1. Chapter 10 Chi-Square Tests and the F-Distribution
Larson/Farber 4th ed

2. Chapter Outline 10.1 Goodness of Fit 10.2 Independence
10.3 Comparing Two Variances
10.4 Analysis of Variance

3. Section 10.1
Goodness of Fit

4. Section 10.1 Objectives
Use the chi-square distribution to test whether a frequency distribution fits a claimed distribution

5. Properties of The Chi-Square Distribution
All chi-square values χ2 are greater than or equal to zero.
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 – Degrees of freedom
The area under each curve of the chi-square distribution equals one.

6. Properties of The Chi-Square Distribution
Chi-square distributions are positively skewed.
chi-square distributions

7. Finding Critical Values for the χ2-Test
Specify the level of significance .
Determine the degrees of freedom d.f. = n – 1.
The critical values for the χ2-distribution are found in Table 6 of Appendix B. To find the critical value(s) for a
right-tailed test, use the value that corresponds to d.f. and .
left-tailed test, use the value that corresponds to d.f. and 1 – .
two-tailed test, use the values that corresponds to d.f. and ½ and d.f. and 1 – ½. 7 7

8. Finding Critical Values for the χ2-Test
Right-tailed
Left-tailed χ2
1 – α χ2
1 – α
Two-tailed χ2
1 – α 8 8

9. Example: Finding Critical Values for χ2
Find the critical χ2-value for a left-tailed test when n = 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. χ2
From Table 6, the critical value is 9 9

10. 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 χ2
From Table 6, the critical values are and 10 10

11. Multinomial Experiments
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.

12. 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 Preferences
Classical 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.

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.

14. Chi-Square Goodness-of-Fit Test
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.

15. 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.

16. 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.

17. 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 8
Country
210
Gospel 72
Oldies 10
Pop 75
Rock
125

18. 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 8
Country
210
Gospel 72
Oldies 10
Pop 75
Rock
125
observed frequency

19. Solution: Finding Observed and Expected Frequencies
Expected Frequency: Ei = npi
Type of music
% of listeners
Observed frequency
Expected frequency
Classical 4% 8
Country
36%
210
Gospel
11% 72
Oldies 2% 10
Pop
18% 75
Rock
29%
125
500(0.04) = 20
500(0.36) = 180
500(0.11) = 55
500(0.02) = 10
500(0.18) = 90
500(0.29) = 145
n = 500

20. Chi-Square Goodness-of-Fit Test
For the chi-square goodness-of-fit test to be used, the following must be true.
The observed frequencies must be obtained by using a random sample.
Each expected frequency must be greater than or equal to 5.

21. Chi-Square Goodness-of-Fit Test
Chapter 10
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
where O represents the observed frequency of each category and E represents the expected frequency of each category.
The test is always a right-tailed test.
Larson/Farber 4th ed

22. Chi-Square Goodness-of-Fit Test
Chapter 10
Chi-Square Goodness-of-Fit Test
In Words In Symbols
Identify the claim. State the null and alternative hypotheses.
Specify the level of significance.
Identify the degrees of freedom.
Determine the critical value.
State H0 and Ha.
Identify .
d.f. = k – 1
Use Table 6 in Appendix B.
Larson/Farber 4th ed

23. Chi-Square Goodness-of-Fit Test
Chapter 10
Chi-Square Goodness-of-Fit Test
In Words In Symbols
Determine the rejection region.
Calculate the test statistic.
Make a decision to reject or fail to reject the null hypothesis.
Interpret the decision in the context of the original claim.
If χ2 is in the rejection region, reject H0. Otherwise, fail to reject H0.
Larson/Farber 4th ed

24. 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 =
Distribution of music preferences
Classical 4%
Country
36%
Gospel
11%
Oldies 2%
Pop
18%
Rock
29%
Survey results (n = 500)
Classical 8
Country
210
Gospel 72
Oldies 10
Pop 75
Rock
125

25. Solution: Performing a Goodness of Fit Test
H0:
Ha:
α =
d.f. =
Rejection Region
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
0.01
6 – 1 = 5
Test Statistic:
Decision:
Conclusion:
0.01 χ2
15.086

26. Solution: Performing a Goodness of Fit Test
Chapter 10
Solution: Performing a Goodness of Fit Test
Type of music
Observed frequency
Expected frequency
Classical 8 20
Country
210
180
Gospel 72 55
Oldies 10
Pop 75 90
Rock
125
145
Larson/Farber 4th ed

27. Solution: Performing a Goodness of Fit Test
H0:
Ha:
α =
d.f. =
Rejection Region
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
0.01
6 – 1 = 5
Test Statistic:
Decision:
χ2 =
Reject H0
0.01 χ2
15.086
There is enough evidence to conclude that the distribution of music preferences differs from the claimed distribution.
22.713

28. Example: Performing a Goodness of Fit Test
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)

29. Example: Performing a Goodness of Fit Test
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
Color
Frequency
Brown 80
Yellow 95
Red 88
Blue 83
Orange 76
Green 78
n = 500

30. Solution: Performing a Goodness of Fit Test
H0:
Ha:
α =
d.f. =
Rejection Region
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
0.10
6 – 1 = 5
Test Statistic:
Decision:
Conclusion:
0.10 χ2
9.236

31. Solution: Performing a Goodness of Fit Test
Chapter 10
Solution: Performing a Goodness of Fit Test
Color
Observed frequency
Expected frequency
Brown 80
83.3
Yellow 95
Red 88
Blue 83
Orange 76
Green 78
Larson/Farber 4th ed

32. Solution: Performing a Goodness of Fit Test
H0:
Ha:
α =
d.f. =
Rejection Region
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
0.01
6 – 1 = 5
Test Statistic:
Decision:
χ2 = 3.016
Fail to Reject H0
0.10 χ2
9.236
There is not enough evidence to dispute the claim that the distribution is uniform.
3.016

33. Section 10.1 Summary
Used the chi-square distribution to test whether a frequency distribution fits a claimed distribution