Question

11.37 The quality control director for a clothing manufac- turer wants to study the effect of operators and machines on the breaking strength (in pounds) of wool serge material A batch of the material is cut into square-yard pieces, and these pieces are randomly assigned, 3 each. to each of the 12 combinations of 4 operators and 3 machines chosen spe- cifically for the experiment. The results. stored in Breakstw are as follows: Machine Operator 109 110 107 110 1 13 1 14 I 03 102 10S 108 115 115 119 108 1 14 105 102 106 10O 103 101 105 107 107 1 14 1 14 109 1 10 106 112 1 15 110 At the 0.05 level of significance, a. is there an interaction between operator and machine? b. is there an effect due to operator? c. is there an effect due to machine? d. Plot the mean breaking strength for each operator for each machine ferences among operators and among machines. machines on breaking strength? Explain Compare your results to those of (c). (e). and e. If appropriate, use the Tukey procedure to examine dif f. What can you conclude about the effects of operators and g. Repeat the analysis, using machines as the only factor.

Answer 1

Answer:

a)

Check whether or not there is an interaction between operator and machine. The hypotheses are given below Null hypothesis Ho: There is no interaction between operator and machine Alternative hypothesis: H: There is an interaction between operator and machine.

From the EXCEL (PHStat in Add-Ins) output in part (a), the p-value is 0.0083 Conclusion Use the significance level α-0.05 Therefore, by the condition, if p-value < α , then reject the null hypothesis H0 Here, the p-value is 0.0083 <0.05 Therefore, the null hypothesis (Ho) is rejected. Hence, it can be concluded that there is interaction between operator and machine.

b)

Check whether or not there is an effect due to operator The hypotheses are given below Null hypothesis That is, there is no effect due to operator. Alternative hypothesis: 4: Not all μ, are eqaul That is, there is an effect due to operator.

Use EXCEL (PHStat in Add-Ins) to analyze the given data EXCEL (PHStat in Add-Ins) procedure: Step 1: Choose PH Stat > Multiple-Sample Tests Two-Way ANOVA with replication Step 2: In Data, enter 0.05 as Level of Significance and select the column of Group Data Cell Range as A1:D13 Step 3: Click OK

EXCEL (PHStat in Add-Ins) output: Two Way ANOVA with replicatlon Machine 1 Machne2 Mahine3 Total SUMLAY 3 3 3 Sum 1008 112 15T500 116 3333333 111 1086666667 23333 Variance 3 3 Sum 115 103333 112 333 110 55556 Variance 13333 260TR 3 3 3110 Sum Variance 2333 23333 120000 3 319 3 Sum Variance

12 12 126: 12 Sum Variance 317 ANONA Faft 3 1003704 2 1669M 434:1 Sample Columns hteracion Within 6 38113 Toral Level of signficonoe From the EXCEL (PHStat in Add-Ins) output, the p-value is 0,0000

Conclusion Use the significance level α-0.05 Therefore, by the condition, if p-value < α , then reject the null hypothesis Ho Here, the p-value is 0.0000 <0.05 Therefore, the null hypothesis (Ho) is rejected. Hence, it can be concluded that there is an effect due to operator.

c)

Check whether or not there is an effect due to machine. The hypotheses are given below Null hypothesis That is, there is no effect due to machine. Alternative hypothesis: Hi : Not all μ, are eqaul That is, there is an effect due to machine.

From the EXCEL (PhStat in Add-ins) output in part (a), the p-value is 「0.00001 Conclusion Use the significance level α = 0.05 Therefore, by the condition, if p-value < α , then reject the null hypothesis Ho Here, the p-value is 0.0000 <0.05 Therefore, the null hypothesis (Ho) is rejected. Hence, it can be concluded that there is an effect due to machine.

