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
	6	7	9
	7	8	11
	8	7	9
	5	6	11
	5	9	8
	6	6	10
	7	8	9
	6	7	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). The p-value is? What is your conclusion?

b. Use Fisher's LSD procedure to compare the comfort levels of nonbrowsers and light browsers. Use . Compute the LSD critical value (to 2 decimals). What is your conclusion?

Answer 1

a)

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

Source	SS	df	MS	F	P value
Between	52.00	2	26.000	22.286	0.000
Within	24.50	21	1.1667
Total	76.50	23

sum of sq;treatment=	52.00
sum of sq; error=	24.50
mean sq;treatment=	26.00
mean square; error=	1.17
test statistic =	22.29
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