Applied Statistics and Probability for Engineers
Applied Statistics and Probability for Engineers
6th Edition
ISBN: 9781118539712
Author: Douglas C. Montgomery
Publisher: WILEY
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Chapter 13, Problem 52SE

a.

To determine

Find the number of levels of factor in the experiment.

a.

Expert Solution
Check Mark

Answer to Problem 52SE

There are 5 levels of factor in the experiment.

Explanation of Solution

Given info:

The data represents the ANOVA table with missing values.

Calculation:

The levels of factors can be obtained as follows:

levels of factor=df(Factor)+1

And, df(Factor)=df(Total)df(Error)

Thus,

df(Factor)=1915=4

Thus,

levels of factor=4+1=5

Thus, there are 5 levels of factor in the experiment.

b.

To determine

Find the numbers of replicates were used in the experiment.

b.

Expert Solution
Check Mark

Answer to Problem 52SE

There are four replicates of each levels are used.

Explanation of Solution

Justification:

From the ANOVA table, it is clear that total degree of freedom is 19. That is the total number of trials is 20(=19+1). From part (a) it is clear that there are 5 levels for each factor. Therefore, there are 4 (=205) replicates of each levels are used.

c.

To determine

Fill the missing information in the ANOVA table. Use Bound for the P-value.

c.

Expert Solution
Check Mark

Answer to Problem 52SE

The complete ANOVA table is given below:

SourceDFSSMSFP-value
Factor4158.739.6753.5330.025<P-value <0.05
Error15167.511.167
Total19326.2
S=3.342RSq=0.4865RSq(adj)=34.96%

Explanation of Solution

Calculation:

From the ANOVA table, df(Error)=15and MSE=167.5

From part (a), df(Factor)=4.

MSE=SSE(a1)=167.515=11.1667

SS(Factor)=SS(Total)SS(Error)=326.2167.5=158.7

MSFactor=SSFactor(a1)=158.74=39.675

F0=MSFactorMSE=39.67511.167=3.553

Rsq=1SSESSTotal=1167.5326.2=10.5135=0.4865

From Appendix table VI, the value F statistic with (4,15) df is 3.06 and corresponding P-value is 0.05. The value F statistic with (4,15) df is 3.8 and corresponding P-value is 0.025. Here, F0=3.553. That is, 3.06<F0 < 3.8. Therefore, 0.025<P-value <0.05.

d.

To determine

Find the conclusion can you draw about difference in the factor-level means if α=0.05. Also find the conclusion if α=0.01.

d.

Expert Solution
Check Mark

Answer to Problem 52SE

There is significant difference in the factor-level means if α=0.05.

There are no significant difference in the factor-level means if α=0.01.

Explanation of Solution

Calculation:

State the hypotheses:

Null hypothesis:

H0:Thereisnodifferenceinfactorlevel

Alternative hypothesis:

H1:Thereisdifferenceinfactorlevel

The level of significance is 0.05. From part (b), it is clear that 0.025<P-value <0.05.

Decision:

If P-valueα, reject the null hypothesis H0.

If P-value>α, fail to reject the null hypothesis H0.

Conclusion:

Here, the P-value is less than the level of significance.

That is, P-value<α(=0.05).

Therefore, the null hypothesis is rejected.

Thus, there is sufficient evidence to conclude that there is significant difference in the factor-level means at α=0.05 level of significance.

Here, 0.025<P-value <0.05 and α=0.01. That is P-value>α. Therefore the null hypothesis is failed to reject. Hence, there is sufficient evidence to conclude that there is no significant difference in the factor-level means at α=0.01 level of significance.

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Chapter 13 Solutions

Applied Statistics and Probability for Engineers

Ch. 13.2 - Prob. 11ECh. 13.2 - Prob. 12ECh. 13.2 - Prob. 13ECh. 13.2 - Prob. 14ECh. 13.2 - Prob. 15ECh. 13.2 - Prob. 16ECh. 13.2 - 13-17 An experiment was run to determine whether...Ch. 13.2 - Prob. 18ECh. 13.2 - Prob. 19ECh. 13.2 - Prob. 20ECh. 13.2 - Prob. 21ECh. 13.2 - Prob. 22ECh. 13.2 - Prob. 23ECh. 13.2 - For each of the following exercises, use the...Ch. 13.2 - Prob. 25ECh. 13.2 - Prob. 26ECh. 13.2 - Prob. 27ECh. 13.2 - Prob. 28ECh. 13.2 - Prob. 29ECh. 13.2 - 13-30. Suppose that four normal populations have...Ch. 13.2 - Prob. 31ECh. 13.2 - Prob. 32ECh. 13.2 - Prob. 33ECh. 13.3 - 13-34. An article in the Journal of the...Ch. 13.3 - Prob. 35ECh. 13.3 - Prob. 36ECh. 13.3 - 13-37. An article in the Journal of Quality...Ch. 13.3 - 13-38. Consider the vapor-deposition experiment...Ch. 13.3 - Prob. 39ECh. 13.3 - 13-40. Reconsider Exercise 13-8 in which the...Ch. 13.3 - Prob. 41ECh. 13.4 - 13-42. Consider the following computer output from...Ch. 13.4 - 13-43. Consider the following computer output from...Ch. 13.4 - Prob. 44ECh. 13.4 - 13-45. Reconsider the experiment of Exercise 13-5....Ch. 13.4 - 13-46. An article in Quality Engineering...Ch. 13.4 - 13-47. In “The Effect of Nozzle Design on the...Ch. 13.4 - 13-48. In Design and Analysis of Experiments, 8th...Ch. 13.4 - Prob. 49ECh. 13.4 - 13-50. An article in the Food Technology Journal...Ch. 13.4 - 13-51. An experiment was conducted to investigate...Ch. 13 - Prob. 52SECh. 13 - 13-53. Consider the following computer...Ch. 13 - 13-54. An article in Lubrication Engineering...Ch. 13 - 13-55. An article in the IEEE Transactions on...Ch. 13 - Prob. 56SECh. 13 - Prob. 57SECh. 13 - Prob. 58SECh. 13 - Prob. 59SECh. 13 - Prob. 60SECh. 13 - Prob. 61SECh. 13 - Prob. 62SECh. 13 - Prob. 63SECh. 13 - Prob. 64SECh. 13 - Prob. 65SECh. 13 - Prob. 66SECh. 13 - Prob. 67SECh. 13 - 13-68. Consider testing the equality of the means...Ch. 13 - Prob. 69SECh. 13 - Prob. 70SECh. 13 - 13-72. Consider the single-factor completely...Ch. 13 - 13-73. Consider the single-factor completely...Ch. 13 - Prob. 74SECh. 13 - Prob. 75SECh. 13 - Prob. 76SECh. 13 - Prob. 77SE
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