chapter 10 chi-square tests and the f- distribution 1

Chapter 10

Chi-Square Tests and the F-Distribution

1

Chapter Outline

• 10.1 Goodness of Fit

• 10.2 Independence

• 10.3 Comparing Two Variances

• 10.4 Analysis of Variance

2

Section 10.1

Goodness of Fit

3

Section 10.1 Objectives

• Use the chi-square distribution to test whether a frequency distribution fits a claimed distribution

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.

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 PreferencesClassical 4% Oldies 2%Country 36% Pop 18%Gospel 11% Rock 29%

Each outcome is classified into categories.

The probability for each possible outcome is fixed.

6

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.

7


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.

8


• 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.

9


• 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.

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.

Survey results

(n = 500)Classical 8Country 210Gospel 72Oldies 10Pop 75Rock 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 8Country 210Gospel 72Oldies 10Pop 75Rock 125

observed frequency

12

Solution: Finding Observed and Expected Frequencies

Expected Frequency: Ei = npi

Type of music

% of listeners

Observed frequency

Expected frequency

Classical 4% 8Country 36% 210Gospel 11% 72Oldies 2% 10Pop 18% 75Rock 29% 125

n = 500

500(0.04) = 20500(0.36) = 180500(0.11) = 55500(0.02) = 10500(0.18) = 90500(0.29) = 145

13

Distribution of music PreferencesClassical 4% Oldies 2%Country 36% Pop 18%Gospel 11% Rock 29%


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.

14


• 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

where O represents the observed frequency of each category and E represents the expected frequency of each category.

22 ( )O E

E The test is always

a right-tailed test.

15


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.

State H0 and Ha.

Identify .

Use Table 6 in Appendix B.

d.f. = k – 1

In Words In Symbols

16


If χ2 is in the rejection region, reject H0. Otherwise, fail to reject H0.

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.

22 ( )O E

E

In Words In Symbols

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.

Survey results (n = 500)

Classical 8Country 210Gospel 72Oldies 10Pop 75Rock 125

Distribution of music preferences

Classical 4%Country 36%Gospel 11%Oldies 2%Pop 18%Rock 29%

18

Solution: Performing a Goodness of Fit Test

• H0:

• Ha:

• α =

• d.f. =

• Rejection Region

• Test Statistic:

• Decision:

• Conclusion:

0.01

6 – 1 = 5

0.01

χ2

0 15.086

music preference is 4% classical, 36% country, 11% gospel, 2% oldies, 18% pop, and 29% rock

music preference differs from the claimed or expected distribution

19


22 ( )O E

E

Type of music

Observed frequency

Expected frequency

Classical 8 20Country 210 180Gospel 72 55Oldies 10 10Pop 75 90Rock 125 145

2 2 2 2 2 2(8 20) (210 180) (72 55) (10 10) (75 90) (125 145)

20 180 55 10 90 14522.713

20


• H0:

• Ha:

• α =

• d.f. =


• Test Statistic:

• Decision:

0.01

6 – 1 = 5

0.01

χ2

0 15.086

music preference is 4% classical, 36% country, 11% gospel, 2% oldies, 18% pop, and 29% rock

music preference differs from the claimed or expected distribution

χ2 = 22.713

22.713

There is enough evidence to conclude that the distribution of music preferences differs from the claimed distribution.

Reject H0

21


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)

22


Color FrequencyBrown 80Yellow 95Red 88Blue 83Orange 76Green 78

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

n = 500


• H0:

• Ha:

• α =

• d.f. =


• Test Statistic:

• Decision:

• Conclusion:

0.10

6 – 1 = 5

0.10

χ2

0 9.236

Distribution of different-colored candies in bags of dark chocolate M&Ms is uniform

Distribution of different-colored candies in bags of dark chocolate M&Ms is not uniform

24


2 2 2 2 2 2(80 83.3) (95 83.3) (88 83.3) (83 83.3) (76 83.3) (78 83.3)

83.3 83.3 83.3 83.3 83.3 83.33.016

ColorObserve

d frequen

cy

Expected

frequency

Brown 80 83.3Yellow 95 83.3Red 88 83.3Blue 83 83.3Orange 76 83.3Green 78 83.3

22 ( )O E

E

25


• H0:

• Ha:

• α =

• d.f. =


• Test Statistic:

• Decision:

0.01

6 – 1 = 5

0.10

χ2

0 9.236

χ2 = 3.016

3.016

There is not enough evidence to dispute the claim that the distribution is uniform.

