Transcript les5e_ppt_10.ppt

Chapter 10
Chi-Square Tests and
the F-Distribution
© 2012 Pearson Education, Inc.
All rights reserved.
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Chapter Outline
•
•
•
•
10.1 Goodness of Fit
10.2 Independence
10.3 Comparing Two Variances
10.4 Analysis of Variance
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Section 10.1
Goodness of Fit
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Section 10.1 Objectives
• Use the chi-square distribution to test whether a
frequency distribution fits a claimed distribution
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Multinomial Experiments
Multinomial experiment
• A probability experiment consisting of a fixed
number of independent trials in which there are more
than two possible outcomes for each trial.
• The probability for each outcome is fixed and each
outcome is classified into categories.
• Recall that a binomial experiment had only two
possible outcomes.
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Multinomial Experiments
• A tax preparation company wants to determine the
proportions of people who used different methods to
prepare their taxes.
• The company can perform a multinomial experiment.
• It wants to test a previous survey’s claim concerning
the distribution of proportions of people who use
different methods to prepare their taxes.
• It can compare the distribution of proportions
obtained in the multinomial experiment with the
previous survey’s specified distribution.
• It can perform a chi-square goodness-of-fit test.
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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.
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Multinomial Experiments
• Results of a survey of tax preparation methods.
Each
outcome is
classified
into
categories.
Distribution of tax preparation
methods
Accountant
25%
By hand
20%
Computer software
35%
Friend/family
Tax preparation service
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5%
The
probability
for each
possible
outcome is
fixed.
15%
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Chi-Square Goodness-of-Fit Test
• To test the previous survey’s claim, a company can
perform a chi-square goodness-of-fit test using the
following hypotheses.
H0: The distribution of tax preparation methods is 25%
by accountant, 20% by hand, 35% by computer
software, 5% by friend or family, and 15 % by tax
preparation service. (claim)
Ha: The distribution of tax preparation methods differs
from the claimed or expected distribution.
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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.
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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.
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Example: Finding Observed and
Expected Frequencies
A tax preparation company
randomly selects 300 adults and
asks them how they prepare their
taxes. The results are shown at the
right. Find the observed frequency
and the expected frequency for each
tax preparation method. (Adapted
from National Retail Federation)
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Survey results
(n = 300)
Accountant
By hand
Computer
software
Friend/family
Tax preparation
service
71
40
101
35
53
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Solution: Finding Observed and
Expected Frequencies
Observed frequency: The number of adults in the
survey naming a particular tax preparation method
Survey results
(n = 300)
Accountant
By hand
Computer
software
Friend/family
Tax preparation
service
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observed frequency
71
40
101
35
53
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Solution: Finding Observed and
Expected Frequencies
Expected Frequency: Ei = npi
Tax preparation
method
Accountant
% of
people
25%
Observed
frequency
71
Expected
frequency
300(0.25) = 75
By hand
20%
40
300(0.20) = 60
Computer Software
35%
101
300(0.35) = 105
5%
35
300(0.05) = 15
15%
53
300(0.15) = 45
Friend/family
Tax preparation
service
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n = 300
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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.
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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.
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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 α.
3. Identify the degrees of
freedom.
d.f. = k – 1
4. Determine the critical
value.
Use Table 6 in
Appendix B.
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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.
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Example: Performing a Goodness of Fit Test
Use the tax preparation method data to perform a chisquare goodness-of-fit test to test whether the
distributions are different. Use α = 0.01.
Distribution of tax preparation
methods
Accountant
25%
By hand
20%
Computer
software
Friend/family
35%
Tax preparation
service
15%
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5%
Survey results
(n = 300)
Accountant
By hand
Computer
software
Friend/family
Tax
preparation
service
71
40
101
35
53
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Solution: Performing a Goodness of Fit Test
The distribution is 25% by accountant, 20% by
• H0: hand, 35% by computer software, 5% by friend/
family, and 15% by tax preparation service. (Claim)
• Ha: The distribution of tax preparation methods
differs from the claimed or expected distribution.
α = 0.01
• d.f. = 5 – 1 = 4
• Rejection Region
• Test Statistic:
• Decision:
• Conclusion:
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Solution: Performing a Goodness of Fit Test
Tax
preparation
method
Accountant
By hand
Computer
hardware
Friend/family
Tax preparation
service
Observed
frequency
Expected
frequency
71
40
101
75
60
105
35
53
15
45
2
2
2
2
2
2
(
O

