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
	9	8	9
	10	9	11
	11	8	9
	8	7	11
	8	10	8
	9	7	10
	10	9	9
	9	8	11

a. Use  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= .05

Compute the LSD critical value (to 2 decimals).

Answer 1

Applying one way ANOVA: (use excel: data: data analysis: one way ANOVA: select Array):

Source	SS	df	MS	F	P value
Between	9.3333	2	4.6667	4.0000	0.034
Within	24.5000	21	1.1667
Total	33.8333	23

a)

sum of sq;treatment=	9.33
sum of sq; error=	24.50
mean sq;treatment=	4.67
mean square; error=	1.17
test statistic =	4.00
p value is less than 0.01
Conclude the treatment mean for the three groups are not all the same

b)

critical value of t with 0.05 level and N-k=21 degree of freedom=	tN-k=	2.080
Fisher's (LSD) for group i and j =(tN-k)*(sp*√(1/ni+1/nj)   =		1.12

Difference	Absolute Value	Conclusion
x1-x2	1.00	not significant difference