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Chapter 11
Chi-Square and Analysis of Variance (ANOVA)
© McGraw-Hill, Bluman, 5th ed., Chapter 11 1
Chapter 11 Overview Introduction 11-1 Test for Goodness of Fit
11-2 Tests Using Contingency Tables
11-3 Analysis of Variance (ANOVA)
Bluman, Chapter 11 2
Chapter 11 Objectives1. Test a distribution for goodness of fit, using
chi-square.
2. Test two variables for independence, using chi-square.
3. Test proportions for homogeneity, using chi-square.
4. Use the ANOVA technique to determine if there is a significant difference among three or more means
Bluman, Chapter 11 3
11.1 Test for Goodness of Fit
The chi-square statistic can be used to see whether a frequency distribution fits a specific pattern. This is referred to as the chi-square goodness-of-fit test.
Bluman, Chapter 11 4
Test for Goodness of FitFormula for the test for goodness of fit:
where
d.f. = number of categories minus 1
O = observed frequency
E = expected frequency
Bluman, Chapter 11 5
2
2
O E
E
Assumptions for Goodness of Fit1. The data are obtained from a random sample.
2. The expected frequency for each category must be 5 or more.
Bluman, Chapter 11 6
Chapter 11Chi-Square and Analysis of Variance (ANOVA)
Section 11-1Example 11-1
Page #592
Bluman, Chapter 11 7
Example 11-1: Fruit Soda FlavorsA market analyst wished to see whether consumers have any preference among five flavors of a new fruit soda. A sample of 100 people provided the following data. Is there enough evidence to reject the claim that there is no preference in the selection of fruit soda flavors, using the data shown previously? Let α = 0.05.
Bluman, Chapter 11 8
Cherry Strawberry Orange Lime Grape
Observed 32 28 16 14 10
Expected 20 20 20 20 20
Step 1: State the hypotheses and identify the claim.H0: Consumers show no preference (claim).H1: Consumers show a preference.
Example 11-1: Fruit Soda Flavors
Bluman, Chapter 11 9
Cherry Strawberry Orange Lime Grape
Observed 32 28 16 14 10
Expected 20 20 20 20 20
Step 2: Find the critical value. D.f. = 5 – 1 = 4, and α = 0.05. CV = 9.488.
Step 3: Compute the test value. 2
2
O E
E
2 2 2 2
2
32 20 28 20 16 20 14 20
20 20 20 20
10 20
20
18.0
Step 4: Make the decision.
The decision is to reject the null hypothesis, since 18.0 > 9.488.
Step 5: Summarize the results.
There is enough evidence to reject the claim that consumers show no preference for the flavors.
Example 11-1: Fruit Soda Flavors
Bluman, Chapter 11 10
Chapter 11Chi-Square and Analysis of Variance (ANOVA)
Section 11-1Example 11-2
Page #594
Bluman, Chapter 11 11
Example 11-2: RetireesThe Russel Reynold Association surveyed retired senior executives who had returned to work. They found that after returning to work, 38% were employed by another organization, 32% were self-employed, 23% were either freelancing or consulting, and 7% had formed their own companies. To see if these percentages are consistent with those of Allegheny County residents, a local researcher surveyed 300 retired executives who had returned to work and found that 122 were working for another company, 85 were self-employed, 76 were either freelancing or consulting, and 17 had formed their own companies. At α = 0.10, test the claim that the percentages are the same for those people in Allegheny County.
Bluman, Chapter 11 12
Example 11-2: Retirees
Bluman, Chapter 11 13
New Company
Self-Employed
Free-lancing
Owns Company
Observed 122 85 76 17
Expected.38(300)=
114.32(300)=
96.23(300)=
69.07(300)=
21
Step 1: State the hypotheses and identify the claim.H0: The retired executives who returned to work are distributed as follows: 38% are employed by another organization, 32% are self-employed, 23% are either freelancing or consulting, and 7% have formed their own companies (claim).
H1: The distribution is not the same as stated in the null hypothesis.
Example 11-2: Retirees
Bluman, Chapter 11 14
New Company
Self-Employed
Free-lancing
Owns Company
Observed 122 85 76 17
Expected.38(300)=
114.32(300)=
96.23(300)=
69.07(300)=
21
Step 2: Find the critical value. D.f. = 4 – 1 = 3, and α = 0.10. CV = 6.251.
Step 3: Compute the test value.
2
2
O E
E
2 2 2 2122 114 85 96 76 69 17 21
114 96 69 21
3.2939
Step 4: Make the decision.
