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

3. Section 10.1
Goodness of Fit
Larson/Farber 4th ed

4. Section 10.1 Objectives
Use the chi-square distribution to test whether a frequency distribution fits a claimed distribution
Larson/Farber 4th ed

5. 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.
Larson/Farber 4th ed

6. 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.
Larson/Farber 4th ed

7. 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.
Larson/Farber 4th ed

8. 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.
Larson/Farber 4th ed

9. 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.
Larson/Farber 4th ed

10. 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.
Larson/Farber 4th ed

11. 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
Larson/Farber 4th ed

12. 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
Larson/Farber 4th ed

13. 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
Larson/Farber 4th ed

14. 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.
Larson/Farber 4th ed

15. 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
Larson/Farber 4th ed

16. 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
Larson/Farber 4th ed

17. 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
Larson/Farber 4th ed

18. 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.
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
Larson/Farber 4th ed

19. 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
Larson/Farber 4th ed

20. 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
Larson/Farber 4th ed

21. 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
Larson/Farber 4th ed

22. 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)
Larson/Farber 4th ed

23. 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
Larson/Farber 4th ed

24. 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
Larson/Farber 4th ed

25. 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
Larson/Farber 4th ed

26. 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
Larson/Farber 4th ed

27. Section 10.1 Summary
Used the chi-square distribution to test whether a frequency distribution fits a claimed distribution
Larson/Farber 4th ed

28. Section 10.2
Independence
Larson/Farber 4th ed

29. Section 10.2 Objectives
Use a contingency table to find expected frequencies
Use a chi-square distribution to test whether two variables are independent
Larson/Farber 4th ed

30. 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.
Larson/Farber 4th ed

31. 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
70 and over
Small / Midsize 42 69
108 60 21
Large 5 18 85
120 22
Larson/Farber 4th ed

32. 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
Larson/Farber 4th ed

33. 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
70 and over
Total
Small / Midsize 42 69
108 60 21
300
Large 5 18 85
120 22
250 47 87
193
180 43
550
marginal totals
Larson/Farber 4th ed

34. Solution: Finding Expected Frequencies
Age
Company size
39 and under
70 and over
Total
Small / Midsize 42 69
108 60 21
300
Large 5 18 85
120 22
250 47 87
193
180 43
550
Larson/Farber 4th ed

35. Solution: Finding Expected Frequencies
Age
Company size
39 and under
70 and over
Total
Small / Midsize 42 69
108 60 21
300
Large 5 18 85
120 22
250 47 87
193
180 43
550
Larson/Farber 4th ed

36. Solution: Finding Expected Frequencies
Chapter 10
Solution: Finding Expected Frequencies
Age
Company size
39 and under
70 and over
Total
Small / Midsize 42 69
108 60 21
300
Large 5 18 85
120 22
250 47 87
193
180 43
550
Larson/Farber 4th ed
Larson/Farber 4th ed

37. 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.
Larson/Farber 4th ed

38. Chi-Square Independence Test
For the chi-square independence 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.
Larson/Farber 4th ed

39. Chi-Square Independence Test
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.
The test is always a right-tailed test.
Larson/Farber 4th ed

40. Chi-Square Independence Test
Chapter 10
Chi-Square Independence 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. = (r – 1)(c – 1)
Use Table 6 in Appendix B.
Larson/Farber 4th ed
Larson/Farber 4th ed

41. Chi-Square Independence Test
Chapter 10
Chi-Square Independence 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
Larson/Farber 4th ed

42. 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 = Expected frequencies are shown in parentheses.
Age
Company size
39 and under
70 and over
Total
Small / Midsize
42 (25.64)
69 (47.45)
108 (105.27)
60 (98.18)
21 (23.45)
300
Large
5 (21.36)
18 (39.55)
85 (87.73)
120 (81.82)
22 (19.55)
250 47 87
193
180 43
550
Larson/Farber 4th ed

43. Solution: Performing a Goodness of Fit Test
H0:
Ha:
α =
d.f. =
Rejection Region
CEOs’ ages are independent of company size
CEOs’ ages are dependent on company size
0.01
(2 – 1)(5 – 1) = 4
Test Statistic:
Decision:
0.01 χ2
13.277
Larson/Farber 4th ed