E
)
(71

75)
(40

60)
(101

105)
(35

15)
(53

45)
2  





E
75
60
105
15
45
 35.121
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Solution: Performing a Goodness of Fit Test
•
•
The distribution is 25% by accountant, 20% by
H0: hand, 35% by computer software, 5% by friend/
family, and 15% by tax preparation service. (Claim)
Ha: The distribution of tax preparation methods
differs from the claimed or expected distribution.
• Test Statistic:
α = 0.01
2 ≈ 35.121
χ
d.f. = 5 – 1 = 4
•
•
• Rejection Region
• Decision: Reject H0
There is enough evidence at the 1%
significance level to conclude that the
distribution of tax preparation methods
differs from the previous survey’s
claimed or expected distribution. 22 of 91
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Example: Performing a Goodness of Fit Test
A researcher 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)
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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
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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
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Solution: Performing a Goodness of Fit Test
• H0: Distribution of different-colored candies in bags
of dark chocolate M&M’s is uniform. (Claim)
• Ha: Distribution of different-colored candies in bags
of dark chocolate M&M’s is not uniform.
• Test Statistic:
• α = 0.10
• d.f. = 6 – 1 = 5
• Rejection Region
• Decision:
0.10
0
9.236
• Conclusion:
χ2
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Solution: Performing a Goodness of Fit Test
2
(
O

E
)
2  
E 2
Color
Brown
Yellow
Red
Blue
Orange
Green
Observed
frequency
80
95
88
83
76
78
Expected
frequency
83.33
83.33
83.33
83.33
83.33
83.33
(80  83.33) (95  83.33)2 (88  83.33)2



83.33
83.33
83.33
(83  83.33)2 (76  83.33)2 (78  83.33)2



83.33
83.33
83.33
 3.016
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Solution: Performing a Goodness of Fit Test
• H0: Distribution of different-colored candies in bags
of dark chocolate M&M’s is uniform. (Claim)
• Ha: Distribution of different-colored candies in bags
of dark chocolate M&M’s 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
9.236
χ2
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There is not enough evidence at
the 10% level of significance to
reject the claim that the
distribution is uniform.
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Section 10.1 Summary
• Used the chi-square distribution to test whether a
frequency distribution fits a claimed distribution
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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.
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Contingency Tables
Example:
• The contingency table shows the results of a random
sample of 2200 adults classified by their favorite way
to eat ice cream and gender. (Adapted from Harris
Interactive)
Favorite way to eat ice cream
Gender
Male
Female
Cup
Cone
600
288
204
24
84
410
340
180
20
50
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Sundae Sandwich Other
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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 
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(Sum of row r)  (Sum of column c)
Sample size
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Example: Finding Expected Frequencies
Find the expected frequency for each cell in the
contingency table. Assume that the variables, favorite
way to eat ice cream and gender, are independent.
Favorite way to eat ice cream
Gender
Male
Female
Total
Cup
Cone
Sundae Sandwich Other
Total
600
288
204
24
84
1200
410
1010
340
628
180
384
20
44
50
134
1000
2200
marginal totals
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Solution: Finding Expected Frequencies
Er,c 
(Sum of row r)  (Sum of column c)
Sample size
Favorite way to eat ice cream
Gender
Male
Female
Total
Cup
Cone
Sundae
Sandwich
Other
Total
600
288
204
24
84
1200
410
1010
340
628
180
384
20
44
50
134
1000
2200
1200 1010
E1,1 
 550.91
2200
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Solution: Finding Expected Frequencies
Favorite way to eat ice cream
Cup
Cone
Sundae
Sandwich
Other
Total
600
288
204
24
84
1200
Female
410
340
180
20
50
1000
Total
1010
628
384
44
134
2200
Gender
Male
E1,2 
1200  628
 342.55
2200
1200  44
E1,4 
 24
2200
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E1,3 
1200  384
 209.45
2200
1200 134
E1,5 
 73.09
2200
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Solution: Finding Expected Frequencies
Favorite way to eat ice cream
Cup
Cone
Sundae
Sandwich
Other
Total
Male
600
288
204
24
84
1200
Female
410
340
180
20
50
1000
Total
1010
628
384
44
134
2200
Gender
E2,1 
1000 1010
1000  628
1000  384
 459.09E2,2 
 285.45E2,3 
 174.55
2200
2200
2200
E2,4 
1000  44
 20
2200
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E2,5 
1000 134
 60.91
2200
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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.
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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.
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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.
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Chi-Square Independence Test
In Words
1. Identify the claim. State the
null and alternative
hypotheses.
2. Specify the level of
significance.
3. Determine the degrees of
freedom.
4. Determine the critical value.
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In Symbols
State H0 and Ha.
Identify α.
d.f. = (r – 1)(c – 1)
Use Table 6 in
Appendix B.
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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.
© 2012 Pearson Education, Inc. All rights reserved.
(O  E) 2
 
