Use the data from Exercise 3.3. Compute a 99% confidence interval for the differ

Question

Use the data from Exercise 3.3. Compute a 99% confidence interval for the difference in response between the average of the three treatment groups (acid, pulp, and salt) and the control group.

silage. The observed moisture contents of the silage are shown below (data from Caro et al. 1990) NaCl 80.5 79.3 79.0 79.6 Grand mean 79.3 c acid Control 76.7 77.2 78.6 77.5 Beet pul Formi 89.1 75.7 81.2 82.0 77.8 79.5 77.0 78.1 Means Compute an analysis of variance table for these data and test the null hypoth- esis that all four treatments vield the same average moisture contents.

Explanation / Answer

The average of 3 treatment groups is 79.9

The average of control group is 77.5

So difference is 2.4 (M1-M2)

SM1-M2 is standard error between sample means.= [(2*MSwithin group)/n]^(1/2) = 2.84605

The 99% confidence interval is

[(M1-M2) - (tcal )*SM1-M2,(M1-M2) + (tcal )*SM1-M2] = [ 2.4-(9.924*2.85),2.4+(9.924*2.85)]

=[-25.8834,30.6834]

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