Since 3.2939 < 6.251, the decision is not to reject the null hypothesis.
Step 5: Summarize the results.
There is not enough evidence to reject the claim. It can be concluded that the percentages are not significantly different from those given in the null hypothesis.
Example 11-2: Retirees
Bluman, Chapter 11 15
Chapter 11Chi-Square and Analysis of Variance (ANOVA)
Section 11-1Example 11-3
Page #595
Bluman, Chapter 11 16
Example 11-3: Firearm DeathsA researcher read that firearm-related deaths for people aged 1 to 18 were distributed as follows: 74% were accidental, 16% were homicides, and 10% were suicides. In her district, there were 68 accidental deaths, 27 homicides, and 5 suicides during the past year. At α = 0.10, test the claim that the percentages are equal.
Bluman, Chapter 11 17
Accidental Homicides Suicides
Observed 68 27 5
Expected 74 16 10
Example 11-3: Firearm Deaths
Step 1: State the hypotheses and identify the claim.
H0: Deaths due to firearms for people aged 1 through 18 are distributed as follows: 74% accidental, 16% homicides, and 10% suicides (claim).
H1: The distribution is not the same as stated in the null hypothesis.
Bluman, Chapter 11 18
Accidental Homicides Suicides
Observed 68 27 5
Expected 74 16 10
Example 11-3: Firearm Deaths
Bluman, Chapter 11 19
Accidental Homicides Suicides
Observed 68 27 5
Expected 74 16 10
Step 2: Find the critical value. D.f. = 3 – 1 = 2, and α = 0.10. CV = 4.605.
Step 3: Compute the test value.
2
2
O E
E
2 2 268 74 27 16 5 10
74 16 10
10.549
Step 4: Make the decision.
Reject the null hypothesis, since 10.549 > 4.605.
Step 5: Summarize the results.
There is enough evidence to reject the claim that the distribution is 74% accidental, 16% homicides, and 10% suicides.
Example 11-3: Firearm Deaths
Bluman, Chapter 11 20
Test for Normality (Optional) The chi-square goodness-of-fit test can be used
to test a variable to see if it is normally distributed.
The hypotheses are:H0: The variable is normally distributed.
H1: The variable is not normally distributed. This procedure is somewhat complicated. The
calculations are shown in example 11-4 on page 597 in the text.
Bluman, Chapter 11 21
11.2 Tests Using Contingency Tables
When data can be tabulated in table form in terms of frequencies, several types of hypotheses can be tested by using the chi-square test.
The test of independence of variables test of independence of variables is used to determine whether two variables are independent of or related to each other when a single sample is selected.
The test of homogeneity of proportions test of homogeneity of proportions is used to determine whether the proportions for a variable are equal when several samples are selected from different populations.
Bluman, Chapter 11 22
Test for Independence The chi-square goodness-of-fit test can be used
to test the independence of two variables. The hypotheses are:
H0: There is no relationship between two variables.
H1: There is a relationship between two variables.
If the null hypothesis is rejected, there is some relationship between the variables.
Bluman, Chapter 11 23
Test for Independence In order to test the null hypothesis, one must
compute the expected frequencies, assuming the null hypothesis is true.
When data are arranged in table form for the independence test, the table is called a contingency tablecontingency table.
Bluman, Chapter 11 24
Contingency Tables
The degrees of freedom for any contingency table are d.f. = (rows – 1) (columns – 1) = (R – 1)(C – 1).
Bluman, Chapter 11 25
Test for IndependenceThe formula for the test for independence:
where
d.f. = (R – 1)(C – 1)
O = observed frequency
E = expected frequency =
Bluman, Chapter 11 26
2
2
O E
E
row sum column sum
grand total
Chapter 11Chi-Square and Analysis of Variance (ANOVA)
Section 11-2Example 11-5
Page #606
Bluman, Chapter 11 27
Example 11-5: College Education and Place of Residence
A sociologist wishes to see whether the number of years of college a person has completed is related to her or his place of residence. A sample of 88 people is selected and classified as shown. At α = 0.05, can the sociologist conclude that a person’s location is dependent on the number of years of college?
Bluman, Chapter 11 28
LocationNo
CollegeFour-Year
DegreeAdvanced
Degree Total
Urban 15 12 8 35
Suburban 8 15 9 32
Rural 6 8 7 21
Total 29 35 24 88
Example 11-5: College Education and Place of Residence
Bluman, Chapter 11 29
Step 1: State the hypotheses and identify the claim.H0: A person’s place of residence is independent of the
number of years of college completed.
