Les5e ppt 10

ChapterChi-Square Tests and the F-Distribution

1 of 91

10

© 2012 Pearson Education, Inc.All rights reserved.

Chapter Outline

• 10.1 Goodness of Fit• 10.2 Independence• 10.3 Comparing Two Variances• 10.4 Analysis of Variance

© 2012 Pearson Education, Inc. All rights reserved. 2 of 91

Section 10.1

Goodness of Fit


Section 10.1 Objectives

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


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.


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.


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.


Multinomial Experiments

• Results of a survey of tax preparation methods.

Distribution of tax preparation methods

Accountant 25%

By hand 20%

Computer software 35%

Friend/family 5%

Tax preparation service 15%

Each outcome is classified into categories.

The probability for each possible outcome is fixed.



• 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.



• 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.



• 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.


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)

Survey results(n = 300)

Accountant 71By hand 40Computer software

101

Friend/family 35Tax preparation service

53


Solution: Finding Observed and Expected Frequencies

Observed frequency: The number of adults in the survey naming a particular tax preparation method



101


53

observed frequency


Solution: Finding Observed and Expected Frequencies

Expected Frequency: Ei = npi

Tax preparation method

% of people

Observed frequency

Expected frequency

Accountant 25% 71 300(0.25) = 75

By hand 20% 40 300(0.20) = 60

Computer Software 35% 101 300(0.35) = 105

Friend/family 5% 35 300(0.05) = 15

Tax preparation service

15% 53 300(0.15) = 45

n = 300© 2012 Pearson Education, Inc. All rights reserved. 14 of 91


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.



• 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.



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 α.

Use Table 6 in Appendix B.

d.f. = k – 1

In Words In Symbols



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


Example: Performing a Goodness of Fit TestUse the tax preparation method data to perform a chi-square goodness-of-fit test to test whether the distributions are different. Use α = 0.01.



101


53

Distribution of tax preparation methods

Accountant 25%By hand 20%Computer software

35%

Friend/family 5%Tax preparation service

15%


Solution: Performing a Goodness of Fit Test

• H0:

• Ha:

α = • d.f. = • Rejection Region

• Test Statistic:

• Decision:

• Conclusion:

0.015 – 1 = 4

The distribution is 25% by accountant, 20% by hand, 35% by computer software, 5% by friend/ family, and 15% by tax preparation service. (Claim)The distribution of tax preparation methods differs from the claimed or expected distribution.



22 ( )O E

E

Tax preparation

method

Observed frequency

Expected frequency

Accountant 71 75By hand 40 60Computer hardware

101 105

Friend/family 35 15Tax preparation service

53 45

2 2 2 2 2(71 75) (40 60) (101 105) (35 15) (53 45)75 60 105 15 45

35.121



• H0:

• Ha:

• α =• d.f. = • Rejection Region

• Test Statistic:

• Decision:

0.015 – 1 = 4

The distribution is 25% by accountant, 20% by hand, 35% by computer software, 5% by friend/ family, and 15% by tax preparation service. (Claim)The distribution of tax preparation methods differs from the claimed or expected distribution.

χ2 ≈ 35.121

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.

Reject H0

© 2012 Pearson Education, Inc. All rights reserved.22 of 91

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)


Example: Performing a Goodness of Fit Test

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

n = 500



• H0:

• Ha:


• Test Statistic:

• Decision:

• Conclusion:

0.106 – 1 = 5

0.10

χ20 9.236

Distribution of different-colored candies in bags of dark chocolate M&M’s is uniform. (Claim)Distribution of different-colored candies in bags of dark chocolate M&M’s is not uniform.



2 2 2

2 2 2

(80 83.33) (95 83.33) (88 83.33)83.33 83.33 83.33

(83 83.33) (76 83.33) (78 83.33)83.33 83.33 83.33

3.016

ColorObserved frequency

Expected frequency

Brown 80 83.33Yellow 95 83.33Red 88 83.33Blue 83 83.33Orange 76 83.33Green 78 83.33

22 ( )O E

E



• H0:

• Ha:


• Test Statistic:

• Decision:

0.016 – 1 = 5

0.10

χ20 9.236

χ2 ≈ 3.016

3.016

There is not enough evidence at the 10% level of significance to reject the claim that the distribution is uniform.

Distribution of different-colored candies in bags of dark chocolate M&M’s is uniform. (Claim)Distribution of different-colored candies in bags of dark chocolate M&M’s is not uniform.

Fail to Reject H0


Section 10.1 Summary

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


Section 10.2

Independence



• Use a contingency table to find expected frequencies• Use a chi-square distribution to test whether two

variables are independent


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.


Contingency TablesExample:• 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 Cup Cone Sundae Sandwich Other

Male 600 288 204 24 84Female 410 340 180 20 50

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 c

r cE


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.


Gender Cup Cone Sundae Sandwich Other Total

Male 600 288 204 24 84 1200Female 410 340 180 20 50 1000Total 1010 628 384 44 134 2200

marginal totals© 2012 Pearson Education, Inc. All rights reserved. 34 of 91

Solution: Finding Expected Frequencies



Male 600 288 204 24 84 1200

Female 410 340 180 20 50 1000Total 1010 628 384 44 134 2200

,(Sum of row ) (Sum of column )

Sample sizer cr cE

1,11200 1010 550.91

2200E





Male 600 288 204 24 84 1200

Female 410 340 180 20 50 1000Total 1010 628 384 44 134 2200

1,21200 628 342.55

2200E

1,31200 384 209.45

2200E

1,41200 44 24

2200E

1,51200 134 73.09

2200E





Male 600 288 204 24 84 1200Female 410 340 180 20 50 1000Total 1010 628 384 44 134 2200

2,21000 628 285.45

2200E

2,41000 44 20

2200E

2,51000 134 60.91

2200E

2,11000 1010 459.09

2200E

2,31000 384 174.55

2200E


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.



