the mcgraw-hill companies, inc., 2000 12-1 chapter 12 analysis of variance (anova)
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© The McGraw-Hill Companies, Inc., Analysis of Variance (ANOVA) analysis of variance (ANOVA) When an F test is used to test a hypothesis concerning the means of three or more populations, the technique is called analysis of variance (ANOVA).TRANSCRIPT
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12-112-1
Chapter 12Chapter 12
Analysis of Variance Analysis of Variance (ANOVA)(ANOVA)
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12-112-1 ObjectivesObjectives Use ANOVA technique to determine
differences among three or more means when data are ratio or interval.
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12-312-3 Analysis of Variance (ANOVA)Analysis of Variance (ANOVA)
When an F test is used to test a hypothesis concerning the means of three or more populations, the technique is called analysis of variance analysis of variance (ANOVA)(ANOVA).
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12-412-4Assumptions for the Assumptions for the FF Test for Test for Comparing Three or More MeansComparing Three or More Means
The populations from which the samples were obtained must be normally or approximately normally distributed.
The samples must be independent of each other and randomly sampled.
The variances of the populations must be equal.
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Although means are being compared in this F test, variances are used in the test instead of the means.
Two different estimates of the population variance are made.
Analysis of VarianceAnalysis of Variance
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Between-group varianceBetween-group variance - this involves computing the variance by using the means of the groups or between the groups.
Within-group varianceWithin-group variance - this involves computing the variance by using all the data and is not affected by differences in the means.
Analysis of VarianceAnalysis of Variance
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The following hypotheses should be used when testing for the difference between three or more means.
H0: = = = … = k
H1: At least one mean is different from the others.
Analysis of VarianceAnalysis of Variance
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d.f.N. = k – 1, where k is the number of groups.
d.f.D. = N – k, where N is the sum of the sample sizes of the groups.
Note: The formulas for this test are tedious to work through, so examples will be done in MINITAB. See text for formulas.
Analysis of VarianceAnalysis of Variance
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A marketing specialist wishes to see whether there is a difference in the average time a customer has to wait in a checkout line in three large self-service department stores. The times (in minutes) are shown on the next slide.
Is there a significant difference in the mean waiting times of customers for each store using = 0.05?
Analysis of VarianceAnalysis of Variance - - Example
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Store A Store B Store C
3 5 1
2 8 3
5 9 4
6 6 2
3 2 7
1 5 3
Store A Store B Store C
3 5 1
2 8 3
5 9 4
6 6 2
3 2 7
1 5 3
Analysis of VarianceAnalysis of Variance - - Example
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Step 1:Step 1: State the hypotheses and identify the claim.
H0: = = H1: At least one mean is different from the others (claim).
Analysis of VarianceAnalysis of Variance - - Example
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Step 2:Step 2: Find the critical value. Since k = 3, N = 18, and = 0.05, d.f.N. = k – 1 = 3 – 1= 2, d.f.D. = N – k = 18 – 3 = 15. The critical value is 3.68.
Step 3:Step 3: Compute the test value. From the SPSS output, F = 2.70. (See your text for computations).
Analysis of VarianceAnalysis of Variance - - Example
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Step 4:Step 4: Make a decision. Since 2.70 < 3.68, the decision is not to reject the null hypothesis.
Step 5:Step 5: Summarize the results. There is not enough evidence to support the claim that there is a difference among the means. The ANOVA summary table is given on the following slides.
Analysis of VarianceAnalysis of Variance - - Example
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Before analyzing the ANOVA table check that there is “homogeneity of variance”.
Below are the results.
Since Sig. (P) is greater than 0.05 the variances are equal (i.e., homogeneous).
Test of Homogeneity of Variances Time
Levene df1 df2 Sig.Statistic0.187 2 15 0.832
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Below is the SPSS output from Compare Means | One-way ANOVA…
Since P = 0.100 (Sig.) is > 0.05 the null hypothesis is not rejected, thus all group means are equal.
ANOVA Time Sum of df Mean F Sig.Squares Square
Between Groups 25.000 2 12.500 2.698 0.100Within Groups 69.500 15 4.633
Total 94.500 17