Question

Use Fisher's LSD procedure to develop a 95% confidence interval estimate of the the difference between the means of Treatments A & B. Interpret the confidence interval.

Treatment Treatment Treatment 32 30 30 26 32 30 6.00 43 36 35 36 40 46 48 45 4.00 Sample mean Sample variance 36 6.50

Write the formula to use: _______

Compute with the formula: ____________

Interval:__________________

Interpretation of the confidence interval: _________________________, ____ Ho. Claim:________________

Answer 1

Solution: For 95% confidence interval, α a/2 0.05/2 0.025 1-0.95-0.05 Formula to compute 95% confidence interval estimate of the difference between the means of Treatments A & B 0.025 Number of treatments k 3 (n-1)s + (n-1)ss+ (n-1 MSE- k(n-1) (5-1) x 6 (5-1) x 4+ (5 1) x 6.50 MSE = 3 x (5 1) MSE -5.50 Degrees of freedom-df-|r-k 15-3 = 12 From t-distribution table, the value of t having an area of (a/2 = 0.025) in its upper tail and for (df = 12) is given by = to.025 2.179 Hence, 95% confidence interval estimate: 15.50-G+I) (30-45) 2.179 -15 2.179 5.50x +

15 t+ 3.232 Lower bound of 95% confidence interval 15-3.232-_18.232 Upper bound of 95% confidence interval-_15 + 3.232 -11.768 Interpretation: we are 95% confident that the difference between the means of Treatments A & B lies between-18.23 2 and-11.768 Since both the lower and upper bounds 95% confidence interval are of same sign, we reject the null hypothesis and hence conclude that there is a significant difference between the means of Treatments A & B