1. 1 Chapter 11 – Test for the Equality of k Population Means nRejection Rule where the value of F  is based on an F distribution with k - 1 numerator d.f. and n T - k denominator d.f. Reject H 0 if p -value <  p -value Approach: Critical Value Approach: Reject H 0 if F > F  nHypotheses H 0 :  1  =  2  =  3  = ... =  k  H a : Not all population means are equal nTest Statistic F = MSTR/MSE

2. 2 Test for the Equality of k Population Means

3. 3 AutoShine, Inc. is considering marketing a long- AutoShine, Inc. is considering marketing a long- lasting car wax. Three different waxes (Type 1, Type 2, and Type 3) have been developed. n Example: AutoShine, Inc. In order to test the durability of these waxes, 5 new In order to test the durability of these waxes, 5 new cars were waxed with Type 1, 5 with Type 2, and 5 with Type 3. Each car was then repeatedly run through an automatic carwash until the wax coating showed signs of deterioration. Testing for the Equality of k Population Means: A Completely Randomized Experimental Design

4. 4 The number of times each car went through the The number of times each car went through the carwash before its wax deteriorated is shown on the next slide. AutoShine, Inc. must decide which wax to market. Are the three waxes equally effective? n Example: AutoShine, Inc. Testing for the Equality of k Population Means: A Completely Randomized Experimental Design Factor... Car wax Treatments... Type I, Type 2, Type 3 Experimental units... Cars Response variable... Number of washes

5. 5 12345273029283133283130302928303231 Sample Mean Sample Variance Observation Wax Type 1 Wax Type 2 Wax Type 3 2.5 3.3 2.5 2.5 3.3 2.5 29.0 30.4 30.0 Testing for the Equality of k Population Means: A Completely Randomized Experimental Design

6. 6 nHypotheses where:  1 = mean number of washes using Type 1 wax  2 = mean number of washes using Type 2 wax  3 = mean number of washes using Type 3 wax H 0 :  1  =  2  =  3  H a : Not all the means are equal Testing for the Equality of k Population Means: A Completely Randomized Experimental Design

7. 7 Because the sample sizes are all equal: Because the sample sizes are all equal: MSE = 33.2/(15 - 3) = 2.77 Within-treatments estimate of σ 2 = {(2.5) + (3.3) + (2.5)}/3 = 33.2 {(29–29.8) 2 + (30.4–29.8) 2 + (30–29.8) 2 }/2=.52 nMean Square Error of Within Treatments nMean Square Between Treatments Testing for the Equality of k Population Means: A Completely Randomized Experimental Design = (29 + 30.4 + 30)/3 = 29.8 Mean Square Between Treatments

8. 8 nRejection Rule where F.05 = 3.89 is based on an F distribution with 2 numerator degrees of freedom and 12 denominator degrees of freedom p -Value Approach: Reject H 0 if p -value <.05 Critical Value Approach: Reject H 0 if F > 3.89 Testing for the Equality of k Population Means: A Completely Randomized Experimental Design

9. 9 nTest Statistic There is insufficient evidence to conclude that the mean number of washes for the three wax types are not all the same. nConclusion F = MSTR/MSE = 2.60/2.77 =.939 The p -value is greater than.10, where F = 2.81. (Excel provides a p -value of.42.) (Excel provides a p -value of.42.) Therefore, we cannot reject H 0. Therefore, we cannot reject H 0. Testing for the Equality of k Population Means: A Completely Randomized Experimental Design

10. 10 Source of Variation Sum of Squares Degrees of Freedom MeanSquares F Treatments Error Total 2 14 5.2 33.2 38.4 12 2.60 2.77.939 nANOVA Table Testing for the Equality of k Population Means: A Completely Randomized Experimental Design p -Value.42

11. 11 nExample: Reed Manufacturing Janet Reed would like to know if there is any Janet Reed would like to know if there is any significant difference in the mean number of hours worked per week for the department managers at her three manufacturing plants (in Buffalo, Pittsburgh, and Detroit). An F test will be conducted using  =.05. Testing for the Equality of k Population Means: An Observational Study

12. 12 nExample: Reed Manufacturing A simple random sample of five managers from A simple random sample of five managers from each of the three plants was taken and the number of hours worked by each manager in the previous week is shown on the next slide. Testing for the Equality of k Population Means: An Observational Study Factor... Manufacturing plant Treatments... Buffalo, Pittsburgh, Detroit Experimental units... Managers Response variable... Number of hours worked

13. 13 1 2 3 4 5 48 54 57 54 62 73 63 66 64 74 51 63 61 54 56 Plant 1 Buffalo Plant 2 Pittsburgh Plant 3 Detroit Observation Sample Mean Sample Variance 55 68 57 26.0 26.5 24.5 Testing for the Equality of k Population Means: An Observational Study

14. 14 H 0 :  1  =  2  =  3  H a : Not all the means are equal where:  1 = mean number of hours worked per week by the managers at Plant 1 week by the managers at Plant 1  2 = mean number of hours worked per  2 = mean number of hours worked per week by the managers at Plant 2 week by the managers at Plant 2  3 = mean number of hours worked per week by the managers at Plant 3 week by the managers at Plant 3 1. Develop the hypotheses. p -Value and Critical Value Approaches p -Value and Critical Value Approaches Testing for the Equality of k Population Means: An Observational Study

15. 15 2. Specify the level of significance.  =.05 p -Value and Critical Value Approaches p -Value and Critical Value Approaches 3. Compute the value of the test statistic. {(55 - 60) 2 + (68 - 60) 2 + (57 - 60) 2 }/2= 49 {(55 - 60) 2 + (68 - 60) 2 + (57 - 60) 2 }/2= 49 = (55 + 68 + 57)/3 = 60 (Sample sizes are all equal.) Mean Square Due to Treatments Testing for the Equality of k Population Means: An Observational Study

16. 16 3. Compute the value of the test statistic. MSE = {(26.0) + (26.5) + (24.5)}/3 = 25.667 Mean Square Due to Error (con’t.) F = MSTR/MSE = 245/25.667 = 9.55 Testing for the Equality of k Population Means: An Observational Study