Question

The following statistics are calculated by sampling from four normal populations whose variances are equal: (You may find it useful to reference the t table and the q table.)

x⎯⎯1x¯1 = 149, n₁ = 10; x⎯⎯2x¯2 = 154, n₂ = 10; x⎯⎯3x¯3 = 143, n₃ = 10; x⎯⎯4x¯4 = 139, n₄ = 10; MSE = 51.3

a. Use Fisher’s LSD method to determine which population means differ at α = 0.01. (Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places. Round your answers to 2 decimal places.)

b. Use Tukey’s HSD method to determine which population means differ at α = 0.01. (If the exact value for n_T – c is not found in the table, then round down. Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places. Round your answers to 2 decimal places.)

c. Do all population means differ?

• No, only some of the population means differ.

• Yes, all population means differ.

• No, none of the population means differ.

Answer 1

x1-x3	6
x1-x4	10
x2-x3	11
x2-x4	15
x3-x4	4

for 36 degree of freedom and 0.01 level ; critical value of t =2.7195

and for Fisher 'LSD =   =8.7108

therefore confidence interval =(1-2) -/+ LSD

hence

	confidence interval
μ1-μ2	-13.71	3.71		No
μ1-μ3	-2.71	14.71		No
μ1-μ4	1.29	18.71		Yes
μ2-μ3	2.29	19.71		Yes
μ2-μ4	6.29	23.71		Yes
μ3-μ4	-4.71	12.71		No

b) for Tukey's HSD =  =10.7110

	confidence interval
μ1-μ2	-15.71	5.71		No
μ1-μ3	-4.71	16.71		No
μ1-μ4	-0.71	20.71		No
μ2-μ3	0.29	21.71		Yes
μ2-μ4	4.29	25.71		Yes
μ3-μ4	-6.71	14.71		No