Question

In a study conducted to investigate browsing activity by shoppers, each shopper was initially classified as a nonbrowser, light browser, or heavy browser. For each shopper, the study obtained a measure to determine how comfortable the shopper was in a store. Higher scores indicated greater comfort. Suppose the following data were collected.

		Light	Heavy
	Nonbrowser	Browser	Browser
	4	3	10
	5	4	12
	6	3	10
	3	2	12
	3	5	9
	4	2	11
	5	4	10
	4	3	12

a. Use a=.05 to test for a difference among mean comfort scores for the three types of browsers.

Compute the values identified below (to 2 decimals, if necessary).

Sum of Squares, Treatment
Sum of Squares, Error
Mean Squares, Treatment
Mean Squares, Error

Calculate the value of the test statistic (to 2 decimals, if necessary).

b. Use Fisher's LSD procedure to compare the comfort levels of nonbrowsers and light browsers. Use a=5.

Compute the LSD critical value (to 2 decimals).

Answer 1

treatment	Nonbrowser	light	Heavy
count, ni =	8	8	8
mean , x̅ i =	4.250	3.25	10.750
std. dev., si =	1.035	1.035	1.165
sample variances, si^2 =	1.071	1.071	1.357
total sum	34	26	86		146	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =	6.08
square of deviation of sample mean from grand mean,( x̅ - x̅̅)²	3.361	8.028	21.778
					TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	26.889	64.222	174.222		265.3333333
SS(within ) = SSW = Σ(n-1)s² =	7.500	7.500	9.500		24.5000

no. of treatment , k =   3
df between = k-1 =    2
N = Σn =   24
df within = N-k =   21
  
mean square between groups , MSB = SSB/k-1 =    132.6667
  
mean square within groups , MSW = SSW/N-k =    1.1667
  
F-stat = MSB/MSW =    113.7143

====================

a)

Sum of Squares, Treatment=265.33
Sum of Squares, Error=24.50
Mean Squares, Treatment=132.67
Mean Squares, Error=1.17

value of test stat = 113.71

b)

Level of significance=	0.0500
no. of treatments,k=	3
DF error =N-k=	21
MSE=	1.1667
t-critical value,t(α/2,df)=	2.0796

Fishers LSD critical value=tα/2,df √(MSE(1/ni+1/nj)) = 1.12

if absolute difference of means > critical value,means are significnantly different ,otherwise not

population mean difference	critical value			result
µ1-µ2	1	1.12			means are not different