expdesign2 - Comparing k Population Means Example: Reed...

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Example: Reed Manufacturing Comparing k Population Means 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). A simple random sample of five managers from each of the three plants was taken and the number of hours worked by each manager for the previous week is shown on the next slide. Conduct an F test using a = .05.
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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 Test for the Equality of k Population Means
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Test for the Equality of k Population Means 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 2 = mean number of hours worked per week by the managers at Plant 2 3 = mean number of hours worked per week by the managers at Plant 3 1. Develop the hypotheses. p -Value and Critical Value Approaches
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2. Specify the level of significance. a = .05 Test for the Equality of k Population Means p -Value and Critical Value Approaches 3. Compute the value of the test statistic. MSTR = 490/(3 - 1) = 245 SSTR = 5(55 - 60) 2 + 5(68 - 60) 2 + 5(57 - 60) 2 = 490 = (55 + 68 + 57)/3 = 60 x (Sample sizes are all equal.) Mean Square Due to Treatments
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3. Compute the value of the test statistic. Test for the Equality of k Population Means MSE = 308/(15 - 3) = 25.667 SSE = 4(26.0) + 4(26.5) + 4(24.5) = 308 Mean Square Due to Error F = MSTR/MSE = 245/25.667 = 9.55 p -Value and Critical Value Approaches
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Treatment Error Total 490 308 798 2 12 14 245 25.667 Source of Variation Sum of Squares Degrees of Freedom Mean Squares 9.55 F Test for the Equality of k Population Means ANOVA Table
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Test for the Equality of k Population Means 5. Determine whether to reject H 0 . We have sufficient evidence to conclude that the mean number of hours worked per week by department managers is not the same at all 3 plant. The p -value < .05, so we reject H 0 . With 2 numerator d.f. and 12 denominator d.f., the p -value is .01 for F = 6.93. Therefore, the p -value is less than .01 for F = 9.55. p Value Approach 4. Compute the p value.
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5. Determine whether to reject H 0 . Because F = 9.55 > 3.89, we reject H 0 . Critical Value Approach 4. Determine the critical value and rejection rule.
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expdesign2 - Comparing k Population Means Example: Reed...

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