1. Consider the golf ball example again using a completely randomized deign with a single factor (see section 10.2). However, this time we only hit each golf ball 5 times instead of 10 times. The golf...

Extracted text: 1. Consider the golf ball example again using a completely randomized deign with a single factor (see section 10.2). However, this time we only hit each golf ball 5 times instead of 10 times. The golf balls are brands A, B, C, D. So, there are 4 brands, (4 treatments), and 20 sample distances. That is, k= 4 and n = 20. The data below gives the average distance means and standard deviations for the sample distances. Brand Sample Mean Sample Standard Deviations A 253.8 9.7570 263.2 5.4037 C 271.0 8.7178 262.0 7.4498 The average mean distance for all the 4 samples is 262.5 (this is x-bar). SST = 743.4 and SSE = 1023.6 Test the hypothesis that all the means are equal (this is the null hypothesis) against the alternative hypothesis that at least two means are different by doing the following: a. Calculate MST b. Calculate MSE С. Calculate F d. Using the table for F, find the rejection region e. Using alpha = 0.05, state whether you reject the null hypothesis or not.