6.84931885 7.85610737 9.67811319 7.7138466 8.94396849 10.4895467 10.9949728 8.50

Question

6.84931885 7.85610737 9.67811319 7.7138466 8.94396849 10.4895467 10.9949728 8.50805805 5.23046643 9.32409728 14.805395 10.8211839 11.2052439 11.9030734 5.68473576 10.9334977 11.0026009 7.95343877 10.0345261 10.6271736 9.21389485 7.87865537 8.06495068 9.01315927 8.17765055 13.9121338 12.138093 13.3387009 11.4062741 10.4782073 10.5057236 6.02776722 7.75310666 12.403886 11.4525798 11.9454416 8.91742015 11.8473953 7.69900528 14.7635879 12.8480554 5.55169289 12.5883694 10.4580738 11.5472542 9.70739245 10.6282493 10.2210482 9.93682782 9.85298659 2.57278884 7.62805861 3.92620491 11.1267535 12.6158516 10.2896294 8.19059468 10.4758448 6.67809774 9.74910124 14.949429 11.1027975 8.64866616 8.72189699 11.9767542 13.1027465 12.6763895 9.11339224 14.2746476 11.9615832 11.9952367 9.22691381 12.3098768 6.80278037 6.35045262 11.5120709 9.86602605 8.14695656 5.49014405 15.2923399 7.30823233 9.94771606 11.3343047 10.2668534 11.4404927 10.39375 7.81410181 8.69926443 2.43339317 12.7587587 8.84909595 14.0725253 12.9560984 9.65611852 12.3036303 9.60792312 14.25966 9.57979783 15.9332836 9.57698988 P1. Choco-Bars Quality Management A chocolate factory has weighed 5 small chocolate bars coming out of production over 20 working days to obtain a dataset with 100 observations, that represent the bars weights in grams. The dataset is named Choco-bars.xlsx Suppose the process has a known mean of 10 and a standard deviation of 3 when it is in control a) Group the data by day into twenty samples (as they appear right now, each column represents a day) and find the sample average and for each sample. Also compute the average of the entire sample. How many standard deviations away from the in-control mean are each of those averages? b) Construct the control limits your tolerance for a false "out of control" detection is -196. Would you declare the process under control based on this sample?

Explanation / Answer

Part (a) Let: Average bar weight = D ; average bar weight for day 1 = D1,for day 2 = D2....day20 = D20 Average bar weight = D ; average bar weight for day 1 = D1,for day 2 = D2....day20 = D20 Number of chocolate bars weighed per day = n ; (where n = 5 for all 20 days) Weight of n chocolate bars per day = x1 to xn = x1, x2, x3, x4, x5 N = n * total number of days = 5*20 = 100 Now, sample averages are D1 = (x1 + x2 + x3 + x4 + x5)/n = (6.84931885+9.21389485+12.8480554+14.949429+7.30823233) / 5 = 10.2337861 Similarly, D2 = 8.467394 ; D3 = 10.06288 ; D4 = 9.234766 ; D5 = 10.41722 ; D6 = 11.52111 ; D7 = 10.85036 ; D8 = 9.976093 ; D9 = 8.656322 ; D10 = 10.87513 ; D11 = 9.745648 ; D12 = 9.55529 ; D13 = 9.630106 ; D14 = 10.37852 ; D15 = 9.68145 ; D16 = 10.85771 ; D17 = 10.44726 ; D18 = 9.600687 ; D19 = 9.167011 ; D20 = 12.00184 Average of entire sample = (D1+D2+D3+....+D20)/20 = (10.2337861+8.467394+10.06288+.....+12.00184) / 20 = 10.0680296 Given that In control mean = 10, Standard deviation away from In control mean for day1 = 10 - D1 = 10 - 10.2337861 = -0.2337861 Standard deviation away from In control mean for day2 = 10 - D2 = 10 - 8.467394 = 1.532606 Standard deviation away from In control mean for day20 = 10 - D20 = 10 - 12.00184 = -2.00184 Part (b) We have now calculated the averages means of weights of chocolate bars for 20 days. To calculate the limits for the sample, we need maximum and minimum of average means for 20 days. Upper limit = D20 = 12.00184 and Lower limit = D2 = 8.467394 Range of sample data = (8.467394,12.00184) As per question, mean of 10 and standard deviation of 3 is considered In control, so Upper limit = 10+3 = 13 and Lower limit = 10-3 = 7 Range for In control = (7,13) Since, the range of sample data (8.467394,12.00184) is a subset of In control range (7,13), our sample data of 20days is considered to be under control.

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