Distribution of different-colored candies in bags of dark chocolate M&Ms is uniform

Distribution of different-colored candies in bags of dark chocolate M&Ms is not uniform

Fail to Reject H0

26

Section 10.1 Summary

• Used the chi-square distribution to test whether a frequency distribution fits a claimed distribution

27

Section 10.2

Independence

28


• Use a contingency table to find expected frequencies

• Use a chi-square distribution to test whether two variables are independent

29

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.

30

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 size

39 and under

40 - 49 50 - 59 60 - 6970 and over

Small / Midsize

42 69 108 60 21

Large 5 18 85 120 22

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

,(Sum of row ) (Sum of column )Expected frequency

Sample sizer cr cE

32

Age

Company size39 and under

40 - 49 50 - 59 60 - 6970 and over

Small / Midsize

42 69 108 60 21

Large 5 18 85 120 22

,

r and c are independent, then:

frequency of E (r AND c)*sample size = p(r)*p(c)*sample size

of row r of column c=( )(

size size

r c

If

Expected p

Total Total

Sample Sample

)* Sample size( of row r)(Total of column c)

= size

Total

Sample

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

39 and under

40 - 49 50 - 59 60 - 6970 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

marginal totals33

Solution: Finding Expected Frequencies

Age

Company size

39 and under

40 - 49 50 - 59 60 - 6970 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

,(Sum of row ) (Sum of column )

Sample sizer cr cE

1,1

300 4725.64

550E

34


Age


40 - 49 50 - 59 60 - 6970 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

1,2

300 8747.45

550E

1,3

300 193105.27

550E

1,4

300 18098.18

550E

1,5

300 4323.45

550E

1,2

300 8747.45

550E

1,3

300 193105.27

550E

1,4

300 18098.18

550E

35


Age


40 - 49 50 - 59 60 - 6970 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

2,2

250 8739.55

550E

2,4

250 18081.82

550E

2,5

250 4319.55

550E

2,1

250 4721.36

550E

2,3

250 19387.73

550E

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.

37


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.

38


• 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

where O represents the observed frequencies and E represents the expected frequencies.

22 ( )O E

E The test is always a

right-tailed test.

39






State H0 and Ha.

Identify .


d.f. = (r – 1)(c – 1)

In Words In Symbols

40


If χ2 is in the rejection region, reject H0. Otherwise, fail to reject H0.





22 ( )O E

E

In Words In Symbols

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


40 - 49 50 - 59 60 - 6970 and over

Total

Small / Midsize

42(25.64)

69(47.45)

108(105.27

)

60(98.18)

21(23.45)

300

Large5

(21.36)18

(39.55)85

(87.73)120

(81.82)22

(19.55)250

Total 47 87 193 180 43 55042


• H0:

• Ha:

• α =

• d.f. =


• Test Statistic:

• Decision:

0.01

(2 – 1)(5 – 1) = 4

0.01

χ2

0 13.277

CEOs’ ages are independent of company size

CEOs’ ages are dependent on company size

43


2 2 2 2 2

2 2 2 2 2

(42 25.64) (69 47.45) (108 105.27) (60 98.18) (21 23.45)

25.64 47.45 105.27 98.18 23.45

(5 21.36) (18 39.55) (85 87.73) (120 81.82) (22 19.55)

21.36 39.55 87.73 81.82 19.5577.9

22 ( )O E

E

44


• H0:

• Ha:

• α =

• d.f. =


• Test Statistic:

• Decision:

0.01

(2 – 1)(5 – 1) = 4

0.01

χ2

0 13.277

CEOs’ ages are independent of company size

CEOs’ ages are dependent on company size

χ2 = 77.9

There is enough evidence to conclude CEOs’ ages are dependent on company size.