44. Solution: Performing a Goodness of Fit Test
Chapter 10
Solution: Performing a Goodness of Fit Test
Larson/Farber 4th ed
Larson/Farber 4th ed

45. Solution: Performing a Goodness of Fit Test
H0:
Ha:
α =
d.f. =
Rejection Region
CEOs’ ages are independent of company size
CEOs’ ages are dependent on company size
0.01
(2 – 1)(5 – 1) = 4
Test Statistic:
Decision:
χ2 = 77.9
Reject H0
0.01 χ2
13.277
There is enough evidence to conclude CEOs’ ages are dependent on company size.
77.9
Larson/Farber 4th ed

46. Section 10.2 Summary
Used a contingency table to find expected frequencies
Used a chi-square distribution to test whether two variables are independent
Larson/Farber 4th ed

47. Comparing Two Variances
Section 10.3
Comparing Two Variances
Larson/Farber 4th ed

48. Section 10.3 Objectives
Interpret the F-distribution and use an F-table to find critical values
Perform a two-sample F-test to compare two variances
Larson/Farber 4th ed

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.
Larson/Farber 4th ed

50. Properties of the F-Distribution
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
F-distributions are positively skewed.
The total area under each curve of an F-distribution is equal to 1.
Larson/Farber 4th ed

51. Properties of the F-Distribution
Chapter 10
Properties of the F-Distribution
F-values are always greater than or equal to 0.
For all F-distributions, the mean value of F is approximately equal to 1.
d.f.N = 1 and d.f.D = 8 F 1 2 3 4
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
Larson/Farber 4th ed
Larson/Farber 4th ed

52. Critical Values for the F-Distribution
Chapter 10
Critical Values for the F-Distribution
Specify the level of significance .
Determine the degrees of freedom for the numerator, d.f.N.
Determine the degrees of freedom for the denominator, d.f.D.
Use Table 7 in Appendix B to find the critical value. If the hypothesis test is
one-tailed, use the  F-table.
two-tailed, use the ½ F-table.
Larson/Farber 4th ed
Larson/Farber 4th ed

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.
Solution:
The critical value is F0 = 2.43.
Larson/Farber 4th ed

54. 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.
Larson/Farber 4th ed

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.
Larson/Farber 4th ed

56. Two-Sample F-Test for Variances
To use the two-sample F-test for comparing two population variances, the following must be true.
The samples must be randomly selected.
The samples must be independent.
Each population must have a normal distribution.
Larson/Farber 4th ed

57. Two-Sample F-Test for Variances
Test Statistic
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
Larson/Farber 4th ed

58. Two-Sample F-Test for Variances
Chapter 10
Two-Sample F-Test for Variances
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.N = n1 – 1 d.f.D = n2 – 1
Use Table 7 in Appendix B.
Larson/Farber 4th ed
Larson/Farber 4th ed

59. Two-Sample F-Test for Variances
Chapter 10
Two-Sample F-Test for Variances
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 F is in the rejection region, reject H0. Otherwise, fail to reject H0.
Larson/Farber 4th ed
Larson/Farber 4th ed

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.
Larson/Farber 4th ed

61. Solution: Performing a Two-Sample F-Test
Because 400 > 256,
H0:
Ha:
α =
d.f.N= d.f.D=
Rejection Region:
σ12 ≤ σ22
σ12 > σ22
Test Statistic:
Decision:
0.10 9 20
Fail to Reject H0
There is not enough evidence to convince the manager to switch to the new system. F
1.96
0.10
1.96
1.56
Larson/Farber 4th ed

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 B
n2 = 30
n1 = 31
s2 = 3.5
s1 = 5.7
Larson/Farber 4th ed

63. Solution: Performing a Two-Sample F-Test
Because 5.72 > 3.52,
H0:
Ha:
½α =
d.f.N= d.f.D=
Rejection Region:
σ12 = σ22
σ12 ≠ σ22
Test Statistic:
Decision:
0. 025 30 29
Reject H0
There is enough evidence to support the claim that one of the two stocks is a riskier investment. F
2.09
0.025
2.09
2.65
Larson/Farber 4th ed

