom for the F-test are
d.f.N = k – 1 and d.f.D = N – k
where k is the number of samples and N is the
sum of the sample sizes.
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Test Statistic for a One-Way ANOVA
In Words
1. Find the mean and
variance of each sample.
2. Find the mean of all
entries in all samples (the
grand mean).
In Symbols
 x 2 (x  x )2
x
s 
n
n 1
x
x
N
3. Find the sum of squares
between the samples.
SS B   ni(xi  x )2
4. Find the sum of squares
within the samples.
SSW  (ni  1)si2
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Test Statistic for a One-Way ANOVA
In Words
In Symbols
5. Find the variance between the
samples.
SS B SS B
MS B 

k  1 d.f.N
6. Find the variance within the
samples
MSW 
7. Find the test statistic.
Larson/Farber 4th ed
SSW
SS
 W
N  k d.f.D
MS B
F
MSW
72
Performing a One-Way ANOVA Test
In Words
1. Identify the claim. State the
null and alternative
hypotheses.
2. Specify the level of
significance.
In Symbols
State H0 and Ha.
Identify .
3. Identify the degrees of
freedom.
d.f.N = k – 1
d.f.D = N – k
4. Determine the critical
value.
Use Table 7 in
Appendix B.
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Performing a One-Way ANOVA Test
In Words
In Symbols
5. Determine the rejection
region.
6. Calculate the test statistic.
7. Make a decision to reject or
fail to reject the null
hypothesis.
8. Interpret the decision in the
context of the original claim.
Larson/Farber 4th ed
F
MS B
MSW
If F is in the
rejection region,
reject H0.
Otherwise, fail to
reject H0.
74
ANOVA Summary Table
• A table is a convenient way to summarize the results in
a one-way ANOVA test.
Variation
Sum of
squares
Degrees of
freedom
Mean
squares
F
MS B
MSW
Between
SSB
d.f.N
SS B
MS B 
d . f .N
Within
SSW
d.f.D
MSW 
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SSW
d. f .D
75
Example: Performing a One-Way ANOVA
A medical researcher wants to determine whether there
is a difference in the mean length of time it takes three
types of pain relievers to provide relief from headache
pain. Several headache sufferers are randomly selected
and given one of the three medications. Each headache
sufferer records the time (in minutes) it takes the
medication to begin working. The results are shown on
the next slide. At α = 0.01, can you conclude that the
mean times are different? Assume that each population
of relief times is normally distributed and that the
population variances are equal.
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Example: Performing a One-Way ANOVA
Medication 1 Medication 2 Medication 3
12
16
14
15
17
12
14
21
15
17
20
15
19
x1 
56
 14
4
s12  6
x2 
85
 17
5
s22  8.5
x3 
66
 16.5
4
s32  7
Solution:
k = 3 (3 samples)
N = n1 + n2 + n3 = 4 + 5 + 4 = 13 (sum of sample sizes)
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Solution: Performing a One-Way ANOVA
• H0: μ1 = μ2 = μ3
• Test Statistic:
• Ha: At least one mean
is different
•
•
•
•
α = 0. 01
d.f.N= 3 – 1 = 2
d.f.D= 13 – 3 = 10
Rejection Region:
• Decision:
0.01
0
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7.56
F
78
Solution: Performing a One-Way ANOVA
To find the test statistic, the following must be calculated.
x
 x 56  85  66

 15.92
N
13
SS B  ni(xi  x ) 2
MS B 

d.f.N
k 1
4(14  15.92)2  5(17  15.92)2  4(16.5  15.92)2

3 1
21.92

 10.96
2
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Solution: Performing a One-Way ANOVA
To find the test statistic, the following must be calculated.
SSW (ni  1)si2
MSW 

d.f.D
N k
(4  1)(6)  (5  1)(8.5)  (4  1)(7)
13  3
73

 7.3
10

F
MS B 10.96
 1.50

MSW
7.3
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Solution: Performing a One-Way ANOVA
• H0: μ1 = μ2 = μ3
• Test Statistic:
MS B
• Ha: At least one mean
F
 1.50
is different
MSW
•
•
•
•
α = 0. 01
d.f.N= 3 – 1 = 2
d.f.D= 13 – 3 = 10
Rejection Region:
0.01
0
1.50
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7.56
F
• Decision: Fail to Reject H0
There is not enough evidence
at the 1% level of significance
to conclude that there is a
difference in the mean length
of time it takes the three pain
relievers to provide relief
from headache pain.
81
Example: Using the TI-83/84 to Perform a
One-Way ANOVA
Three airline companies offer flights between Corydon
and Lincolnville. Several randomly selected flight times
(in minutes) between the towns for each airline are
shown on the next slide. Assume that the populations of
flight times are normally distributed, the samples are
independent, and the population variances are equal. At
α = 0.01, can you conclude that there is a difference in
the means of the flight times? Use a TI-83/84.
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Example: Using the TI-83/84 to Perform a
One-Way ANOVA
Larson/Farber 4th ed
Airline 1
Airline 2
Airline 3
122
119
120
135
133
158
126
143
155
131
149
126
125
114
147
116
124
164
120
126
134
108
131
151
142
140
131
113
136
141
83
Solution: Using the TI-83/84 to Perform a
One-Way ANOVA
• H0: μ1 = μ2 = μ3
• Ha: At least one mean is different
• Store data into lists L1, L2, and L3
• Decision: P-value < α
Reject H0
There is enough evidence to support the claim. You can
conclude that there is a difference in the means of the flight
times.
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