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	39	32	28
2	43	30	25
3	31	24	19
4	33	20	13
ANOVA: Two-Factor without Replication SUMMARY Count Sum Average Variance Method 1 3 99 33.0000 31.0000 Method 2 3 98 32.6667 86.3333 Method 3 3 74 24.6667 36.3333 Method 4 3 66 22.0000 103.0000 Salesman A 4 146 36.50 30.3333 Salesman B 4 106 26.50 30.3333 Salesman C 4 85 21.25 44.2500 ANOVA Source of Variation SS df MS F P-Value F crit Rows 281.5833 3 93.8611 16.98 .0025 4.7571 Columns 480.1667 2 240.0833 43.43 .0003 5.1433 Error 33.1667 6 5.52778 Total 794.9167 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 = 16.98, p-value = .0025; (Click to select)Do not rejectReject H0: there is (Click to select)no differencea difference 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 = 43.43, p-value = .0003; (Click to select)Do not rejectReject H0: salesman (Click to select)dodo not 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: [, ] *Values above are incorrect*

Answer 1

Solution

Part (a)

Since F (16.98) > Fcrit (4.7571) or equivalently p-value (0.025) < α ( .05),

We Reject H₀ Answer 1

And conclude that there is difference in effects of the sales methods (treatments) on mean weekly sales. Answer 2

Part (b)

Since F ( = 43.43) > Fcrit (= 5.1433) or equivalently, p-value (= .0003) < α (= .05);

We Reject H₀: Answer 3

And conclude salesman do have an effect on sales Answer 4

Part (c)

Tukey simultaneous 95 percent confidence intervals to make pair-wise comparisons of the sales method effects on mean weekly sales.

Back-up Theory

100(1 - α)) Confidence Limits: d(i, j) ± [(q_{α, k, N – k}/√2){σ_cap√(2/n)}],

Where

d(i, j) = difference between the two means under comparison

q_{α, k, N – k} = α% critical value of Studentised Range with k (total number of means under comparison) and degrees of freedom (N - k)

N = total number of observations

σ_cap = SE = √(MSE of ANOVA)

n = number of observations the means under comparison are based on [this must be common for all means]

Now, to work out the limits,

α = 0.05 [implied by given 95% confidence limits]

k = 4

N = 12

n = 3

σ_cap = √5.52778 [from ANOVA Table]

q_{α, k, N – k} = q_{0.054 k,8} = 4.5288 [from Standard Studentised Range Tables]

then,

[(q_{α, k, N – k}/√2){σ_cap√(2/n)}] = 6.1475

Finally, the Confidence Limits

pair (I, j)	Mean i	Mean j	d	±	LL	UL
M1 - M2	33	32.6667	0.3333	6.1475	-5.8142	6.4808
M1 - M3	33	24.6667	8.3333	6.1475	2.1858	14.4808
M1 - M4	33	22	11	6.1475	4.8525	17.1475
M2 - M3	32.6667	24.6667	8	6.1475	1.8525	14.1475
M2 - M4	32.6667	22	10.6667	6.1475	4.5192	16.8142
M3 - M4	24.6667	22	2.6667	6.1475	-3.4808	8.8142

Since only in first and last pairs, the confidence limits hold zero, only these two pairs are not significant. All other pairs are significant. Answer 5

DONE