E
2
If χ2 is in the
rejection region,
reject H0.
Otherwise, fail to
reject H0.
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Example: Performing a χ2 Independence Test
Using the gender/favorite way to eat ice cream
contingency table, can you conclude that the adults
favorite ways to eat ice cream are related to gender? Use
α = 0.01. Expected frequencies are shown in
parentheses.
Favorite way to eat ice cream
Gender
Cup
Cone
Sundae
Sandwich
Other
Total
Male
600
288
204
(550.91) (342.55) (209.45)
24
(24)
84
(73.09)
1200
Female
410
340
180
(459.09) (285.45) (174.55)
20
(20)
50
(60.91)
1000
44
134
2200
Total
1010
628
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384
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Solution: Performing a Goodness of Fit Test
• H0: The adults’ favorite ways to eat ice cream are
independent of gender.
• Ha: The adults’ favorite ways to eat ice cream are
dependent on gender. (Claim)
• α = 0.01
• Test Statistic:
• d.f. = (2 – 1)(5 – 1) = 4
• Decision:
• Rejection Region
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Solution: Performing a Goodness of Fit Test
2
(
O

E
)
2  
E
(600  550.91)2 (288  342.55)2 (204  209.45)2 (24  24)2 (84  73.09)2





550.91
342.55
209.45
24
73.09
(410  459.09)2 (340  285.45)2 (180  174.55)2 (20  20)2 (50  60.91)2





459.09
285.45
174.55
20
60.91
 32.630
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Solution: Performing a Goodness of Fit Test
•
•
•
•
•
The adults’ favorite ways to eat ice cream are
H0: independent of gender.
Ha: The adults’ favorite ways to eat ice cream are
dependent on gender. (Claim)
α = 0.01
d.f. = (2 – 1)(5 – 1) = 4
• Test Statistic:
χ2 ≈ 32.630
Rejection Region
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• Decision: Reject H0
There is enough evidence at the
1% level of significance to
conclude that the adults’
favorite ways to eat ice cream
and gender are dependent. 46 of 91
Section 10.2 Summary
• Used a contingency table to find expected frequencies
• Used a chi-square distribution to test whether two
variables are independent
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Section 10.3
Comparing Two Variances
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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
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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.
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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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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
2
3
4
F-Distributions
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Finding Critical Values for the F-Distribution
1. Specify the level of significance α.
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 α F-table.
b. two-tailed, use the ½α F-table.
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Example: Finding Critical F-Values
Find the critical F-value for a right-tailed test when
α = 0.10, d.f.N = 5 and d.f.D = 28.
Solution:
The critical value is F0 = 2.06.
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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
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Solution: Finding Critical F-Values
½α = 0.025, d.f.N = 4 and d.f.D = 8
The critical value is F0 = 5.05.
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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.
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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.
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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 α.
3. Determine 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.
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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.
© 2012 Pearson Education, Inc. All rights reserved.
In Symbols
s12
F 2
s2
If F is in the
rejection region,
reject H0.
Otherwise, fail to
reject H0.
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Example: Performing a Two-Sample F-Test
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.
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Solution: Performing a Two-Sample F-Test
•
•
•
•
•
Because 400 > 256, s12  400 and s22  256
H0: σ12 ≤ σ22
• Test Statistic:
s12 400
Ha: σ12 > σ22 (Claim)
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:
at the 10% level of
significance to convince the
0.10
manager to switch to the new
system.
0
1.561.96
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F
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Example: Performing a Two-Sample F-Test
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
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Stock B
n1 = 31
s1 = 5.7
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Solution: Performing a Two-Sample F-Test
•
•
•
•
•
Because 5.72 > 3.52, s12  5.72 and s22  3.52
H0: σ12 = σ22
• Test Statistic:
s12 5.72
Ha: σ12 ≠ σ22 (Claim)
F  2  2  2.652
s2 3.5
½α = 0. 025
d.f.N= 30 d.f.D= 29
• Decision: Reject H0
There is enough evidence at
Rejection Region:
the 5% level of significance
to support the claim that one
0.025
of the two stocks is a riskier
investment.
0
2.09 2.652 F
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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
Test statistic 
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.
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
5. Find the variance between the
samples.
6. Find the variance within the
samples
7. Find the test statistic.
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In Symbols
SS B  ni(xi  x)2
MS B 