77.9

Reject H0

45

Age


40 - 49 50 - 59 60 - 6970 and over

Small / Midsize

42 69 108 60 21

Large 5 18 85 120 22


• Used a contingency table to find expected frequencies

• Used a chi-square distribution to test whether two variables are independent

47

Section 10.3

Comparing Two Variances

48


• Interpret the F-distribution and use an F-table to find critical values

• Perform a two-sample F-test to compare two variances

49

F-Distribution

• Let represent the sample variances of two different populations.

• If both populations are normal and the population variances are equal, then the sampling distribution of

is called an F-distribution.

2 21 2 and s s

2 21 2 and σ σ

2122

sF

s

50

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.

51

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

F1 2 3 4

52

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.

53

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.

The critical value is F0 = 2.43.

54

Solution:

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 right-tailed critical value. •You must remember to use the ½α table.

1(0.05) 0.025

2

1

2

55

Solution: Finding Critical F-Values

½α = 0.025, d.f.N = 4 and d.f.D = 8

The critical value is F0 = 5.05.

56

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.

57


• Test Statistic2122

sF

s

where represent the sample variances with

• The degrees of freedom for the numerator is d.f.N = n1 – 1 where n1 is the size of the sample having variance

• The degrees of freedom for the denominator is d.f.D = n2 – 1, and n2 is the size of the sample having variance

2 21 2 and s s

2 21 2.s s

21 .s

22.s

58






State H0 and Ha.

Identify .


d.f.N = n1 – 1 d.f.D = n2 – 1

In Words In Symbols

59


If F is in the rejection region, reject H0. Otherwise, fail to reject H0.





2122

sF

s

In Words In Symbols

60

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.

61

Solution: Performing a Two-Sample F-Test

• H0:

• Ha:

• α =

• d.f.N= d.f.D=

• Rejection Region:

• Test Statistic:

• Decision:

σ12 ≤ σ2

2

σ12 > σ2

2

0.10

9 20

0 F1.96

0.10

Because 400 > 256, 2 21 2400 and 256s s

21

22

4001.56

256

sF

s

There is not enough evidence to convince the manager to switch to the new system.

1.961.56

Fail to Reject H0

62

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 Stock Bn2 = 30 n1 = 31

s2 = 3.5 s1 = 5.7

64

Solution: Performing a Two-Sample F-Test

• H0:

• Ha:

• ½α =

• d.f.N= d.f.D=


• Test Statistic:

• Decision:

σ12 = σ2

2

σ12 ≠ σ2

2

0. 025

30 29

0 F2.09

0.025

Because 5.72 > 3.52, 2 2 2 21 25.7 and 3.5s s

2 21

2 22

5.72.65

3.5

sF

s

There is enough evidence to support the claim that one of the two stocks is a riskier investment.

2.092.65

Reject H0

65


• Interpreted the F-distribution and used an F-table to find critical values

• Performed a two-sample F-test to compare two variances

66

Section 10.4

Analysis of Variance

67


• Use one-way analysis of variance to test claims involving three or more means

• Introduce two-way analysis of variance

68

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.

69

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.

70

One-Way ANOVA

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.

Variance between samplesVariance

T

est statiwithin sa

stics

mple

71

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 B

W

MSF

MS

• 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.

72

Test Statistic for a One-Way ANOVA

1. Find the mean and variance of each sample.

2. Find the mean of all entries in all samples (the grand mean).

3. Find the sum of squares between the samples.

4. Find the sum of squares within the samples.

22 ( )

1x x xx s

n n

xxN

2( ) B i iSS n x x

2( 1) W i iSS n s

In Words In Symbols

73

Test Statistic for a One-Way ANOVA

5. Find the variance between the samples.

6. Find the variance within the samples

7. Find the test statistic. B

W

MSF

MS

N1 d.f.B B

BSS SS

MSk

Dd.f.W W

WSS SS

MSN k

In Words In Symbols

74

Performing a One-Way ANOVA Test





State H0 and Ha.

