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
	7	8	5
	8	9	7
	9	8	5
	6	7	7
	6	10	4
	7	7	6
	8	9	5
	7	8	7

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).

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 a= .05.

Compute the LSD critical value (to 2 decimals).

What is your conclusion?

Answer 1

By using SPSS we have

choose data> analyze> compare means > one way anova.

a]

Sum of Squares, Treatment=25.33
Sum of Squares, Error=24.5
Mean Squares, Treatment=12.67
Mean Squares, Error=1.17

The value of test statistic = MS (TREATMENT)MS(ERROR)= 12.67/1.17=10.86

Pvalue= less than .01

Since Pvalue less than the level of significance reject the null hypothesis H0.

Conclude that the mean comfort scores are not all the same for the browser groups

b] Critical LSD VALUE

LSD = 2.08 √ (1.167 * (2/8)) = 1.12

Conclusion: THERE IS NO DIFFERENCE BETWEEN MEANS OF NON BROWSER AND LIGHT BROWSER