H1: A person’s place of residence is dependent on the number of years of college completed (claim).
Step 2: Find the critical value.The critical value is 4.605, since the degrees offreedom are (2 – 1)(3 – 1) = 2.
Example 11-5: College Education and Place of Residence
Bluman, Chapter 11 30
LocationNo
CollegeFour-Year
DegreeAdvanced
Degree Total
Urban15 12 8
35
Suburban 8 15 9
32
Rural 6 8 7
21
Total 29 35 24 88
Compute the expected values. row sum column sum
grand totalE
1,1
35 2911.53
88 E
(11.53)
(10.55)
(6.92)
(13.92)
(12.73)
(8.35)
(9.55)
(8.73)
(5.73)
Example 11-5: College Education and Place of Residence
Bluman, Chapter 11 31
Step 3: Compute the test value. 2
2
O E
E
2 2 2
2 2 2
2 2 2
15 11.53 12 13.92 8 9.55
11.53 13.92 9.55
8 10.55 15 12.73 9 8.73
10.55 12.73 8.73
6 6.92 8 8.35 7 5.73
6.92 8.35 5.73
3.01
Example 11-5: College Education and Place of Residence
Bluman, Chapter 11 32
Step 4: Make the decision.Do not reject the null hypothesis, since 3.01<9.488.
Step 5: Summarize the results.There is not enough evidence to support the claim that a person’s place of residence is dependent on the number of years of college completed.
Chapter 11Chi-Square and Analysis of Variance (ANOVA)
Section 11-2Example 11-6
Page #608
Bluman, Chapter 11 33
Example 11-6: Alcohol and GenderA researcher wishes to determine whether there is a relationship between the gender of an individual and the amount of alcohol consumed. A sample of 68 people is selected, and the following data are obtained. At α = 0.10, can the researcher conclude that alcohol consumption is related to gender?
Bluman, Chapter 11 34
Gender
Alcohol Consumption
TotalLow Moderate High
Male 10 9 8 27
Female 13 16 12 41
Total 23 25 20 68
Example 11-6: Alcohol and GenderStep 1: State the hypotheses and identify the claim.
H0: The amount of alcohol that a person consumes is independent of the individual’s gender.
H1: The amount of alcohol that a person consumes is dependent on the individual’s gender (claim).
Step 2: Find the critical value.
The critical value is 9.488, since the degrees of freedom are (3 – 1 )(3 – 1) = (2)(2) = 4.
Bluman, Chapter 11 35
Example 11-6: Alcohol and Gender
Bluman, Chapter 11 36
Gender
Alcohol Consumption
TotalLow Moderate High
Male10 9 8
27
Female13 16 12
41
Total 23 25 20 68
Compute the expected values. row sum column sum
grand totalE
1,1
27 239.13
68 E
(9.13) (9.93) (7.94)
(13.87) (15.07) (12.06)
Example 11-6: Alcohol and GenderStep 3: Compute the test value.
Bluman, Chapter 11 37
2
2
O E
E
2 2 2
2 2 2
10 9.13 9 9.93 8 7.94
9.13 9.93 7.94
13 13.87 16 15.07 12 12.06
13.87 15.07 12.06
0.283
Example 11-6: Alcohol and GenderStep 4: Make the decision.
Do not reject the null hypothesis, since 0.283 < 4.605.
.
Step 5: Summarize the results.
There is not enough evidence to support the claim that the amount of alcohol a person consumes is dependent on the individual’s gender.
Bluman, Chapter 11 38
Test for Homogeneity of Proportions
Homogeneity of proportions test Homogeneity of proportions test is used when samples are selected from several different populations and the researcher is interested in determining whether the proportions of elements that have a common characteristic are the same for each population.
Bluman, Chapter 11 39
Test for Homogeneity of Proportions
The hypotheses are:H0: p1 = p2 = p3 = … = pn
H1: At least one proportion is different from the others.
When the null hypothesis is rejected, it can be assumed that the proportions are not all equal.
Bluman, Chapter 11 40
Assumptions for Homogeneity of Proportions
1. The data are obtained from a random sample.
2. The expected frequency for each category must be 5 or more.
Bluman, Chapter 11 41
Chapter 11Chi-Square and Analysis of Variance (ANOVA)
Section 11-2Example 11-7
Page #610
Bluman, Chapter 11 42
Example 11-7: Lost LuggageA researcher selected 100 passengers from each of 3 airlines and asked them if the airline had lost their luggage on their last flight. The data are shown in the table. At α = 0.05, test the claim that the proportion of passengers from each airline who lost luggage on the flight is the same for each airline.
