Question

In: Statistics and Probability

A student, asked to give a class demonstration of the use of a confidence interval for...

A student, asked to give a class demonstration of the use of a confidence interval for comparing two
treatment means, proposed to construct a 99 percent confidence interval for the pairwise comparison
D = µ5 −µ3, where there are a total of five factor levels. The student selected this particular comparison
because the estimated treatment means Y¯
5. and Y¯
3. are the largest and smallest, respectively, and stated:
“This confidence interval is particularly useful. If it does not contain zero, it indicates, with significance
level α = .01, that the factor level means are not equal.”
a. Explain why the student’s assertion is not correct.
b. How should the confidence interval be constructed so that the assertion can be made with significance
level α = .01?

Solutions

Expert Solution

A student, asked to give a class demonstration of the use of a confidence interval for comparing two treatment means, proposed to construct a 99 percent confidence interval for the pairwise comparison D = µ5 −µ3, where there are a total of five factor levels. The student selected this particular comparison

because the estimated treatment means Y¯5. and Y¯3. are the largest and smallest, respectively, and stated:

“This confidence interval is particularly useful. If it does not contain zero, it indicates, with significance level α = .01, that the factor level means are not equal.”

a. Explain why the student’s assertion is not correct.

There are 5 groups. The total number of comparisons is 5*4/2 = 10 comparisons. Therefore the

total error rate of 1-(0.99)5 =0.049 and not the 0.01 level.

Therefore our interpretation should be

If it does not contain zero, it indicates, with significance level α = .049, that the factor level means are not equal.”

Therefore the student’s assertion is not correct

b. How should the confidence interval be constructed so that the assertion can be made with significance level α = .01?

There are 5 groups. The total number of comparisons is 5*4/2 = 10 comparisons.

Therefore construct interval with α=0.01/10 = 0.001 level. Then we can say that

factor level means are not equal at 0.01 level of significance.


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