Identify .


d.f.N = k – 1 d.f.D = N – k

In Words In Symbols

75

Performing a One-Way ANOVA Test

If F is in the rejection region, reject H0. Otherwise, fail to reject H0.





B

W

MSF

MS

In Words In Symbols

76

ANOVA Summary Table

• A table is a convenient way to summarize the results in a one-way ANOVA test.

d.f.DSSWWithin

d.f.NSSBBetween

FMean squares

Degrees of freedom

Sum of squares

Variation

. .B

BN

SSMS

d f

. .W

WD

SSMS

d f

B

W

MSMS

77

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.

78

Example: Performing a One-Way ANOVA

Medication 1 Medication 2 Medication 3

12 16 14

15 14 17

17 21 20

12 15 15

19

1

5614

4x 2

8517

5x 3

6616.5

4x

21 6s 2

2 8.5s 23 7s

Solution: k = 3 (3 samples)N = n1 + n2 + n3 = 4 + 5 + 4 = 13 (sum of sample sizes)

79

Solution: Performing a One-Way ANOVA

• H0:

• Ha:

• α =

• d.f.N=

• d.f.D=


• Test Statistic:

• Decision:

μ1 = μ2 = μ3

At least one mean is different

0. 01

3 – 1 = 2

13 – 3 = 10

0 F7.56

0.01

80


To find the test statistic, the following must be calculated.

xxN 56 85 66 15.92

13

2

N

( ) d.f. 1

i iBB

n x xSSMS

k

2 2 24(14 15.92) 5(17 15.92) 4(16.5 15.92)

3 121.92 10.962

81


To find the test statistic, the following must be calculated.

2

D

( 1) d.f.

W i iW

SS n sMS

N k

(4 1)(6) (5 1)(8.5) (4 1)(7)13 3

73 7.310

B

W

MSF

MS 10.96

1.507.3

82


• H0:

• Ha:

• α =

• d.f.N=

• d.f.D=


• Test Statistic:

• Decision:

μ1 = μ2 = μ3


0. 01

3 – 1 = 2

13 – 3 = 10

0 F7.56

0.01

1.50B

W

MSF

MS

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.

1.50

Fail to Reject H0

83

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.

84

Example: Using the TI-83/84 to Perform a One-Way ANOVA

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

85

Solution: Using the TI-83/84 to Perform a One-Way ANOVA

• H0:

• Ha:

• Store data into lists L1, L2, and L3

μ1 = μ2 = μ3


• Decision:There is enough evidence to support the claim. You can conclude that there is a difference in the means of the flight times.

P-value < α Reject H0

86

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.

87

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.

Males taking type I Females taking type I

Males taking type II Females taking type II

Males taking type III Females taking type III

GenderMale Female

I

II

III

88

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.

89


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 oftime 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.

90


Hypotheses for interaction effects:

• H0: There is no interaction effect between gender and typeof medication on the mean length of time it takes a pain reliever to provide relief.

• Ha: There is an interaction effect between gender and typeof medication on the mean length of time it takes a pain reliever to provide relief.

91

Two-Way ANOVA

• A two-way ANOVA test calculates an F-test statistic for each hypothesis.

• It is possible to reject none, one, two, or all of the null hypotheses.

• Can use a technology tool such as MINITAB to perform a two-way ANOVA test.

92


• Used one-way analysis of variance to test claims involving three or more means

• Introduced two-way analysis of variance

93

94© 2012 Pearson Education, Inc.

Elementary Statistics:

Picturing the World

Fifth Edition

by Larson and Farber

Chapter 10: Chi-Square Tests and the F-Distribution


The distribution of favorite ice cream flavor is shown in the graph. The results from a random sample of college students is shown below. Find the expected frequency for the Vanilla category.

