Statistic analysis of an exporting apple company - Essay Example

Problem Set An apple exporting company is currently retrenching and would like to reduce the number of packers in one of their processing plants from 3 packers to only 2. In finding out the most efficient packers, they conducted a 8 hour study for 6 days based on their speed in packing apples. Below are six study results for the three packers indicating the number of boxes packed in 8 hours. Which packer is best Pat Mark Sheri 157 180 156 148 175 158 150 165 145 140 195 151 160 185 166 167 170 161 Sum of Variation Sum of Squares d.f. Mean P value F Between 2212 2 1106 .000621 12.57 Error 1319 15 87.97 Total 3532 17 The probability of this result, assuming the null hypothesis, is 0.001 PAT: Mean = 154. 95% confidence interval for Mean: 145.5 thru 161.8 Standard Deviation = 9.61 Hi = 167. Low = 140. Median = 154. Average Absolute Deviation from Median = 7.67 MARK: Mean = 178. 95% confidence interval for Mean: 170.2 thru 186.5 Standard Deviation = 10.8 Hi = 195. Low = 165. Median = 178. Average Absolute Deviation from Median = 8.33 SHERI: Mean = 156. 95% confidence interval for Mean: 148.0 thru 164.3 Standard Deviation = 7.41 Hi = 166. Low = 145. Median = 157. Average Absolute Deviation from Median = 5.50 The F ratio 12.57 is larger than the F crit value 3.68. Mark is the best packer followed by Sheri. The difference between him and the other two packers is statistically significant because with this, he is sure to maintain his position as a packer. Problem Set 2 An industrial psychologist is interested in brainstorming among groups as a means of solving complex problems and she decides to manipulate two types of problem "sets" or attitudes. She selects 6 groups of four people to participate in the experiment. Three of the groups are given problem "set" 1 and three of the groups are given problem "set" 2. In addition, however, two of the participants in each group are males and two are females. She measures number of problems solved by each individual after group discussions at the end of each of three sessions (max = 30). Examine all interesting effects, present important data, and consider problems in the analysis. Total Problem "set" 1 G11 Males S1 8 S2 7 Females S3 27 S4 24 G12 Males S5 20 S6 24 Females S7 27 S8 28 G13 Males S9 14 S10 18 Females S11 27 S12 26 Problem "set" 2 G24 Males S13 26 S14 30 Females S15 4 S16 8 G25 Males S17 26 S18 29 Females S19 15 S20 18 G26 Males S21 28 S22 28 Females S23 8 S24 12 1) sH0 : AProblemSet 1 = 2 G/A 1 = 2 = 3 = 4 = 5 = 6 BGender M = F (A)B 1M = 2M = 1F = 2F sHa : Not sH0 2) Between Subjects Hierarchical S2(G3B2/A2) 2-tailed (A): (1,4) = 7.71 (G/A): (4,12) = 3.26 (B): (1,4) = 7.71 (AB): (1,4) = 7.71 (GB/A): (4,12) = 3.26 3) = .05 4) Final Source Table: Source DF Sum of Squares Mean Square F-Value F-crit A Problem Set 1 13.50 13.50 .29 7.71 G/A Groups 4 187.83 46.95 10.25* 3.26 B Gender 1 48.17 48.17 1.36 7.71 AB Problem Set*Gender 1 1204.17 1204.17 34.12* 7.71 (GB/A) 4 141.17 35.29 7.70* 3.26 S(GB/A) 12 55.00 4.58 T 23 1649.83 A Problem Set, B Gender, and AB Problem Set*Gender F values are different from SAS output. Why 1 - First, have to test to determine proper error term to use; Fcrit (4, 12) = 3.26 , = .05 G/A / S(GB/A) = 46.96 / 4.58 = 10.25* so must use G/A to test A. F ratio for A = 13.50 / 46.95 = .29, NS Fcrit (4, 12) = 3.26 , = .05 GB/A / S(GB/A) = 35.29 / 4.58 = 7.71* so must use GB/A to test B and AB F ratio for B = 48.17 / 35.29 = 1.36, NS F ratio for AB = 1204.17 / 35.29 = 7.70* significant! (Didn't really need to do this because the group error terms were significant at .05 and cannot be pooled) Subsequent Tests: LSDAB = 2.78 [2(35.29) / 6] = 9.53 M Female-P1 - M Female-P2 = 26.50 - 10.83 = 15.67* M Male-P1 - M Male-P2 = 15.17 - 27.83 = -12.66* 5) The data indicate there was no significant main effect for Problem Set, F(1,4) = 0.29, MSe = 46.95, or for Gender, F(1,4) = 1.36, MSe = 35.29. However, there was a significant interaction between Problem Set and Gender, F(1,4) = 34.12, MSe = 35.29. As shown in Figure 1, females scored significantly higher on number of problems solved in Problem Set 1, however males scored significantly higher on Problem Set 2. These differences were confirmed by a Fisher's LSD of 9.53. Problem Set 3 An experiment was conducted in a small supermarket on the sales of a single brand of dog food involving three levels of shelf height: knee, waist an eye level. The sales, in hundreds of dollars, for a period of 8 days were as follows: Knee waist eye level level level 77 88 85 82 94 85 86 93 87 78 90 81 81 91 80 86 94 79 77 90 87 81 87 93 Source of Variation Sum of Squares d.f. Mean F Between 399.2 2 199.6 14.52 Error 288.8 21 13.75 Total 688.0 23 The probability of this result, assuming the null hypothesis, is 0.000 Knee Level: Mean = 81.0 95% confidence interval for Mean: 78.27 thru 83.73 Standard Deviation = 3.63 Hi = 86.0 Low = 77.0 Median = 81.0 Average Absolute Deviation from Median = 2.75 Waist Level: Mean = 90.9 95% confidence interval for Mean: 88.15 thru 93.60 Standard Deviation = 2.64 Hi = 94.0 Low = 87.0 Median = 90.5 Average Absolute Deviation from Median = 2.12 Eye Level: Mean = 84.6 95% confidence interval for Mean: 81.90 thru 87.35 Standard Deviation = 4.60 Hi = 93.0 Low = 79.0 Median = 85.0 Average Absolute Deviation from Median = 3.38 This is statistically significant for this indicates that in promoting slow moving dog products, these items will be placed on the waist level shelves. This also applies for goods that need to be sold immediately like old stocks and products approaching expiration dates. Through this, inventory and the First-In-First-Out products will be controlled. Read More