Question

A marketing organization wishes to study the effects of four sales methods on weekly sales of a product. The organization employs a randomized block design in which three salesman use each sales method. The results obtained are given in the following table, along with the Excel output of a randomized block ANOVA of these data. Salesman, j Sales Method, i A B C 1 38 30 25 2 43 28 26 3 31 24 20 4 32 19 17 ANOVA: Two-Factor without Replication SUMMARY Count Sum Average Variance Method 1 3 93 31.0000 43.0000 Method 2 3 97 32.3333 86.3333 Method 3 3 75 25.0000 31.0000 Method 4 3 68 22.6667 66.3333 Salesman A 4 144 36.00 31.3333 Salesman B 4 101 25.25 23.5833 Salesman C 4 88 22.00 18.0000 ANOVA Source of Variation SS df MS F P-Value F crit Rows 194.9167 3 64.9722 16.36 .0027 4.7571 Columns 429.5000 2 214.7500 54.06 .0001 5.1433 Error 23.8333 6 3.97222 Total 648.2500 11 (a) Test the null hypothesis H0 that no differences exist between the effects of the sales methods (treatments) on mean weekly sales. Set α = .05. Can we conclude that the different sales methods have different effects on mean weekly sales? F = 16.36, p-value = .0027; H0: there is in effects of the sales methods (treatments) on mean weekly sales. (b) Test the null hypothesis H0 that no differences exist between the effects of the salesmen (blocks) on mean weekly sales. Set α = .05. Can we conclude that the different salesmen have different effects on mean weekly sales? F = 54.06, p-value = .0001; H0: salesman have an effect on sales (c) Use Tukey simultaneous 95 percent confidence intervals to make pairwise comparisons of the sales method effects on mean weekly sales. Which sales method(s) maximize mean weekly sales? (Round your answers to 2 decimal places. Negative amounts should be indicated by a minus sign.) Method 1 – Method 2: [ , ] Method 1 – Method 3: [ , ] Method 1 – Method 4: [ , ] Method 2 – Method 3: [ , ] Method 2 – Method 4: [ , ] Method 3 – Method 4: [ , ]

Answer 1

MSE=			3.972
df(error)=			6
number of treatments =			4
pooled standard deviation=Sp =√MSE=			1.993

critical q with 0.05 level and k=4, N-k=6 df=			4.90
Tukey's (HSD) =(q/√2)*(sp*√(1/ni+1/nj)         =

			Lower bound	Upper bound	differ
	(xi-xj )	ME	(xi-xj)-ME	(xi-xj)+ME
μ1-μ2	-1.33	5.64	-6.97	4.31	not significant difference
μ1-μ3	6.00	5.64	0.36	11.64	significant difference
μ1-μ4	8.33	5.64	2.69	13.97	significant difference
μ2-μ3	7.33	5.64	1.69	12.97	significant difference
μ2-μ4	9.67	5.64	4.03	15.31	significant difference
μ3-μ4	2.33	5.64	-3.31	7.97	not significant difference