Chapter 9.2, Problem 16E

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 four sample means given in Exercise 14. 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 5% level, while the Tukey–Kramer method does not find any pair of means to differ at the 5% level.

14.    In an experiment to determine the effect of curing time on the compressive strength of a certain type of concrete, the mean strengths, in MPa, for specimens cured for each of four curing times were ${\bar{X}}_1=1316$, ${\bar{X}}_2=1326$, ${\bar{X}}_3=1375$, and ${\bar{X}}_4=1389$. Assume that four specimens were cured for each curing time.

1. a. If MSE = 875.2, compute the value of the F statistic for testing the null hypothesis that all four curing times have the same mean strength. Can this null hypothesis be rejected at the 5% level?
2. b. Use the Tukey–Kramer method to determine which pairs of curing times, if any, may be concluded to differ at the 5% level.