Question

could you do this problem by using R

13.1. An experiment was performed to investigate the capa- bility of a measurement system. Ten parts were randomly selected, and two randomly selected operators measured each part three times. The tests were made in random order, and the data are shown in Table P13.1. TABLE P13.1 Measurement Systems Data for Problem 13.1 Operator 2 Measurements Operator 1 Measurements Part No 2 3 1 2 1 3 50 50 48 1 50 49 51 51 51 2 52 52 51 51 54 52 3 53 50 50 51 48 50 4 49 51 50 51 48 5 49 48 48 49 48 50 50 52 50 6 52 50 50 51 50 7 51 51 51 48 50 53 50 49 8 52 48 49 51 50 50 9 51 47 48 46 46 49 10 47 (a) Analyze the data from this experiment (b) Estimate the variance components using the ANOVA method

Answer 1

Terminal x Console - C:/Users/parekh ku/OneDrive - Merck Sharp & Dohme, Corp/Desktop/ part-rep (1:10 , 6) part [1] 1 [17] 7 [33] 3 4 [49] op c(rep ("OP1",30) , rep("OP2",30)) op > 2 3 4 5 6 9 10 1 2 3 4 5 8 9 10 1 2 4 5 6 7 8 9 10 1 5 6 7 8 9 10 1 2 3 4 5 6 7 8 5 6 7 8 9 10 1 2 3 4. 9 10 "OP1" "OP1' "OP1" "оP1" [1] [9] "OP1" "OP1" "OP1" [17] "OP1" "OP1" "OP1" "oP1" "oP1" "oP1" "oP1" "oP1" [25] "OP1" "OP1" [33] "OP2" "OP2" "OP2" "oP2" "OP2" "oP2" "OP2" "oP2" [41] [49] [571 "OP2" "OP2" "OP2" "OP2" data-c (50 "ОP1" "ОP1" "ОP1" "ОP1" "ОP1" "ОP1" "OP1" "ОP1" "ОP1" "OP1" "OP1" "OP1" "OP1" "OP2" "OP2" "ОP2" "ОP2" "ОP2" "ОP2" "ОP2" "OP2" "OP2" "ОP2" "OP2" "OP2" "OP2" "OP2" "OP2" "ОP2" "OP2" "OP2" 51, 49, 50, 51, 53 50, 48, 50, 52, 52 50, 51. 51, 51. 49 50, 47 50, 48 50, 51 51, 49, 51, 50, 52. 50, 50, 50, 51, 48, 51. 51, 49, 48, 51. 46, 49, 54. 48 48,52, 48, 52, 50, 50,52,50,49, 48) 53, 50, 51,48 50,51 49, 47, 46 length(data) [1] 60 > data [1] [351 50 50 51 51 51 51 50 50 52 50 49 53 48 50 50 51 50 51 48 49 47 46 49 46 47 48 op as.factor (op) part-as.factor (part) datal-cbind (data,op,part) str (datal) num [1:60, 1:3] 50 49 50 50 48 51 52 52 51 51 ... - attr(*, "dimnames")-List of 2 ..$: NULL ..$ : chr [1:3] "data" "op" "part" > datal > 50 49 50 50 48 51 52 52 51 51 51 51 53 50 50 54 52 51 49 51 50 48 50 51 48 49 48 48 49 48 52 50 50 52 data op part [1,] [2, [3,] [4, [5, [6,] [7, [8, [9, [10, [11, [12, [13,] [14,] [15, [16, [17,] [18, [19, [20, 50 1 1 49 1 2 50 1 50 1 48 1 5 51 1 6 52 1 7 52 1 51 1 51 1 51 1 1 51 1 2 53 1 50 1 4 50 1 5 54 1 6 52 1 7 51 1 4' 51 1 10 LO N 00 N N00 0O NCO m st. CO O OH N m Ln oN 00 OO

[20,] [21,] [22, [23, [24, [25, [26, [27, [28, [29,] [30, [31,] [32, [33, [34,] [35, [36, [37, [38, [39, [40, [41, [42, [43, [44,] [45, [46, [47, [48, [49, [50, [51, [52, [53,] [54,] [55, [56, [57, [58, [59, [60,] data1- as. data.frame (datal) >str (datal) data. frame' : data num S op $ part num mode12wayanova-aov (data-op+part,data=data1) 51 1 10 50 1 1 48 1 2 50 1 51 1 48 1 5 49 1 6 48 1 7 48 1 49 1 48 1 10 52 2 1 50 2 2 50 2 52 2 50 2 5 50 2 6 51 2 7 51 2 51 2 2 51 10 50 2 1 50 2 52 2 4. 50 2 49 2 5 53 2 6 7 48 2 2 50 50 2 2 51 50 2 1 51 2 2 48 2 4. 49 2 2 47 5 46 2 6 7 49 2 2 46 8 47 2 9 48 10 60 obs. of 3 variables: 50 49 50 50 48 51 52 52 51 51 : num 1 1 1 1 1 1 1 1 11... 1 2 3 4 5 67 8 9 10. . . HNMmt un LoN HNm ONCO O OHN m N m n o

mode12wayanova-aov (data-op+part,data-datal) summary (model2wayanova) Df Sum Sq Mean sq F value Pr >F) 1 2. 82 2. 817 1.002 0.321 op part Residuals mode12wayanova-aov (data-op*part, data-datal) 1 1. 88 1.879 0.669 0.417 57 160.15 2. 810 summary (model2wayanova) Df Sum Sq Mean sq F value Pr >F) 1 2.82 2. 817 0.996 0. 323 op part op: part Residuals 1 1.88 1.879 0.664 0.418 1 1.76 1.758 0.622 0.434 2. 829 56 158.40

Analyzed data below

mean(data1$data[data1$op==1]) [1] 50.16667 mean(data1$data[data1$op==2]) [1] 49.73333 > #above mean of op1 and 2 are 50.17 and 49.73 respectively sd(data1$data[data1$op==1]) [1] 1.5775 sd(data1$data[data1$op-=2]) [1] 1.760355 > #Same way std deviation above which is also not varying much >

Performed ANOVA test above and since we have got the p value of the test

1st without interaction effect and 2nd after interaction effect in all the cases as more than 0.05 with 95% conf interval we can say that the mean of all the groups are same.

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