Question

A computer manufacturer has production lines that utilize three different assembly methods. These methods differ in the order in which tasks are performed and the consequent layout of the production lines. QC randomly observes 30 computers assembled under each of the three methods and records the assembly times. The resulting data set is the proj2-1.txt file on BlackBoard. Use ANOVA to test the null hypothesis that all the three methods have the same assembly speed.

Method 1'       'Method 2'      'Method 3'
22.579999923706055      27.959999084472656      23.600000381469727
27.3799991607666        26.739999771118164      19.579999923706055
25.8799991607666        24.5    21.34000015258789
28.18000030517578       28.65999984741211       19.65999984741211
21.200000762939453      23.600000381469727      21.31999969482422
26.600000381469727      21.780000686645508      23.6200008392334
23.979999542236328      24.079999923706055      23.399999618530273
26.6200008392334        25.6200008392334        20.780000686645508
25.360000610351562      25.059999465942383      23.579999923706055
25.18000030517578       26.860000610351562      20.3799991607666
25.040000915527344      24.579999923706055      23.520000457763672
23.420000076293945      23.139999389648438      21.760000228881836
27.360000610351562      25.219999313354492      20.719999313354492
25.799999237060547      26.239999771118164      20.280000686645508
25.059999465942383      25.31999969482422       25.6200008392334
22.799999237060547      23.260000228881836      24.280000686645508
24.299999237060547      23.139999389648438      21.520000457763672
25.739999771118164      27.15999984741211       21.420000076293945
25.059999465942383      24.18000030517578       23.420000076293945
24.34000015258789       24.040000915527344      24.700000762939453
24.799999237060547      22.219999313354492      21.100000381469727
21.18000030517578       28.6200008392334        22.34000015258789
27.139999389648438      27.860000610351562      23.440000534057617
23.260000228881836      25.84000015258789       26.360000610351562
26.559999465942383      25.299999237060547      22.760000228881836
25.360000610351562      24.719999313354492      24.540000915527344
25.760000228881836      25.31999969482422       20.079999923706055
23.520000457763672      24.040000915527344      21.579999923706055
27.760000228881836      21.440000534057617      22.280000686645508
23.3799991607666        27.219999313354492      25.239999771118164

Answer 1

Answer: ---- Date: ----22/5/2019

Excel add-in (Mega Stat) procedure Step 1: In EXCEL, Select Add-Ins> Mega Stat>Descriptive Statistics. Step 2: In Input range, select the cell range Al:C31. Step 3: Check Select Defaults Step 4: Choose Box plot Step 5: Click OK. Excel add-in (Mega Stat) output riptive statistics Method 1 Method 2 Method 3 30 30 30 BoxPlot 20 21 23 24 25 26 27 28 29 Method 1

BoxPlot 20 24 26 28 30 Method 2 BoxPlot 15 17 19 21 23 25 27 Method 3

Null hypothesis: That is, there is no difference in assembly times. Alternative hypothesis: Ha At least one u, is different That is, there is a difference in assembly times Rejection rule: If p-values α (-0.05) . then reject the null hypothesis Ho Excel add-in (Mega Stat) procedure: Step 1: In EXCEL, Select Add-Ins> Mega Stat> Analysis of Variance. Step 2: Choose One-Factor ANOVA. Step 3: In Input Range, enter the column of Al:C31 Step 4: In Post-Hoc Analysis, Choose When p < 0.05. Step 5: Click OK. Excel add-in (Mega Stat) output:

One factor ANOVA Mean Std. Dev 25.020 25.124 1.8200 Method 1 30 30 1.9350 Method 2 30 90 22.474 1.8370 Method 3 24.206 2.2175 Total ANOVA table Source df MS value Treatment 135.1543 302.5036 67.57716 19.44 1.05E-07 3.47705 87 89 Error 437.6580 Total Post hoc analysis p-values for pairwise t-tests Method 3 22.474 Method 1 Method 2 25,020 25.124 Method 3 22.474 Method 1 25.020 9.09E-07 Method 2 25.124 3.69E-07 8295 Thus the p-value is 0 Conclusion

Use a significance level, α 0.05 Here, p-Value is 0, which is lesser than the value of α-0.05- That is, p-value(-0)<a(= 0.05) Therefore, by the rejection rule, reiect the null hvpothesis Ho Interpretation There is sufficient evidence at level of significance α-: 0.05 That is, there is a difference in assembly times. From Post hoc result, method 1 and 3 are faster. Method 2 and 3 are faster.