For some data sets, the F statistic will reject the null hypothesis of no differ

Question

For some data sets, the F statistic will reject the null hypothesis of no difference in mean yields, but the Tukey-Kramer, method will not find any pair of means that can be concluded to differ. For the sample means X bar_1 = 89.85, X bar_2 = 89.53, X bar_3 = 86.99, and X bar_4 = 85.78, assuming a sample size of 4 for each treatment, find a value of MSE so that the F statistic rejects the null hypothesis of no difference at the 1% level, while the Tukey-Kramer method does not find any pair of means to differ at the l% level. Round intermediate calculations and the final answer to three decimal places, where necessary.

Explanation / Answer

Fcritical Here for alpha = 0.01 and dFfactors = 4-1 =3 and dF error = (6-1) * 4 = 20

so Fcritical = 4.94

so MSB/ MSE > 4.94 for rejection of F- test of ANOVA

Here Grand average = (89.95 + 89.52 + 87.03 + 85.76)/4 = 88.065

so MSB = 6 [ (89.95 - 88.065)2 + (89.52 - 88.065)2 + (87.03 - 88.065)2 + (85.76 - 88.065)2]

= 72.327

so MSE < MSB/4.94

MSE< 72.327/ 4.94 => MSE < 14.641

this is the first lower limit of MSE

Now we will calculate it for Tukey HSD post Hoc test .

Here maximum mean Mmax= 89.95

minimum mean Mmin= 85.76

Qcritical for alpha = 0.01 and dFerror = 20 and number of treatments = 4

Qcritical = 5.02

so critical Q value must be greater than any Q value for means so we can reject null hypothesis for Tukey post hoc test.

so Qcritical > (Mmax - Mmin)/ sqrt [ MSE/n]

5.02 > (89.95 - 85.76)/ sqrt [ MSE/6]

sqrt [ MSE/6] > 0.8366

MSE > 4.18

so 4.18 < MSE < 14.64

these are the criticical values of MSE.

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