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
	3	4	7
	4	5	9
	5	4	7
	2	3	9
	2	6	6
	3	3	8
	4	5	7
	3	4	9

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).

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).

Answer 1

A -

P.S - All of this experiment is a two-tailed experiment since we are only checking for difference in mean comfort level and not if one group's comfort level is higher/lower than the other.

You have to create 3 treatment groups in excel or by hand like below and calculate the summary statistics of different groups as follows -

Summary of Data
	Treatments
	Non browser	Light browser	Heavy browser	Total
N	8	8	8	24
∑X	26	34	62	122
Mean	3.25	4.25	7.75	5.083
∑X2	92	152	490	734
Std.Dev.	1.0351	1.0351	1.165	2.2247

Result Details
Source	Sum of Squares	df	Mean squares
Between-treatments (Treatments)	89.3333	2	44.6667	F = 38.28571
Within-treatments (Error)	24.5	21	1.1667
Total	113.8333	23

The f-ratio value is 38.28571.

The p-value is < .00001.

The result is significant at p < .05.

Hence, we can reject the null hypothesis that - "There is no significant difference among mean comfort scores for three browsers"

B -

Calculate fisher's lsd result in s same manner as we did for ANOVA just that we will consider only two treatment groups now instaed of three.

Fishers LSD Result :
	Sum of squares	degrees of freedom	Mean squares	F value
Between	4	1	4	3.7334
Within	15	14	1.0714
Total	19	15

LSD = t_0.05/2DFW *√(MSW(1/n_a + 1/n_b)

Step 1: Find the t-critical value. The t-critical value for α=0.05, dfw = 14 is 2.145 (I used online t value table). Make sure you are using the Within groups DF from your results!

Step 2: Insert the given values, the MSE from your results (I used 1.0714 from Within groups on the above table) the t-distribution value from Step 2 into the least significant difference formula:
LSD = 2.145 √ (1.0714 * (2/8))

= 1.11

Put this number aside for a moment,

Step 3: Calculate mean1 - mean2 from the results. For this example, we get -1

Step 4: Compare Step 2 and Step 3. If |mean1 - mean2| ≥ LSD then you can reject the null hypothesis that there is no significant difference among mean comfort scores for non browsers and light browsers

That’s it!

Conclusion - We reject the null hypothesis there is no significant difference among mean comfort scores for non browsers and light browsers

Tip: The LSD will only make sense if you have a significant result from ANOVA (i.e. if you reject the null hypothesis). Therefore, you shouldn’t run the test if you do not get a significant result from ANOVA.