A marketing organization wishes to study the effects of four sales methods on weekly sales of a product. The organization employs a randomized block design in which three salesman use each sales method. The results obtained are given in the following table, along with the JMP output of a randomized block ANOVA of these data. Salesman, 1' Sales Method, 1'. A B C 1 32 29 30 2 32 30 23 3 23 25 23 4 25 24 24 Analysis of Variance Sum of Mean Seurce DF Squares Square 1' Ratio Model 5 106.50000 21.3000 19.6615 Error 6 6.50000 1.0833 Prob > F c. Total 11 113.00000 22.3033 0.0012* Effect II'ests Sum of Source Nparm 131" Squares 1" Ratio Prob > 5' Method 3 3 87.000000 26.7692 0.0007* Salesman 2 2 19.500000 9.0000 0.0150 Least Squares Means Estimates Method Estimate Salesman Estimate 1 30.333333 A 29.250000 2 30.000000 B 27.000000 3 25.333333 C 26.250000 4 24.333333 Tukey HSD All Pairwise Comparisons Quantile = 3.46171, Adjusted DF = 60, Adjustment = Tukey Method -Method Difference std Error t Ratio Prob) |t| Lower 95% Upper 95:. 1 2 0.333333 0.3498366 0.39 0.9777 2.60855 3.275220 1 3 5.000000 0.3498366 5.38 0.0043\" 2.05812 7.941830 1 4 6.000000 0.3498366 7.06 0.0017* 3.05812 8.941880 2 3 4.666667 0.3498366 5.49 0.0061* 1.72478 7.608550 2 4 5.666667 0.3498366 6.67 0.0023\" 2.72478 8.608550 3 4 1.000000 0.3498366 1.18 0.6613 l.94188 3.941830 (3) Test the null hypothesis so that no differences exist between the effects of the sales methods (treatments) on mean weekly sales. Set a = .05. Can we conclude that the different sales methods have different effects on mean weekly sales? 9 Answer is complete and correct. F = 26.7692, p-value = 0 0007. H0: there is a difference a in sales methods. Reject 0 (b) Test the null hypothesis an that no differences exist between the effects of the salesmen {blocks} on mean weekly sales. Set a = .05. Can we conclude that the different salesmen have different effects on mean weekly sales? 0 Answer is complete and correct. F = 9.0000. p-value = 0 0156' Reject 9 H0: salesman do 9 have an effecton sales. (c) Use Tukey simultaneous 95 percent confidence intervals to make pairwise comparisons of the sales method effects on mean weekly sales. Which sales methodis) maximize mean weekly sales? {Round your answers to 3 decimal places. Negative amounts should be indicated by a minus sign.) Method 1 Method 2: Method 1 Method 3: [ [ Method 1 Method 4: [ Method 2 Method 3: [ i [ Method 2 Method 4: Method 3 Method 4: A consumer preference study compares the effects of three different bottle designs (A, B, and C.) on sales of a popular fabric softener. A completely randomized design is employed. Specically, 15 supermarkets of equal sales potential are selected, and 5 of these supermarkets are randomly assigned to each bottle design. The number of bottles sold in 24 hours at each supermarket is recorded. The data obtained are displayed in the following table. Bottle Design Study Data A B C 19 32 26 14 31 21 19 31 2? 18 29 23 13 32 23 The Excel output of a one-way ANOVA of the Bottle Design Study Data is shown below. SUMMARY Groups Count Sum Average Variance Design A 5 83 16.6 8.3 Design B 5 155 31.0 1.5 Design C 5 120 24.0 6.0 ANOVA Source of. Variation SS df Ms 1' PValue F orit Between Groups 513.5333 2 259.266".II 49.23 1.64EOE 3.38529 Within Groups 63.2 12.0 5.2630 Total 531.7333 14 (a) Test the null hypothesis that #11. mg, and ye are equal by setting a = .05. Based on this test, can we conclude that bottle designs A. B. and C have different effects on mean daily sales? (Round your answers to 2 decimal places} 9 Answer is not complete. p~va|ue = 0.000 @ Reject 0 HO: bottle design does 0 have an impact on sales. (b) Consider the pairwise differences ,uB M, #c ,uA , and m;- ,uB. Find a point estimate of and a Tukey simultaneous 95 percent confidence interval for each pairwise difference. Interpret the results in practical terms. Which bottle design maximizes mean daily sales? (Round your answers to 2 decimal places. Negative amounts should be Indicated by a minus sign.) 9 Answer is not complete. Point estimate Condence Interval pB pA: . [ . ] p0 pA: . [ . 