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
	10	9	6
	11	10	8
	12	9	6
	9	8	8
	9	11	5
	10	8	7
	11	10	6
	10	9	8

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 -value is - Select your answer -less than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 6

What is your conclusion?

- Select your answer -Conclude that the mean comfort scores are not all the same for the browser groupsDo not reject the assumption that the mean comfort scores are equal for the browser groupsItem 7

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?

- Select your answer -Conclude that nonbrowsers and light browsers have different mean comfort scoresCannot conclude that nonbrowsers and light browsers have different mean comfort scoresItem 9

Answer 1

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	26.00	22.29	0.0000
Within	24.50	21	1.17
Total	76.50	23

a)

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

Cannot conclude that nonbrowsers and light browsers have different mean comfort scores