Find an individual 95 percent confidence interval for where _C = 51.5, _B = 55.8,and MSE = 6.125.There were 4 treatments and 24 observations total and the number of observations were equal in each group. A)(-7.25 -1.35) B)(-6.41 -2.19) C)(-5.35 -3.25) D)(-7.28 -1.32)

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Quizplus · Accepted Answer

We can use the formula for the confidence interval of the difference between two means: CI = (xbar1 - xbar2) ± t(alpha/2, df) * sqrt(MSE(1/n1 + 1/n2)) where: - xbar1 and xbar2 are the sample means of the two groups being compared - t(alpha/2, df) is the t-score from the t-distribution with df degrees of freedom and an alpha level of (1-confidence level) - MSE is the mean square error from the ANOVA table - n1 and n2 are the sample sizes of the two groups being compared In this case, we are interested in the difference between the means of _C and _B. Therefore, xbar1 = 51.5 and xbar2 = 55.8. The degrees of freedom for the MSE is (n-k), where n is the total number of observations and k is the number of treatments. Therefore, df = 20. Using a 95% confidence level, alpha/2 = 0.025 and t(0.025, 20) = 2.086. Plugging in the values, we get: CI = (51.5 - 55.8) ± 2.086 * sqrt(6.125 * (1/12 + 1/12)) CI = -4.3 ± 2.086 * 0.8102 CI = -7.28 to -1.32 Therefore, the correct choice is D.

Find an Individual 95 Percent Confidence Interval for Where