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 253.8 9.7570 263.2 5.4037 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 c. Calculate F с. d. Using the table for F, find the rejection region e. Using alpha = 0.05, state whether you reject the null hypothesis or not.

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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.
Transcribed Image 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.
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