Question

Suppose the salinity of water samples taken from 3 new and separate sites in the Bimini Lagoon was measured with the results given below:

Site 1: 45.44 45.46 50.81 46.00 42.87 40.47 36.81 45.48 45.75 35.27 45.68
Site 2: 62.53 62.54 67.62 63.06 60.08 57.81 54.35 62.56 62.82 52.89 62.75 61.89
Site 3: 53.00 53.01 57.43 53.46 50.87 48.89 45.87

Basic Statistics for the 3 samples:

Level	N	Mean	SD
1	11	43.64	4.50
2	12	60.91	4.08
3	7	51.79	3.69

One-way ANOVA output:

Source	DF	SS	MS	F	P
Factor	2	1715.69	857.85	49.56	0.0000
Error	27	467.31	17.31
Total	29	2183

Using the data above, construct simultaneous (1−α*)100% confidence intervals for the pairwise differences in means using the Bonferroni procedure. Use an overall experiment-wide α = 0.05.

α^*: 0.0167

critical t-value has a student's t distribution with 27 degrees of freedom

What is the critical t-value? You can use the invcdf command in Minitab or the qt() function in R.  Is it possible to calculate this by hand? If so, how? [2 pt(s)]

Calculate the confidence interval for the difference in means for Levels 1 and 2, i.e., μ₁−μ₂. Lower bound:  Upper bound: [3 pt(s)]

What is your conclusion about the relative size of the means for Levels 1 and 2?
μ₁ > μ₂
μ₁ < μ₂
μ₁ = μ₂
μ₁ ≠ μ₂
[3 pt(s)]

Calculate the confidence interval for the difference in means for Levels 1 and 3, i.e., μ₁−μ₃. Lower bound:  Upper bound: [3 pt(s)]

What is your conclusion about the relative size of the means for Levels 1 and 3?
μ₁ ≥ μ₃
μ₁ ≠ μ₃
μ₁ < μ₃
μ₁ = μ₃
[3 pt(s)]

What is your conclusion about the relative size of the means for Levels 2 and 3? Calculate the confidence interval for the difference in means for Levels 2 and 3, i.e., μ₂−μ₃. Lower bound:  Upper bound: [3 pt(s)]

μ₂ = μ₃
μ₂ ≥ μ₃
μ₂ ≤ μ₃
μ₂ ≠ μ₃
[3 pt(s)]

Answer 1

Critical value t = 2.0518

The confidence interval for the difference in means for Levels 1 and 2, i.e., μ₁−μ₂.

Margin of error = t*standard error = 2.0518*sqrt(17.31*(1/11+1/12)) = 3.5634

Lower bound = (43.64 - 60.91) - 3.5634 = -20.8334

Upper bound = (43.64 - 60.91) + 3.5634 = -13.7066

Conclusion : µ1≠ µ2

The confidence interval for the difference in means for Levels 1 and 2, i.e., μ₁−μ₃.

Margin of error = t*standard error = 2.0518*sqrt(17.31*(1/11+1/7)) = 4.1274

Lower bound = (43.64 - 51.79) - 4.1274 = -12.2774

Upper bound = (43.64 - 51.79) + 4.1274 = -4.0226

Conclusion : µ1≠ µ3

The confidence interval for the difference in means for Levels 1 and 2, i.e., μ₂−μ₃.

Margin of error = t*standard error = 2.0518*sqrt(17.31*(1/12+1/7)) = 4.05995

Lower bound = (43.64 - 51.79) - 4.05995 = -12.20995

Upper bound = (43.64 - 51.79) + 4.05995 = -4.09005

Conclusion : µ2 ≠ µ3