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Name ___________________________ Stat 252 Summer 2001 Exam 21. Students at five business colleges entered a competition in which they began their own businesses. Each team hadeleven members and part of the competition involved an evaluation of each member’s communication skills. The team members made individual presentation and a panel of judges scored each. The results by college are analyzed in the MINITAB output that follows. Answer all questions with specific regard to this output and situation.Analysis of Variance for Rating Source DF SS MS F PCollege 4 2.094 0.523 3.76 0.010Error 50 6.967 0.139Total 54 9.061 Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev ------+---------+---------+---------+Wharton 11 3.3364 0.4523 (------*-------) Wisconsin 11 3.4909 0.3448 (------*-------) Columbia 11 3.3000 0.3715 (-------*-------) Manhattan 11 3.0636 0.2203 (------*-------) Cal Poly 11 3.6455 0.4321 (-------*------) ------+---------+---------+---------+Pooled StDev = 0.3733 3.00 3.30 3.60 3.90Tukey's pairwise comparisons Family error rate = 0.0500Individual error rate = 0.00671Intervals for (column level mean) - (row level mean) Wharton Wisconsin Columbia Manhattan Wisconsin -0.6048 0.2957 Columbia -0.4138 -0.2593 0.4866 0.6411 Manhattan -0.1775 -0.0229 -0.2138 0.7229 0.8775 0.6866 Cal Poly -0.7593 -0.6048 -0.7957 -1.0320 0.1411 0.2957 0.1048 -0.1316Fisher's pairwise comparisons Family error rate = 0.277Individual error rate = 0.0500Intervals for (column level mean) - (row level mean) Wharton Wisconsin Columbia Manhattan Wisconsin -0.4743 0.1652 Columbia -0.2834 -0.1289 0.3561 0.5107 Manhattan -0.0470 0.1075 -0.0834 0.5925 0.7470 0.5561 Cal Poly -0.6289 -0.4743 -0.6652 -0.9016 0.0107 0.1652 -0.0257 -0.2620a) Is this an observational study or a designed experiment? Explain. (2)b) For parts i) through v), choose the most appropriate response from the following choices (the same answer may be given more than once): (5)(1) Presenters (2) Rating of a presenter’s communication skills(3) Colleges(4) Wharton, Wisconsin, Columbia , Manhattan, Cal Poly (5) None of the previousi) What is (are) the factor(s)? ii) What is (are) the factor level(s)? iii) What is (are) the experimental unit(s)?iv) What is (are) the treatment(s)? v) What is (are) the response(s)? c) What is (are) the parameter(s) of interest in this problem? (2)d) Give the hypotheses of interest. (2)e) What is the decision rule for this test? Use a level of significance of 0.05. (2)f) What is the value of the test statistic? (2)g) True or false: (5)i) The null hypothesis would be rejected.ii) The decision is made with 95% confidence.iii) The decision implies that all five colleges have different mean scores.iv) For this procedure to be valid, it is necessary that the scores be normally distributed.v) For this procedure to be valid, it is necessary that the five colleges have equal variances in their scores.h) Which pairs of means appear to be significantly different according to Tukey’s HSD? (2)i) What is the probability that any one particular Tukey confidence interval is incorrect? (2)j) Which pairs of means appear to be significantly different according to Fisher’s LSD? (2)k) What is the probability that any one particular Fisher confidence interval is incorrect? (2)l) Use the underline notation to summarize the results from Tukey's procedure. (3)m) Use the underline notation to summarize the results from Fisher's procedure. (3)n) Why do the results from Tukey and Fisher's procedures differ? (Hint: To answer, explain the difference between liberal and conservative multiple comparison procedures, indicating which type Tukey's and Fisher'sare.) (3)2. The MINITAB output below was used by the manager of a cannery to see if she could predict the number of pieces of fruit on a plant based on the weight of the plant.Regression Analysis: N of Fruit versus Wt.Plant KgThe regression equation isN of Fruit = 22.1 + 3.41 Wt.Plant KgPredictor Coef SE Coef T PConstant 22.094 5.144 4.30 0.000Wt.Plant 3.4144 0.9122 3.74 0.001S = 13.27 R-Sq = 24.6% R-Sq(adj) = 22.8%Analysis of VarianceSource DF SS MS F PRegression 1 2467.5 2467.5 14.01 0.001Residual Error 43 7573.7 176.1Total 44 10041.2Predicted Values for New ObservationsNew Obs Fit SE Fit 95.0% CI 95.0% PI1 39.17 1.99 ( 35.16, 43.17) ( 12.10, 66.23) 2 42.58 2.11 ( 38.33, 46.83) ( 15.48, 69.68) 3 46.00 2.57 ( 40.82, 51.17) ( 18.73, 73.26) 4 49.41 3.23 ( 42.90, 55.92) ( 21.87, 76.95) 5 52.82 3.99 ( 44.78, 60.86) ( 24.88, 80.77)Values of Predictors for New ObservationsNew Obs Wt.Plant1 5.002 6.003 7.004 8.005 9.00a) From the output, read the values of the following. (5)i) The quantity that the least squares line minimizes.ii) The estimate of 2.iii) The standard deviation in the estimate of the slope.iv) The quantity that measures the proportion of error removed from the estimation of the number of pieces of fruit by using the weight of the plant rather than using the mean number of pieces of fruit per plant.v) The estimate of the average change in the number of pieces of fruit for an increase of one kilogram in the weight of a plant.b) Interpret the coefficient of determination from this output. You can obtain the correlation coefficient by taking the square root of the coefficient of determination. How do you determine whether the correlation is positive or negative? (2)c) Test to see if there is a linear relationship between the

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