Bluman, Chapter 11 43
Airline 1 Airline 2 Airline 3 Total
Yes 10 7 4 21
No 90 93 96 279
Total 100 100 100 300
Example 11-7: Lost LuggageStep 1: State the hypotheses.
H0: p1 = p2 = p3 = … = pn
H1: At least one mean differs from the other.
Step 2: Find the critical value.
The critical value is 5.991, since the degrees of freedom are (2 – 1 )(3 – 1) = (1)(2) = 2.
Bluman, Chapter 11 44
Example 11-7: Lost Luggage
Bluman, Chapter 11 45
Compute the expected values. row sum column sum
grand totalE
1,1
21 1007
300 E
Airline 1 Airline 2 Airline 3 Total
Yes 10 7 4
21
No 90 93 96
279
Total 100 100 100 300
(7) (7) (7)
(93) (93) (93)
Example 11-7: LuggageStep 3: Compute the test value.
Bluman, Chapter 11 46
2
2
O E
E
2 2 2
2 2 2
10 7 7 7 4 7
7 7 7
90 93 93 93 96 93
93 93 93
2.765
Example 11-7: Lost LuggageStep 4: Make the decision.
Do not reject the null hypothesis, since 2.765 < 5.991.
.
Step 5: Summarize the results.
There is not enough evidence to reject the claim that the proportions are equal. Hence it seems that there is no difference in the proportions of the luggage lost by each airline.
Bluman, Chapter 11 47
11-3 Analysis of Variance (ANOVA) The F test, used to compare two variances,
can also be used to compare three of more means.
This technique is called analysis of varianceanalysis of variance or ANOVAANOVA.
For three groups, the F test can only show whether or not a difference exists among the three means, not where the difference lies.
Bluman, Chapter 11 48
Analysis of Variance (ANOVA) When an F test is used to test a
hypothesis concerning the means of three or more populations, the technique is called analysis of variance analysis of variance (commonly abbreviated as ANOVAANOVA).
Although the t test is commonly used to compare two means, it should not be used to compare three or more.
Bluman, Chapter 11 49
Assumptions for the F TestThe following assumptions apply when using the F test to compare three or more means.
1. The populations from which the samples were obtained must be normally or approximately normally distributed.
2. The samples must be independent of each other.
3. The variances of the populations must be equal.
Bluman, Chapter 11 50
The F Test In the F test, two different estimates of
the population variance are made. The first estimate is called the between-between-
group variancegroup variance, and it involves finding the variance of the means.
The second estimate, the within-group within-group variancevariance, is made by computing the variance using all the data and is not affected by differences in the means.
Bluman, Chapter 11 51
The F Test If there is no difference in the means, the
between-group variance will be approximately equal to the within-group variance, and the F test value will be close to 1—do not reject null hypothesis.
However, when the means differ significantly, the between-group variance will be much larger than the within-group variance; the F test will be significantly greater than 1—reject null hypothesis.
Bluman, Chapter 11 52
Chapter 11Chi-Square and Analysis of Variance (ANOVA)
Section 11-3Example 11-8
Page #620
Bluman, Chapter 11 53
Example 11-8: Lowering Blood PressureA researcher wishes to try three different techniques to lower the blood pressure of individuals diagnosed with high blood pressure. The subjects are randomly assigned to three groups; the first group takes medication, the second group exercises, and the third group follows a special diet. After four weeks, the reduction in each person’s blood pressure is recorded. At α = 0.05, test the claim that there is no difference among the means.
Bluman, Chapter 11 54
Example 11-8: Lowering Blood Pressure
Bluman, Chapter 11 55
Step 1: State the hypotheses and identify the claim.H0: μ1 = μ2 = μ3 (claim)H1: At least one mean is different from the others.
Example 11-8: Lowering Blood Pressure
Bluman, Chapter 11 56
Step 2: Find the critical value.Since k = 3, N = 15, and α = 0.05,d.f.N. = k – 1 = 3 – 1 = 2d.f.D. = N – k = 15 – 3 = 12The critical value is 3.89, obtained from Table H.
Example 11-8: Lowering Blood Pressure
Bluman, Chapter 11 57
Step 3: Compute the test value.a.Find the mean and variance of each sample (these were provided with the data).
b.Find the grand meangrand mean, the mean of all values in the samples.
c. Find the between-group variancebetween-group variance, .
