Hwk9SolnGdDrtSp06 - HW 9 SOLUTIONS Question 1 Midway...

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HW 9 SOLUTIONS Question 1. Midway through the oft-delayed construction of the new Starbucks in Ithaca, corporate headquarters in Seattle changed the design of new stores in order to eliminate bottlenecks in the ordering process. They claim that the new design is especially efficient during peak periods, and therefore, as more orders are made, more total time is saved. In order to convince the store owner that it is worth further delaying the project and the additional costs, they send a data file showing the total amount of orders placed during any 4 hours during peak-periods and the change in the average amount of time it took to process those orders from a store with the older design. Data appears in Hwk9Q1DatSp06. Unless otherwise stated, use α = 0.05 or 95% confidence level, where applicable. Regression Analysis: Change in total versus Number of Orders The regression equation is Change in total processing time = 36.8 - 0.474 Number of Orders Predictor Coef SE Coef T P Constant 36.791 4.482 8.21 0.000 Number of Orders -0.47448 0.02269 -20.91 0.000 S = 11.5621 R-Sq = 88.3% R-Sq(adj) = 88.1% Analysis of Variance Source DF SS MS F P Regression 1 58450 58450 437.23 0.000 Residual Error 58 7754 134 Total 59 66203 A. Does the new system save time over the older system as the number of orders increases (state the hypotheses, compute the test statistic and corresponding p- value, and state conclusions)? Ho: β 1 = 0 Ha: β 1 < 0 * BE CAREFUL: This is an awkwardly phrased question. The change in time represents negative numbers (it’s not old-new processing time, which would give positive values), but rather the decrease in magnitude from old to new. We want to prove that the change in the total processing time as we go from the old to the new system decreases (it increases in absolute terms but since the change is in negative numbers it is technically decreasing) as the number of orders increases. This is why we want the slope to be negative. TS: β 1 - β 0 /SE(β 1 )= -0.47448-0/0.02269 = -20.91 p-value: P(t 58 <20.91) <.001 We have a very high p-value so we can not reject the null hypothesis and conclude that the new system probably does save more time over the older system as the number of orders increases.
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H ADM 201 Homework IX Spring 2006 B. What is the likelihood that the new system reduces mean order processing time by more than one minute per order (state the hypothesis, compute the test statistic and p-value, and state conclusion)? Ho: β 1 = -1 Ha: β 1 < -1 TS: β 1 - β 0 /SE(β 1 )= -0.47448-(-1)/0.02269 = 0.52552/0.02269 = 23.16 p-value: P(t 58 < 23.16) > 0.999 We have a very high p-value so we think that it is unlikely that the new system can reduce the mean ordering process time by more than one minute per order. C. What is the likelihood that the new system reduces mean order processing time by less than 0.5 minutes (state the hypothesis, compute the test statistic and p- value, and state the conclusion)? Ho: β 1 = -0.5 Ha: β 1 > -0.5 TS: β 1 - β 0 /SE(β 1 )= -0.47448-(-0.5)/0.02269 = 0.02552/0.02269= 1.12 Student's t distribution with 58 DF x P( X <= x ) 1.12 0.866335 Using minitab: p-value: P(t 58 > 1.12) = 1- 0.866 = .134 Using table: p-value: P(t 58 >1.12) = 1- P(t 58 <1.12)=1-( >0.1)= .134.
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