d.f.N
k 1
SSW (ni  1)si2
MSW 

d.f.D
N k
MS B
F
MSW
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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 α.
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.
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F
MS B
MSW
If F is in the
rejection region,
reject H0.
Otherwise, fail to
reject H0.
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ANOVA Summary Table
• A table is a convenient way to summarize the results in
a one-way ANOVA test.
Variation
Between
Within
Sum of
squares
Degrees of
freedom
Mean
squares
F
d.f.N
SS B
MS B 
d.f.N
MS B
MSW
d.f.D
SSW
MSW 
d.f.D
SSB
SSW
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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
x1 
56
 14
4
s12  6
14
21
15
19
x2 
85
 17
5
s22  8.5
17
20
15
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. (Claim)
•
•
•
•
α = 0. 01
d.f.N= 3 – 1 = 2
d.f.D= 13 – 3 = 10
Rejection Region:
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• Decision:
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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
21.9232

 10.9616
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 10.9616
 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. (Claim)
MSW
•
•
•
•
α = 0.01
d.f.N= 3 – 1 = 2
d.f.D= 13 – 3 = 10
Rejection Region:
1.50
7.56
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• 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.
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Example: Using the TI-83/84 to Perform a
One-Way ANOVA
A researcher believes that the mean earnings of top-paid
actors, athletes, and musicians are the same. The
earnings (in millions of dollars) for several randomly
selected from each category are shown in the table in
next slide. Assume that the populations are normally
distributed, the samples are independent, and the
population variances are equal. At α = 0.10, can you
conclude that the mean earnings are the same for the
three categories? Use a technology tool. (Source:
Forbes.com LLC)
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Example: Using the TI-83/84 to Perform a
One-Way ANOVA
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Solution: Using the TI-83/84 to Perform a
One-Way ANOVA
• H0: μ1 = μ2 = μ3 (Claim)
• Ha: At least one mean is different.
• Store data into lists L1, L2, and L3
• Decision: P ≈ 0.06 so P < α
Reject H0
There is enough evidence at the 10% level of significance
to reject the claim that the mean earnings are the same.
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Two-Way ANOVA
Two-way analysis of variance
• A hypothesis-testing technique that is used to test the
effect of two independent variables, or factors, on
one dependent variable.
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Two-Way ANOVA
Example:
• Suppose a medical researcher wants to test the effect
of gender and type of medication on the mean length
of time it takes pain relievers to provide relief.
Gender
Type of
medication
Male
I Males taking type I
II Males taking type II
III Males taking type III
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Female
Females taking type I
Females taking type II
Females taking type III
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Two-Way ANOVA Hypotheses
Main effect
• The effect of one independent variable on the
dependent variable.
Interaction effect
• The effect of both independent variables on the
dependent variable.
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Two-Way ANOVA Hypotheses
Hypotheses for main effects:
• H0: Gender has no effect on the mean length of time it
takes a pain reliever to provide relief.
• Ha: Gender has an effect on the mean length of time it
takes a pain reliever to provide relief.
• H0: Type of medication has no effect on the mean length of
time it takes a pain reliever to provide relief.
• Ha: Type of medication has an effect on the mean length of
time it takes a pain reliever to provide relief.
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Two-Way ANOVA Hypotheses
Hypotheses for interaction effects:
• H0: There is no interaction effect between gender and type
of medication on the mean length of time it takes a
pain reliever to provide relief.
• Ha: There is an interaction effect between gender and type
of medication on the mean length of time it takes a
pain reliever to provide relief.
Perform a two-way ANOVA test, calculating the F-statistic for
each hypothesis. It is possible to reject none, one, two, or all of
the null hypotheses. The statistics involved with a two-way
ANOVA test is beyond the scope of this course. You can use a
technology tool such as MINITAB to perform the test.
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Section 10.4 Summary
• Used one-way analysis of variance to test claims
involving three or more means
• Introduced two-way analysis of variance
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