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	5	10
	7	6	12
	8	5	10
	5	4	12
	5	7	9
	6	4	11
	7	6	10
	6	5	12

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).

Answer 1

H0: Null Hypothesis: (There is no difference among mean comfort scores for the three types of browsers)

HA: Alernative Hypothesis: (At least one mean is different from other 2 means) (There is a difference among mean comfort scores for the three types of browsers) (Claim)

From the given data, the following Table is calculated:

	Nonbrowser	Light Browser	Heavy Browser	Total
N	8	8	8	24
	50	42	86	178
Mean	6.25	5.25	10.75	7.417
	320	228	934	1482
Std. Dev.	1.0351	1.0351	1.165	2.6526

From the above Table, ANOVA Table is calculated as follows:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F Stat
Treatments	137.3333	2	137.3333/2=68.6667	68.6667/1.1667=58.86
Error	24.5	21	24.5/21=1.1667
Total	161.8333	23

F Stat = 68.6667/1.1667=58.86

By Technology, P Value= 0.0000

Since P - value is less than , the difference is significant. Reject null hypothesis.

Conclusion:
The data support the claim that there is a difference among mean comfort scores for the three types of browsers.

Answers to Questions asked:

Sum of squares, Treatment	137.3333
Sum of squares, Error	24.5
Mean square, Treatment	68.6667
Mean square, Error	1.1667

The value of the test statistic = 58.86