1 p0 -pB: . [ - 1 0 Answer is complete and correct. Bottle design B o maximlzes sales. (c) Find a 95 percent condence interval for each of the treatment means #5113, and pic. Interpret these intervals. [Round your answers to 2 decimal places. Negative amounts should be indicated by a minus sign.) 9 Answer is not complete. Condence Interval HA: [ pB: [ . p0: [ . A telemarketing firm has studied the effects of two factors on the response to its television advertisements. The rst factor is the time ofdayt at which the ad is run, while the second is the position of the ad within the hour. The data in the following table. which were obtained by using a completely randomized experimental design. give the number of calls placed to an 800 number following a sample broad cast of the advertisement. If we use Excel to analyze these data. we obtain the output in the table below. The Telemarketing Date and the Excel Output of e TwoIla)! ANDVA Position of Advertisement Time of Day On the Hour On the HalfHour Early in Program Late in Program 10:00 morning 45 35 60 53 35 41 6'? 46 39 40 63 49 4:00 afternoon 61 53 3? 66 59 64 31 61 5? 54 32 64 9:00 evening 104 96 130 105 93 95 124 101 104 103 123 110 ANDVA: Two-Factor with Replication Summary Hour HalfHour Early Late Total Morning Count 3 3 3 3 12 Sun 119 116 190 143 5'13 Average 39.6? 33.67 63.33 49.33 47.15 Variance 25.33 10.33 12.33 12.33 113.20 Afternoon Count 3 3 3 3 12 Sum 1'1"? 176 250 191 3'94 Average 59.00 53.6? 33.33 63.6]II 66.17 Variance 4.00 25.33 10.33 6.33 119.19 Evening Count 3 3 3 3 12 Sum 306 294 332 316 1,293 Average 102.00 93.00 121.33 105.33 103.1\":r Variance 12.00 19.00 9.33 20.33 151.9? Total Count 9 9 9 9 Sum 602 536 322 655 Average 66.39 65.11 91.33 12.73 . I69? .11 Variance 773.36 312.00 644.44 AND\" Source of Variation SS df HS F P'Value F crit Sample 23,013.39 2 11,506.69 826.33 .0001] 3.403 Columns 3,333.64 3 1,294.55 93.02 .0000 3.009 Interaction 11.94 6 11.99 .36 .5367 2.503 Error 334.00 24 13.911I Total 27.302. 9? 35 150 M orning 100 w lift 2 rnoon 5|] \"fr/HE Evening Hour Half Early Late Hou' {h} Test the signicance of time of clayilI effects with a = .05. 9 Answer Is complete and correct. F = 325.33, p-value = less than .05; reject 0 H0: time of day is important (c) Test the significance of position of advertisement effects with a = .05. Answer is complete and correct. F = 93.02, p-value = less than .001; reject |HO: position of the ads is important (d) Make pairwise comparisons of the morning, afternoon, and evening times by using Tukey simultaneous 95 percent confidence intervals. (Round your answers to 2 decimal places. Negative amounts should be indicated by a minus sign.) Tukey q.05 = 3.53, MSE = 13.917 UM - HA UM - HE HA - HE (e) Make pairwise comparisons of the four ad positions by using Tukey simultaneous 95 percent confidence intervals. (Round your answers to 2 decimal places. Negative amounts should be indicated by a minus sign.) H1 - H2 H1 - H3 H1 - H4 H2 - H3 H2 - 14 13 - 14 (f) Which time of day and advertisement position maximizes consumer response? Compute a 95 percent (individual) confidence interval for the mean number of calls placed for this time of day/ad position combination. (Round your answers to 2 decimal places.) x Answer is not complete. 9:00 evening early in the program Confidence interval =A study compared three display panels used by air traffic controllers. Each display panel was tested for four different simulated emergency conditions. Twenty-four highly trained air traffic controllers were used in the study. Two controllers were randomly assigned to each display panel-emergency condition combination. The time (in seconds) required to stabilize the emergency condition was recorded. The following table gives the resulting data and the JMP output of a two-way ANOVA of the data. Emergency Condition Display Panel 1 2 3 4 A 17 26 31 14 14 24 34 13 B 15 22 28 9 12 19 31 10 21 29 32 15 24 28 37 19 Least Squares Means Estimates Panel Estimate Condition Estimate A 21 . 625000 17. 166670 B 18 . 250000 24 . 666670 25 . 625000 32. 166670 13.333300 Analysis of Variance Sum of Mean Source DF Squares Square F Ratio Mode] 11 1, 488.333 135.303 31. 8360 Error 12 51 . 0000 4. 250 Prob > F C. Total 23 1, 539.3333 <. effect tests sum of source nparm df squares f ratio prob> F Panel 2 2 218.0833 25 . 6569 <. condition panel .2500 tukey hsd all pairwise comparisons quantile="2.66776," adjusted df="12.0," adjustment="Tukey" difference std error t ratio prob> / t Lower 953 Upper 95 B 3. 37500 1. 030776 3.27 0 . 0169 0 . 6251 6. 12486 -4. 00000 1. 030776 -3. 88 0 . 0057* -6. 7499 -1. 25014 -7. 37500 1. 030776 -7.15 lt Lower 95% Upper 95% -7. 5000 1.190238 -6.30 0 . 0002* -11. 0336 -3.9664 W N -15 . 0000 1. 190238 -12 . 60 . 0001* -18.5336 -11 . 4664 3.8333 1. 190238 3.22 0. 0323 0 . 2997 7.3669 -7.5000 1. 190238 -6. 30 0 . 0002* -11. 0336 -3. 9664 WNNH 11.3333 1. 190238 9.52 .0001 7. 7997 14 . 8669 18.8333 1. 190238 15 . 82