(cl) Construct the graph of average breaking strength for each operator for each machine by using EXCEL (PHStat in Add-Ins). EXCEL (PHStat in Add-Ins) procedure Step 1: Choose PH Stat > Multiple-Sample Tests Two-Way ANOVA with replication Step 2: In Data, enter 0.05 as Level of Significance and select the column of Group Data Cell Range as A1:D13 Step 3: Select Cell Means Plot Step 4: Click OK

EXCEL (PHStat in Add-Ins) output: Cell Means Plot 120 115 110 105 100 95 90 Machine 1 Machine 2 Machine 3

From part (a), the null hypothesis is rejected. Hence, the Tukey-Kramer procedure is appropriate to determine which operators significantly differ with respect to mean breaking strength Find the value of Q-statistic From Table A.9 "CRITICAL VALUES OF THE STUDENTIZED RANGE Q", the value of Q-statistic 30 denominator degrees of freedom is 3.84 freedom is 3.79 3.84+3.79 -3.81 Determine which operators significantly differ with respect to mean breaking strength by using EXCEL (PHStat in Add-ns) EXCEL (PHStat in Add-Ins) procedure Step 1: Choose PH Stat > Multiple-Sample Tests> One-Way ANOVA. Step 2: In Data, enter the value 0.05 in Level of Significance and select the column of Group Data Cell Range as A1:D10 Step 3: Select Tukey-Kramer Procedure.

Step 4: Click OK Step 5: Enter Q-Statistic value as 3.81 EXCEL (PHStat in Add-Ins) output: Stze 1EA 2: B 3: C 4: D Go1b Gop2 1414 132600981 5024 Hka sae te L0M Levald 3.8 Statistic Conclusion From the output, it can be concluded that the operator"A" is significantly higher in mean breaking strength when compared to the other operators

From part (C), the null hypothesis is rejected. Hence, the Tukey-Kramer procedure is appropriate to determine which machines significantly differ with respect to mean breaking strength Find the value of Q-statistic From Table A.9 "CRITICAL VALUES OF THE STUDENTIZED RANGE Q", the value of Q-statistic 30 denominator degrees of freedom is 3.49 freedom is 3.44 Thus, the value of Q-statistic with 3 numerator degrees of freedom and 33 denominator degrees of freedom is 3.49+3.44-3.465 Determine which operators significantly differ with respect to mean breaking strength by using EXCEL (PHStat in Add-I EXCEL (PHStat in Add-Ins) procedure Step 1: Choose PH Stat > Multiple-Sample Tests> One-Way ANOVA.

Step 2: In Data, enter the value 0.05 in Level of Significance and select the column of Group Data Cell Range as A1:D10 Step 3: Select Tukey-Kramer Procedure Step 4: Click OK. Step 5: Enter Q-Statistic value as 3.465. EXCEL (PHStat in Add-Ins) output: Sre 1: Lacne 1 2 ucine 2 Hdine3 105 Leaof e clf 14 4.8 Q Statistic 3.485 Conclusion From the output, it can be concluded that the machine 1 is significantly higher in mean breaking strength when compared to the other machines.

Conclusion From the results, it is clear that there is an effect due to operator and machine. Moreover, there

mean breaking strength The hypotheses are given below Null hypothesis That is, there is no efflect due to operator in mean breaking strength Alternative hypothesis: 4: Not all μ, are eqaul That is, there is an effect due to operator in mean breaking strength

Use EXCEL (PHStat in Add-Ins) to analyze the given data EXCEL (PHStat in Add-Ins) procedure: Step 1: Choose PH Stat > Multiple-Sample Tests> One-Way ANOVA. Step 2: In Data, enter the value 0.05 in Level of Significance and select the column of Group Data Cell Range as A1;D10 Step 3: Click OK

EXCEL (PHStat in Add-Ins) output: Calaulatīons ANOVAsnge Factor 3:6 Coat Average Variance 112 15 T50 9 1003 9 995 110 5555556 26I2T8 9 939 103333333 120COO R36 105555556 9R8 ANOVA Source of Variation ss Between Groupsm1111 Mhin Grou ps FPyoue Ferit MMS To a 807 555G 5 level ofsfgMficotceOLnS From the EXCEL (PHStat in Add-Ins) output, the p-value is 0.0017

Conclusion Use the significance level α-0.05 Therefore, by the condition, if p-value <a, then reject the null hypothesis Ho Here, the p-value is 0.0017 <0.05 Therefore, the null hypothesis (Ho) is rejected. Hence, it can be concluded that there is an effect due to operator in mean breaking strength. If the operators were randomly selected from the population of operators, then it results in rejection of null hypothesis. Hence, the conclusions are similar.