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ANOVA Analysis of Temporary Employee Work Duration Across Different Industries

Solved assignment conducting ANOVA to compare temporary employee work duration across various industries.

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    ANOVA Testing The manager of an agency providing temporary employees to city offices is analyzing the number of days temporary hires typically work in various types of industries. The data are as follows: a. Legal clerical: 2, 1, 4, 4, 2, 5, 6 b. Accounting firms: 3, 6, 4, 5, 5, 7, 8 c. Insurance: 5, 4, 7, 9, 9, 8, 11 Using the data above, answer the following questions: Are there significant differences in the length of time temporary employee’s work in the different industries? How much of the differences can be explained by the industry? Which groups are significantly different from others? Why would a manager be focused on measuring the number of days that a temporary works each week? SUMMARY FOR SINGLE FACTORS INDUSTRY TYPES TOTAL SUM MEAN VARIANCE Legal Clerical 7 24 3.428571429 3.2857 Accounting Firms 7 38 5.428571429 2.9524 Insurance 7 53 7.571428571 5.9524 Legal Clerical 2 - 3.428571429 = - 1.428571429 ( - 1.428571429 x - 1.428571429) = 2.040 1 - 3.428571429 = - 2.428571429 ( - 2.428571429 x - 2.428571429) = 5.897 4 - 3.428571429 = 0.571428571 (0.571428571 x 0.571428571) = 0.326 4 - 3.428571429 = 0.571428571 (0.571428571 x 0.571428571) = 0.326 2 - 3.428571429 = - 1.428571429 ( - 1.428571429 x - 1.428571429) = 2.040 5 - 3.428571429 = 1.571428571 (1.571428571 x 1.571428571) = 2.469 6 - 3.428571429 = 2.571428571 (2.571428571 x 2.571428571) = 6.612 19.71

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    Variance s2=Σ(x−M)2n−1 S2 = 19.91728571/6 S2 = 3.2857 Accounting Firms 3 - 5.428571429 = - 2.428571429 ( - 2.428571429 x - 2 .428571429) = 5.897 6 - 5.428571429 = 0.571428571 (0.571428571 x 0.571428571) = 0.326 4 - 5.428571429 = - 1.428571429 ( - 1.428571429 x - 1.428571429) = 2.040 5 - 5.428571429 = - 0.428571429 ( - 0.428571429 x - 0.428571429) = 0.18 5 - 5.428571429 = - 0.428571429 ( - 0.428571429 x - 0.428571429) = 0.18 7 - 5.428571429 = 1.571428571 (1.571428571 x 1.571428571) = 2.469 8 - 5.428571429 = 2.571428571 (2.571428571 x 2.571428571) = 6.612 17. 71 Variance s2=Σ(x−M)2n−1 s2 = 17.71428571/6 s2 = 2.9524 Insurance 5 - 7.571428571 = - 2.571428571 ( - 2.571428571 x - 2.571428571) = 6. 612 4 - 7.571428571 = - 3.571428571 ( - 3.571428571 x - 3.571428571) = 12.75 7 - 7.571428571 = - 0.571428571 ( - 0.571428571 x - 0.571428571) = 0.326 9 - 7.571428571 = 1.428571429 (1.428571429 x 1.428571429) = 2.040 9 - 7.571428571 = 1.428571429 (1.428571429 x 1.428571429) = 2.040 8 - 7.571428571 = 0.428571429 (0.428571429 x 0.428571429) = 0.18 11 - 7.571428571 = 3.428571429 (3.428571429 x 3.428571429) = 11.75 35.71 Varianc e s2=Σ(x−M)2n−1

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