10 12 9 4 1167.73
15 15
GM
XX
N
2Bs
2
2
1
i i GM
B
n X Xs
k
Example 11-8: Lowering Blood Pressure
Bluman, Chapter 11 58
Step 3: Compute the test value. (continued)c. Find the between-group variancebetween-group variance, .
d.Find the within-group variancewithin-group variance, .2Ws
22 1
1
i iB
i
n ss
n
4 5.7 4 10.2 4 10.3 104.808.73
4 4 4 12
2 2 2
2 5 11.8 7.73 5 3.8 7.73 5 7.6 7.73
3 1160.13
80.072
Bs
2Bs
Step 3: Compute the test value. (continued)
e. Compute the F value.
Step 4: Make the decision.
Reject the null hypothesis, since 9.17 > 3.89.
Step 5: Summarize the results.
There is enough evidence to reject the claim and conclude that at least one mean is different from the others.
Example 11-8: Lowering Blood Pressure
Bluman, Chapter 11 59
2
2 B
W
sF
s80.07
9.178.73
ANOVA The between-group variance is sometimes
called the mean square, MSmean square, MSBB.
The numerator of the formula to compute MSB is called the sum of squares between sum of squares between groups, SSgroups, SSBB.
The within-group variance is sometimes called the mean square, MS mean square, MSWW.
The numerator of the formula to compute MSW is called the sum of squares within groups, sum of squares within groups, SSSSWW.
Bluman, Chapter 11 60
ANOVA Summary Table
Bluman, Chapter 11 61
Source Sum of Squares
d.f. MeanSquares
F
Between
Within (error)
SSB
SSW
k – 1
N – k
MSB
MSW
Total
MS
MSB
W
ANOVA Summary Table for Example 11-8
Bluman, Chapter 11 62
Source Sum of Squares
d.f. MeanSquares
F
Between
Within (error)
160.13
104.80
2
12
80.07
8.73
9.17
Total 264.93 14
Chapter 11Chi-Square and Analysis of Variance (ANOVA)
Section 11-3Example 11-9
Page #622
Bluman, Chapter 11 63
Example 11-9: Toll Road EmployeesA state employee wishes to see if there is a significant difference in the number of employees at the interchanges of three state toll roads. The data are shown. At α = 0.05, can it be concluded that there is a significant difference in the average number of employees at each interchange?
Bluman, Chapter 11 64
Example 11-9: Toll Road Employees
Bluman, Chapter 11 65
Step 1: State the hypotheses and identify the claim.H0: μ1 = μ2 = μ3 H1: At least one mean is different from the others (claim).
Example 11-9: Toll Road Employees
Bluman, Chapter 11 66
Step 2: Find the critical value.Since k = 3, N = 18, and α = 0.05,d.f.N. = 2, d.f.D. = 15The critical value is 3.68, obtained from Table H.
Example 11-9: Toll Road Employees
Bluman, Chapter 11 67
Step 3: Compute the test value.a.Find the mean and variance of each sample (these were provided with the data).
b.Find the grand meangrand mean, the mean of all values in the samples.
c. Find the between-group variancebetween-group variance, .
7 14 32 11 1528.4
15 18
GM
XX
N
2Bs
2
2
1
i i GM
B
n X Xs
k
Example 11-9: Toll Road Employees
Bluman, Chapter 11 68
Step 3: Compute the test value. (continued)c. Find the between-group variancebetween-group variance, .
d.Find the within-group variancewithin-group variance, . 2Ws
22 1
1
i iB
i
n ss
n
5 81.9 5 25.6 5 29.0 682.545.5
4 4 4 15
2 2 2
2 6 15.5 8.4 6 4 8.4 6 5.8 8.4
3 1459.18
229.592
Bs
2Bs
Step 3: Compute the test value. (continued)
e. Compute the F value.
Step 4: Make the decision.
Reject the null hypothesis, since 5.05 > 3.68.
Step 5: Summarize the results.
There is enough evidence to support the claim that there is a difference among the means.
Example 11-9: Toll Road Employees
Bluman, Chapter 11 69
2
2 B
W
sF
s229.59
5.0545.5
ANOVA Summary Table for Example 11-9
Bluman, Chapter 11 70
Source Sum of Squares
d.f. MeanSquares
F
Between
Within (error)
459.18
682.5
2
15
229.59
45.5
5.05
Total 1141.68 17