A. 33.75

B. 22.5

C. 8.1

D. 111.11

Chocolate Vanilla Strawberry Other

62 30 21 12


The distribution of favorite ice cream flavor is shown in the graph. The results from a random sample of college students is shown below. Find the expected frequency for the Vanilla category.

A. 33.75

B. 22.5

C. 8.1

D. 111.11


62 30 21 12


The distribution of favorite ice cream flavor is shown in the graph. The results from a random sample of college students is shown below. Find the critical value for α = 0.05.

A. 9.488

B. 7.815

C. 2.353

D. 1.645


62 30 21 12

98

The distribution of favorite ice cream flavor is shown in the graph. The results from a random sample of college students is shown below. Find the critical value for α = 0.05.

A. 9.488

B. 7.815

C. 2.353

D. 1.645


62 30 21 12: categories

df = k-1

k number of


The distribution of favorite ice cream flavor is shown in the graph. The results from a random sample of college students is shown below. Find the value of the test statistic.

A. 0.949

B. 0.919

C. 37.125

D. 0.010


62 30 21 12

100

The distribution of favorite ice cream flavor is shown in the graph. The results from a random sample of college students is shown below. Find the value of the test statistic.Chocolate Vanilla Strawberry Other

62 30 21 12


Calculate the expected frequency for the highlighted cell E2,3.

A. 20.62

B. 18.33

C. 16.92

D. 15.38

Favorite Pastime

Gender Reading Television Music

Male 12 32 19

Female 10 20 17


Calculate the expected frequency for the highlighted cell E2,3.

A. 20.62

B. 18.33

C. 16.92

D. 15.38

Favorite Pastime


Male 12 32 19

Female 10 20 17

( )( )rowTotal columnTotal

TotalAllCells

OR


Find the critical value for α = 0.05.

A. 5.991

B. 12.592

C. 2.920

D. 1.645

Favorite Pastime


Male 12 32 19

Female 10 20 17


Find the critical value for α = 0.05.

A. 5.991

B. 12.592

C. 2.920

D. 1.645

Favorite Pastime


Male 12 32 19

Female 10 20 17

( 1)( 1) (1)(2) 2df r c


Find the value of the test statistic.

A. 16.437

B. 15.798

C. 0.751

D. 0.018

Favorite Pastime


Male 12 32 19

Female 10 20 17



A. 16.437

B. 15.798

C. 0.751

D. 0.018

Favorite Pastime


Male 12 32 19

Female 10 20 17


Find the critical F-value for the following situation:

Claim: α = 0.10

n1 = 9; n2 = 8

A. 2.75

B. 3.73

C. 3.39

D. 3.50

2 21 2

. ,s21 25 32 . ,s2

2 19 78

108

Find the critical F-value for the following situation:

Claim: α = 0.10

n1 = 9; n2 = 8

A.A. 2.752.75

B. 3.73

C. 3.39

D. 3.50

2 21 2

. ,s21 25 32 . ,s2

2 19 78

d.f.N = n1 – 1 d.f.D = n2 – 1 d.f.N = 9 – 1=8 d.f.D = 8 – 1=7



Claim: α = 0.10

n1 = 9; n2 = 8

A. 1.64

B. 1.13

C. 0.78

D. 1.28

2 21 2

. ,s21 25 32 . ,s2

2 19 78

110


Claim: α = 0.10

n1 = 9; n2 = 8

A. 1.64

B. 1.13

C. 0.78

D. 1.28

2 21 2

. ,s21 25 32 . ,s2

2 19 78


A professor wants to test if the mean score on an exam is the same in all three of his elementary statistics classes.

True or False. The appropriate hypotheses would be

H0:Ha:

A. True

B. False

1 2 3

1 2 3

chapter 10 chi-square tests and the f- distribution 1

Documents

frequency distribution

expected distribution

chisquare distribution

specified distribution

claimed distribution

hypothesized distribution

observed frequency o

frequency e