Question

Colleen is the marketing manager for Virtually Viral, an entertainment company that collects viral videos from around the Internet and aggregates them on their website. Whether it’s videos of cats or unusual marriage proposals, Virtually Viral collects them all. Almost all of Virtually Viral’s revenue comes from clicks on advertisements surrounding the videos. To maximize profits, Colleen tries to match ad content to video content. For example, for the ‘Wacky Weddings’ section of the website, most advertisements link to wedding planners and invitation/paper product suppliers. As part of this effort, Colleen contracted a web design firm to put together a new look for the website, with the goal of improving the amount of time visitors spend on the website. They produced four different versions, each arranging the videos and advertisements differently. Colleen is unsure which of these designs would result in the greatest amount of time spent on the site. To solve this problem, Colleen designs an experiment. She sets up a system to randomly assign visitors to the website to experience one of the four designs, recording the number of seconds that they spend on the site. She wants to compare the groups with each other and see if the different designs result in different lengths of time viewing the website. Whichever results in the longest visits will become the new design for the site in general. She knows from Chapter 7 that she has a research question and that this calls for some type of hypothesis testing. In Chapter 9, she learned that treating groups differently and comparing them means that she has independent data. But the independent-samples t-test only compares two groups with each other and she has four. Should she run multiple independent-samples t-tests? Or is there a better way?

Also complete an ANOVA and post-hoc test.

webdesign	seconds
1	55
1	71
1	72
1	62
1	67
2	115
2	86
2	98
2	120
2	115
2	103
3	86
3	108
3	66
3	37
3	90
4	71
4	62
4	48
4	69
4	55
4	57

Answer 1

here we should use one-way anova with

null hypothesis H0:design1=design2=design3=design4

alternate hypothesis Ha: atleast one design is different from others.

since the p-value=0.0004 of the between groups is less than alpha=0.05, so we reject the null hypothesis and conclude that

atleast one design is different from others. Now we use post-hoc test for multiple comparison.

we use here fisher least significant difference (LSD)=sqrt(MSE*(1/r1+1/r2)*t(alpha/2,error df)

for design 1 and design 2 the LSD=SQRT(241.2537*(1/5+1/6))*TINV(0.05,18)=SQRT(241.2537*(1/5+1/6))*2.1001=19.7598

similary for other pair is given as

comparison		difference	LSD	remarks
design1	design2	40.76666667	19.7598	signifianct
	design3	12	20.63844	non-signifianct
	design4	5.066666667	19.7598	non-signifianct
design2	design3	28.76666667	19.7598	signifianct
	design4	45.83333333	18.84023	signifianct
design3	design4	17.06666667	19.7598	non-signifianct

following one-way anova information has been generated

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
design1	5	327	65.4	49.3
design2	6	637	106.1667	166.1667
design3	5	387	77.4	732.8
design4	6	362	60.33333	76.66667
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	7428.024	3	2476.008	10.2631	0.0004	3.1599
Within Groups	4342.567	18	241.2537
Total	11770.59	21