please use minitab A mechanical engineer at a manufacturing plant keeps a close

Question

please use minitab

A mechanical engineer at a manufacturing plant keeps a close watch on the performance and condition of the machines. The following data are the weight losses (n milligrams) of certain machine parts due to friction when used with three different lubricants. (a) Do the data provide sufficient evidence that there is a significant difference in weight loss among the three lubricants? (i) State the null and alternative hypotheses tested with the Anova. ii) State all assumptions. iii) Use the critical value approach at alpha = 0.05. Can you accept/reject the null hypothesis tested. with the Anova? (iv) Use the p-value Do the data provide sufficient evidence that there is a significant difference in average weight loss among the three lubricants? (b) Compute a 95% confidence interval for the average weight loss for the lubricant A (c) Using the Minitab output, find a 95% confidence interval for the difference in lubricants A and B (d) Using Tukey's yardstick measure, is there any pair of lubricants which has significant difference in the means?

Explanation / Answer

Part-a-i

Null hypothesis H0:µA= µB= µC

Alternative hypothesis Ha: Atleast one pair of mean differ significantly.

Part-a-ii

The data for each should be normally distributed and homogeneity of variance assumptions should be satisfied in addition to randomness and independence.

Part-a-iii

Results of ANOVA are as follows from where degree of freedom =(2,21)

So critical F0.05(2,21)=3.467

As calculated F=1.43<3.467, we do not reject the null hypothesis

One-way ANOVA: Lubricant A, Lubricant B, Lubricant C

Method

Null hypothesis         All means are equal

Alternative hypothesis At least one mean is different

Significance level      = 0.05

Equal variances were assumed for the analysis.

Factor Information

Factor Levels Values

Factor       3 Lubricant A, Lubricant B, Lubricant C

Analysis of Variance

Source DF Adj SS Adj MS F-Value P-Value

Factor   2   16.33   8.167     1.43    0.261

Error   21 119.62   5.696

Total   23 135.96

Part-a-iv: As p-value=0.261>0.05, we do not reject the null hypothesis and conclude that there is not a signfiant difference in the average weight loss among the three lubricants.

Part-b

From following results 95% confidence interval for lubricant A mean =(7.120, 10.630)

Means

Factor       N    Mean StDev       95% CI

Lubricant A 8   8.875 2.031 (7.120, 10.630)

Lubricant B 8   8.875 2.475 (7.120, 10.630)

Lubricant C 8 10.625 2.615 (8.870, 12.380)

Part-c

From following results 95% confidence intevral for difference in A and B=B-A=(-3.00, 3.00)

Tukey Simultaneous Tests for Differences of Means

                           Difference       SE of                          Adjusted

Difference of Levels         of Means Difference      95% CI     T-Value   P-Value

Lubricant B - Lubricant A        0.00        1.19 (-3.00, 3.00)     0.00     1.000

Lubricant C - Lubricant A        1.75        1.19 (-1.25, 4.75)     1.47     0.327

Lubricant C - Lubricant B        1.75        1.19 (-1.25, 4.75)     1.47     0.327

Part-d

There is no pair difference as all p-values are >0.05.

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