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How long are resale houses on the market? Suppose that a survey reported that in Thunder Bay, resale houses are on the market an average of 112 days. Of course, the length of time varies by market. Suppose random samples of 13 houses in Thunder Bay and 11 houses in Moncton that are up for resale are traced. The data shown here represent the number of days each house was on the market before being sold. Use the given data and a 1% level of significance to determine whether the population variances for the number of days until resale are different in Thunder Bay than in Moncton. Assume the numbers of days that resale houses are on the market are normally distributed. Thunder Moncton Bay 132 126 118 56 138 94 85 69 131 161 113 67 127 133 81 54 99 119 94 137 126 88 93 134 Find the F-score to make a conclusion. Round your answer to 2 decimal places. Observed F = 1.71 The decision is to fail to reject the null hypothesis +Is there more variation in the output of one shift in a manufacturing plant than in another shift? In an effort to study this question, plant managers gathered productivity reports from the 3 AM. to 4 RM. shift for eight days. The reports indicated that the following numbers of units were produced on each day for this shift. 5,503 4,709 5,112 5.380 4,918 4.763 5,101 5.106 Productivity information was also gathered from seven days for the 4 PM. to midnight shift, resulting in the following data. 4,325 4.024 4,3?2 4.633 3,952 4.754 4,116 Use these data and a = 0.01 to test to determine whether the variances of productivity for the two shifts are the same. Assume productivity is normally distributed in the population. (Round the values of mean to 4 decimal places, e.g. 1895. 1539. Round the values of 512 and 522 to the nearest integer, e.g. 18 9515. Round your answer to 2 decimal places, es. 035.] Observed F = | Thedecision is to I #:l- There is l t l in the variances of the shifts. One of the new areas of quality-control management is to examine the process by which a product is produced. This approach also applies to paperwork. In industries where large long-term projects are undertaken, days and even weeks may elapse as a change order makes its way through a maze of approvals before receiving final approval. This process can result in long delays, stretching schedules to the breaking point. Suppose a quality-control consulting group claims that it can significantly reduce the number of days required for such paperwork to receive approval. In an attempt to prove its case, the group selects five jobs for which it revises the paperwork system. The following data show the number of days required for a change order to be approved before the group intervened and the number of days required for a change order to be approved after the group instituted a new paperwork system. Before After 12 CO 7 10 CO 15 8 4 Use a = 0.01 to determine whether there was a significant drop in the number of days required to process paperwork to approve change orders. Assume that the differences in days are normally distributed. (Round the intermediate values to 4 decimal places, e.g. 1.7589. Round your answer to 2 decimal places, e.g. 1.75.) Observed t = The decision is toCompute a one-way ANDVA on the following data. 1 14 13 10 2 10 9 12 9 10 3 11 12 13 12 4 1s 17 14 16 17 5 14 12 13 13 12 14 Appendix A Statistical Tat Determine the observed Fvalue. lCompare the observed F value with the critical table F value and decide whether to reject the null hypothesis. Use o = .01. [Round the quantitative answers to 2 decimal places, ea. 1.25.] Observed F: | Critical F= ' I : the null hypothesis. Fall to reject Reject Compute a one-way ANDVA on the following data. 1 2 27 22 31 27 31 2 5 29 23 30 26 27 27 28 23 Determine the observed F value. Compare it with the critical table F value and decide whether to reject the null hypothesis. Perform a ttest for independent measures on the data. [Round the quantitative answers to 2 decimal places, e.g. 1.75.} Observed F: | $1 the null hypothesis. Compare the t and F values.Are the results different? Use a: 0.05 Family transportation costs are usually higher than most people believe because those costs include car payments, insurance, fuel costs, repairs, parking, and public transportation. Twenty randomly selected families in four major cities are asked to use their record to estimate a monthly figure for transportation cost. Use the data obtained and ANOVA to test whether there is a significant difference in monthly transportation costs for families living in these cities. Assume that a = 0.05. Edmonton Toronto Vancouver Halifax $650 $250 $850 $540 480 525 700 450 550 300 950 675 600 175 780 550 675 500 600 600 (Round the quantitative answers to 2 decimal places, e.g. 1.75.) Observed F = Critical F = the null hypothesis. There v significant difference. is a is no Appendix A Statistical TablBusiness is very good for a chemical company. In fact, it is so good that workers are averaging more than 40 hours per week at each of the chemical company's five plants. However, management is not certain whether there is a difference between the five plants in the average number of hours worked per week per worker. Random samples of data are taken at each of the five plants. The data are analyzed using Excel. The results follow. Determine whether there is an overall significant difference between the means at a = 0.05? What are the values of the means? B C D 1 ANOVA: Single Factor 2 3 SUMMARY 4 Groups Count Sum Average Variance 5 Plant 11 636.5577 57.87 63.5949 6 Plant 2 12 601.7648 50.15 62.4813 7 Plant 3 8 491.7352 61 47 47.4772 Plant 4 5 246.0172 49.20 65.6072 Plant 5 7 398.6368 56.95 140.3540 B 9 10 ANOVA 11 Source of Variation SS df MS F P value F crit 12 Between Groups 900.0863 4 225.0216 3.10 0.026595 2.62 13 Within Groups 2760.136 38 72.63516 14 15 Total 3660.223 42 There significant difference. Plant 1: Plant 2: Plant 3: Plant 4: Plant 5