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 4 11 10 6 12 9 4 9 8 6 9 11 3 10 8 5 11 10 4 10 9 6 a. Use .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 .10 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 groups b. Use Fisher's LSD procedure to compare the comfort levels of nonbrowsers and light browsers. Use .05 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 scores

Answer 1

(a)

Following table shows the group totals:

	Nonbrowser, G1	Light Browser, G2	Heavy browser, G3
	10	9	4
	11	10	6
	12	9	4
	9	8	6
	9	11	3
	10	8	5
	11	10	4
	10	9	6
Total	82	74	38

And following table shows the grand total and total of square of observations:

	G	G^2
	10	100
	11	121
	12	144
	9	81
	9	81
	10	100
	11	121
	10	100
	9	81
	10	100
	9	81
	8	64
	11	121
	8	64
	10	100
	9	81
	4	16
	6	36
	4	16
	6	36
	3	9
	5	25
	4	16
	6	36
Total	194	1730

So we have

Now

Now

Now

Since there are 3 different groups so we have k=3. Therefore degree of freedoms are:

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Now

F test statistics is

P-value:

p-value is less than 0.005

Conclusion:

Conclude that not all treatment means are equal

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The fisher LSD procedure to test whether there is a significant difference between the means:

Critical value of t for level of significance 0.05 and degree of freedom of error is df = 21 is 2.080.

And we have

Since sample sizes are same so LSD will be

Following table shows the differences:

| Nonbrowser, G1-Light Browser, G2 |	8
| Nonbrowser, G1-Heavy browser, G3 |	44
| Light Browser, G2-Heavy browser, G3 |	36

Since absolute difference between G1 and G3 is greater than 1.12 so these are significantly differ.

Since absolute difference between G2 and G3 is greater than 1.12 so these are significantly differ.

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Since absolute difference between G1 and G2 is not greater than 1.12 so these are not significantly differ.