4:
Here we have 5 different paints so k=5
Total number of data values is n = 4*5 = 20
Degree of freedom of error: df = 20 -5 = 15
Critical value for , df=15 and k=5 is
So Tukey's HSD will be
Following table shows the absolute difference between all possible pairs of mean:
Samples | Samples mean | Absolute Difference | ||
1 | 2 | 462 | 502.8 | 40.8 |
1 | 3 | 462 | 427.5 | 34.5 |
1 | 4 | 462 | 469.3 | 7.3 |
1 | 5 | 462 | 532.1 | 70.1 |
2 | 3 | 502.8 | 427.5 | 75.3 |
2 | 4 | 502.8 | 469.3 | 33.5 |
2 | 5 | 502.8 | 532.1 | 29.3 |
3 | 4 | 427.5 | 469.3 | 41.8 |
3 | 5 | 427.5 | 532.1 | 104.6 |
4 | 5 | 469.3 | 532.1 | 62.8 |
Following table shows the differences that are greater than Tukey's HSD:
Samples | Samples mean | Absolute Difference | ||
1 | 5 | 462 | 532.1 | 70.1 |
2 | 3 | 502.8 | 427.5 | 75.3 |
3 | 5 | 427.5 | 532.1 | 104.6 |
4 | 5 | 469.3 | 532.1 | 62.8 |
Correct options are:
X1 and X5
X2 and X3
X3 and X5
X4 and X5
No. 4 Sprint int . 8:01 PM С 46% An experiment to compare the spreading rates...
10.1 (1) An experiment to compare the spreading rates of five different brands of yellow interior latex paint available in a particular area used 4 gallons (J = 4) of each paint. The sample average spreading rates (ft2/gal) for the five brands were x1. = 462.0, x2. = 512.8, x3. = 427.5,x4. = 469.3, and x5. = 532.1. The computed value of F was found to be significant at level α = 0.05. With MSE = 360.8, use Tukey's procedure...
6. -/13 POINTS DEVORESTATS 10.8.011. MY NOTES An experiment to compare the spreading rates of five different brands of yellow interior latex paint available in a particular area used 4 gallons () - 4) of each paint. The sample average spreading rates (ft/gal) for the five brands were 1. - 462.0, X2-512.8, X3. - 437.5,4 = 469.3, and s. - 532.1. The computed value of F was found to be significant at level a = 0.05. With MSE - 380.8,...
Consider an experiment in which 45 zirconium disks were divided into five groups of 9 each. Then a different contamination/cleaning protocol was used for each group. Use Tukey's procedure on the data below to identify differences in true average bond strengths (MPa) among the five protocols. (Use α = 0.05.) Treatment: 1 2 3 4 5 Sample mean 10.5 14.8 15.7 16.0 21.6 Grand mean = 15.7 Sample sd 4.2 6.8 6.2 6.9 6 MSE = 37.186 Find the value...