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Chapter 13Chapter 13

Analysis of Variance Analysis of Variance (ANOVA)(ANOVA)

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13-213-2 OutlineOutline

12-1 Analysis of Variance (ANOVA)

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13-313-3 ObjectivesObjectives

Use ANOVA technique to determine a difference among three or more means.

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13-413-413-1 Analysis of Variance (ANOVA)13-1 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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13-513-513-1 Assumptions for the 13-1 Assumptions for the F-F-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.

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.

13-1 Analysis of Variance13-1 Analysis 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.

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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.

13-1 Analysis of Variance13-1 Analysis 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.

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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?

13-1 Analysis of Variance13-1 Analysis 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

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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).

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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 MINITAB output, F = 2.70. (See your text for computations).

13-1 Analysis of Variance13-1 Analysis 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 next slide.

13-1 Analysis of Variance13-1 Analysis of Variance --Example

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