For some data sets, the F statistic will reject the null hypothesis of no in mea
ID: 3231885 • Letter: F
Question
For some data sets, the F statistic will reject the null hypothesis of no in mean yields, but the Tukey-Kramer method will method any pair of means that can be concluded to differ. For the sample means X_1 = 89.88, X_2 = 89.46, X_3 = 87.01, and X_4 = 85.78, assuming a sample size of 4 for each treatment, find a value of MSE so that the 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 1% level. Round intermediate calculations and the final answer to three decimal places, where necessary.Explanation / Answer
The mean of the numbers 89.88, 89.46, 87.01, 85.78 is 88.0325
SSTR = 4 * [(89.88 – 88.0325)^2 + (89.46– 88.0325)^2 + (87.01– 88.0325)^2 + (85.78– 88.0325)^2) = 46.281
Degree of freedom of SSTR = 4-1 = 3
MSTR = SSTR/3 = 46.281/3 = 15.427
N = 4 * 4 = 16 (4 in each 4 groups)
Degree of freedom of SSE = N – r = 16 – 4 = 12
Critical value of F at 99% significance and df=3,12 is 5.952
F = MSTR/MSE
To reject the null hypothesis, F > critical value
F > 5.952
MSTR/MSE > 5.952
15.427/MSE > 5.952
MSE < 15.427/5.952
MSE < 2.592
For Tukey Kramer test, we need to show that the pair of means are not different,
The critical value of test statistic for df = 12 and number of treatments = 4 is 5.5
To show that the pair of means are not different
Difference in means/MSE < 5.5
MSE > Difference in means/5.5
The largest difference in means = 89.88 – 85.78 = 4.1
MSE > 4.1/5.5 = 0.745
So, 0.745 < MSE < 2.592