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.



• 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.





3. Determine the degrees of freedom.


State H0 and Ha.

Identify α.


d.f. = (r – 1)(c – 1)

In Words In Symbols



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





22 ( )O E

E

In Words In Symbols


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.



Male 600(550.91)

288(342.55)

204(209.45)

24(24)

84(73.09)

1200

Female 410(459.09)

340(285.45)

180(174.55)

20(20)

50(60.91)

1000

Total 1010 628 384 44 134 2200



• H0:

• Ha:


• Test Statistic: • Decision:

0.01(2 – 1)(5 – 1) = 4

The adults’ favorite ways to eat ice cream are independent of gender. The adults’ favorite ways to eat ice cream are dependent on gender. (Claim)



2 2 2 2 2

2 2 2 2 2

(600 550.91) (288 342.55) (204 209.45) (24 24) (84 73.09)550.91 342.55 209.45 24 73.09

(410 459.09) (340 285.45) (180 174.55) (20 20) (50 60.91)459.09 285.45 174.55 20 60.91

32.630

22 ( )O E

E



• H0:• Ha:


• Test Statistic:

• Decision:

0.01(2 – 1)(5 – 1) = 4

The adults’ favorite ways to eat ice cream are independent of gender.The adults’ favorite ways to eat ice cream are dependent on gender. (Claim)

χ2 ≈ 32.630

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.

Reject H0



• Used a contingency table to find expected frequencies• Used a chi-square distribution to test whether two

variables are independent


Section 10.3

Comparing Two Variances



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

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


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

sFs


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.


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


F-Distributions

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.


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.

The critical value is F0 = 2.06.

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) 0.0252

12


Solution: Finding Critical F-Values

½α = 0.025, d.f.N = 4 and d.f.D = 8

The critical value is F0 = 5.05.


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.



• Test Statistic2122

sFs

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





3. Determine the degrees of freedom.


State H0 and Ha.

Identify α.


d.f.N = n1 – 1 d.f.D = n2 – 1

In Words In Symbols



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





2122

sFs

In Words In Symbols


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.


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 (Claim)0.10

9 20

0 F1.96

0.10

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

21

22

400 1.56256

sF

s

There is not enough evidence at the 10% level of significance to convince the manager to switch to the new system.

1.961.56

Fail to Reject H0


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 = 31s2 = 3.5 s1 = 5.7


Solution: Performing a Two-Sample F-Test

• H0: • Ha:• ½α = • d.f.N= d.f.D=• Rejection Region:

• Test Statistic:

• Decision:

σ12 = σ2

2

σ12 ≠ σ2

2 (Claim)0. 02530 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

5.7 2.6523.5

sFs

There is enough evidence at the 5% level of significance to support the claim that one of the two stocks is a riskier investment.

2.09 2.652

Reject H0



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

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


Section 10.4

Analysis of Variance



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

• Introduce two-way analysis of variance


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.


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.


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 samplesVarianceT est stati within sastic s mple


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.


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 ( ) 1

x x xx sn n

xx N

SSB ni(xi x)2

SSW (ni 1)si2

In Words In Symbols


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

MSB

SSBd.f.N

ni(xi x)2

k 1

MSW

SSWd.f.D

(ni 1)si

2 N k

In Words In Symbols


Performing a One-Way ANOVA Test



3. Identify the degrees of freedom.


State H0 and Ha.

Identify α.


d.f.N = k – 1 d.f.D = N – k

In Words In Symbols


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


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 squaresVariation

MSB

SSBd.f.N

MSW

SSWd.f.D

B

W

MSMS


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.


Example: Performing a One-Way ANOVA

Medication 1 Medication 2 Medication 312 16 1415 14 1717 21 2012 15 15

19

156 144

x 285 175

x 366 16.54

x

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)


Solution: Performing a One-Way ANOVA

• H0: • Ha:

• α = • d.f.N=

• d.f.D=• Rejection Region:

• Test Statistic:

• Decision:

μ1 = μ2 = μ3

At least one mean is different. (Claim)

0. 013 – 1 = 213 – 3 = 10



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

xx N 56 85 66 15.9213

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 1

21.9232 10.96162



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 10.9616 1.507.3



• H0: • Ha:

• α = • d.f.N=

• d.f.D=• Rejection Region:

• Test Statistic:

• Decision:

μ1 = μ2 = μ3

At least one mean is different. (Claim)

0.013 – 1 = 213 – 3 = 10

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


7.56

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)


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


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 (Claim)At least one mean is different.

• Decision:There is enough evidence at the 10% level of significance to reject the claim that the mean earnings are the same.

P ≈ 0.06 so P < α Reject H0


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.


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 IMales taking type II Females taking type IIMales taking type III Females taking type III

GenderMale Female

III

III


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.


Two-Way ANOVA HypothesesHypotheses 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.


Two-Way ANOVA HypothesesHypotheses 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 typeof 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.


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

• Introduced two-way analysis of variance
