Chi-Square Tests and the F-Distribution 1 Chapter 10.

Chi-Square Tests and the F-Distribution

1

Chapter 10

Chapter Outline

2

10.1 Goodness of Fit10.2 Independence10.3 Comparing Two Variances10.4 Analysis of Variance

Goodness of Fit

3

Section 10.1

Section 10.1 Objectives

4

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

Properties of The Chi-Square Distribution1. All chi-square values χ2 are greater than or

equal to zero.

2. The chi-square distribution is a family of curves, each determined by the degrees of freedom. To find the critical values, use the χ2-distribution with degrees of freedom equal to one less than the sample size.

• d.f. = n – 1 Degrees of freedom

3. The area under each curve of the chi-square distribution equals one.

5

Properties of The Chi-Square Distribution

4. Chi-square distributions are positively skewed.

chi-square distributions6

Finding Critical Values for the χ2-Test

1. Specify the level of significance .

2. Determine the degrees of freedom d.f. = n – 1.

3. The critical values for the χ2-distribution are found in Table 6 of Appendix B. To find the critical value(s) for aa. right-tailed test, use the value that corresponds to

d.f. and .b. left-tailed test, use the value that corresponds to

d.f. and 1 – .

c. two-tailed test, use the values that corresponds to d.f. and ½ and d.f. and 1 – ½.

7

Finding Critical Values for the χ2-Test

χ2

12

2L 2

R

12

1 – α

χ2

20

1 – α

χ2

20

1 – α

Right-tailed Left-tailed

Two-tailed

8

Example: Finding Critical Values for χ2

Find the critical χ2-value for a left-tailed test whenn = 11 and = 0.01.

Solution:•Degrees of freedom: n – 1 = 11 – 1 = 10 d.f. •The area to the right of the critical value is 1 – = 1 – 0.01 = 0.99.

From Table 6, the critical value is .

20 2.558 χ2

0.01

20

20 2.558

9

Example: Finding Critical Values for χ2

Find the critical χ2-value for a two-tailed test when n = 13 and = 0.01.

Solution:•Degrees of freedom: n – 1 = 13 – 1 = 12 d.f. •The areas to the right of the critical values are

From Table 6, the critical values are and

02

051

.0

11 9

20.9 5

2L 2

Rχ2

1 0.0052

1 0.0052

2 3.074L 2 28.299R 2 3.074L

2 28.299R 10

Multinomial Experiments

11

Multinomial experimentA 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.

Multinomial Experiments

12

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.

Chi-Square Goodness-of-Fit Test

13

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.

Chi-Square Goodness-of-Fit Test

14

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.

Chi-Square Goodness-of-Fit Test

15

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.

Chi-Square Goodness-of-Fit Test

16

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

17

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

Solution: Finding Observed and Expected Frequencies

18

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

Solution: Finding Observed and Expected Frequencies

19

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

Chi-Square Goodness-of-Fit Test

20

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.

Chi-Square Goodness-of-Fit Test

21

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.

Chi-Square Goodness-of-Fit Test

22

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

Chi-Square Goodness-of-Fit Test

23

22 ( )O E

E

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.

In Words In Symbols

Example: Performing a Goodness of Fit Test

24

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%

Solution: Performing a Goodness of Fit Test

25

• 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

Solution: Performing a Goodness of Fit Test

26

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

Solution: Performing a Goodness of Fit Test

27

• H0:

• Ha:

• α =

• d.f. =

• Rejection Region

• 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

Example: Performing a Goodness of Fit Test

28

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)

Example: Performing a Goodness of Fit Test

29

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

Solution: Performing a Goodness of Fit Test

30

• H0:

• Ha:

• α =

• d.f. =

• Rejection Region

• 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

Solution: Performing a Goodness of Fit Test

31

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

ColorObserved frequency

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

Solution: Performing a Goodness of Fit Test

32

• H0:

• Ha:

• α =

• d.f. =

• Rejection Region

• 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

Section 10.1 Summary

33

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