64. Section 10.3 Summary
Interpreted the F-distribution and used an F-table to find critical values
Performed a two-sample F-test to compare two variances
Larson/Farber 4th ed

65. Section 10.4
Analysis of Variance
Larson/Farber 4th ed

66. Section 10.4 Objectives
Use one-way analysis of variance to test claims involving three or more means
Introduce two-way analysis of variance
Larson/Farber 4th ed

67. One-Way ANOVA One-way analysis of variance
Chapter 10
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.
Larson/Farber 4th ed
Larson/Farber 4th ed

68. One-Way ANOVA In a one-way ANOVA test, the following must be true.
Each sample must be randomly selected from a normal, or approximately normal, population.
The samples must be independent of each other.
Each population must have the same variance.
Larson/Farber 4th ed

69. One-Way ANOVA
The variance between samples MSB measures the differences related to the treatment given to each sample and is sometimes called the mean square between.
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.
Larson/Farber 4th ed

70. 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
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.
Larson/Farber 4th ed

71. Test Statistic for a One-Way ANOVA
Chapter 10
Test Statistic for a One-Way ANOVA
In Words In Symbols
Find the mean and variance of each sample.
Find the mean of all entries in all samples (the grand mean).
Find the sum of squares between the samples.
Find the sum of squares within the samples.
Larson/Farber 4th ed
Larson/Farber 4th ed

72. Test Statistic for a One-Way ANOVA
Chapter 10
Test Statistic for a One-Way ANOVA
In Words In Symbols
Find the variance between the samples.
Find the variance within the samples
Find the test statistic.
Larson/Farber 4th ed
Larson/Farber 4th ed

73. Performing a One-Way ANOVA Test
Chapter 10
Performing a One-Way ANOVA 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.N = k – 1 d.f.D = N – k
Use Table 7 in Appendix B.
Larson/Farber 4th ed
Larson/Farber 4th ed

74. Performing a One-Way ANOVA Test
Chapter 10
Performing a One-Way ANOVA 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 F is in the rejection region, reject H0. Otherwise, fail to reject H0.
Larson/Farber 4th ed
Larson/Farber 4th ed

75. Chapter 10
ANOVA Summary Table
A table is a convenient way to summarize the results in a one-way ANOVA test.
d.f.D
SSW
Within
d.f.N
SSB
Between F
Mean squares
Degrees of freedom
Sum of squares
Variation
Larson/Farber 4th ed
Larson/Farber 4th ed

76. 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.
Larson/Farber 4th ed

77. Example: Performing a One-Way ANOVA
Medication 1
Medication 2
Medication 3 12 16 14 15 17 21 20 19
Solution:
k = 3 (3 samples)
N = n1 + n2 + n3 = = 13 (sum of sample sizes)
Larson/Farber 4th ed

78. Solution: Performing a One-Way ANOVA
H0:
Ha:
α =
d.f.N=
d.f.D=
Rejection Region:
μ1 = μ2 = μ3
At least one mean is different
Test Statistic:
Decision:
0. 01
3 – 1 = 2
13 – 3 = 10 F
7.56
0.01
Larson/Farber 4th ed

79. Solution: Performing a One-Way ANOVA
To find the test statistic, the following must be calculated.
Larson/Farber 4th ed

80. Solution: Performing a One-Way ANOVA
To find the test statistic, the following must be calculated.
Larson/Farber 4th ed

81. Solution: Performing a One-Way ANOVA
H0:
Ha:
α =
d.f.N=
d.f.D=
Rejection Region:
μ1 = μ2 = μ3
At least one mean is different
Test Statistic:
Decision:
0. 01
3 – 1 = 2
13 – 3 = 10
Fail to Reject H0
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. F
7.56
0.01
1.50
Larson/Farber 4th ed

82. 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.
Larson/Farber 4th ed

83. 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
125
114
147
116
124
164
134
108
151
142
140
113
136
141
Larson/Farber 4th ed

84. Solution: Using the TI-83/84 to Perform a One-Way ANOVA
Ha:
Store data into lists L1, L2, and L3
μ1 = μ2 = μ3
At least one mean is different
Decision:
P-value < α Reject H0
There is enough evidence to support the claim. You can conclude that there is a difference in the means of the flight times.
Larson/Farber 4th ed