Question

The authors of a paper titled "Age and Violent Content Labels Make Video Games Forbidden Fruits for Youth" carried out an experiment to determine if restrictive labels on video games actually increased the attractiveness of the game for young game players.

Participants read a description of a new video game and were asked how much they wanted to play the game. The description also included an age rating. Some participants read the description with an age restrictive label of 7+, indicating that the game was not appropriate for children under the age of 7. Others read the same description, but with an age restrictive label of 12+, 16+, or 18+.

The summary data shown below are the ratings of 12- to 13-year-old boys for how much they wanted to play the game on a scale of 1 to 10. You can assume that boys were assigned at random to one of the four age label treatments (7+, 12+, 16+, 18+) and that ratings were normally distributed.

Group	Sample Size	Sample Mean	Sample Std Dev
18+ label	22	7.2	1.33
16+ label	21	6.2	1.7
12+ label	18	6	1.5
7+ label	19	5	2.03

(f) Complete the ANOVA table. Note that you've already found two of the three sums of squares. Use three decimals for the sums of squares and means squares, then round the F test statistic to 2 decimals.

Source of Variation	df	Sum of Squares	Mean Square	F
Treatment
Error				X
Total			X	X

(g) Do the data provide convincing evidence that the mean rating associated with the game description by 12- to 13-year-old boys is not the same for all four restrictive rating labels? Test the appropriate hypotheses using a significance level of 0.05.

test statistic:

Now we can test each mean against each other mean, and use a Bonferroni correction to control the family-wise error rate at α=0.05. Recall from above that:

$(x)^^\_$ 18+ = 7.2

$(x)^^\_$ 16+ = 6.2

$(x)^^\_$ 12+ = 6

$(x)^^\_$ 7+ = 5

For each comparison, calculate the difference in sample means, the t-test statistic (rounded to three decimal places), and the p-value. The standard error and degrees of freedom for each comparison are provided.

Parameter	difference in sample means	standard error of difference	t-test statistic	degrees of freedom	two sided p-value
μ18+− μ16+		0.50		41
μ18+− μ12+		0.53		38
μ18+− μ7+		0.52		39
μ16+− μ12+		0.53		37
μ16+− μ7+		0.52		38
μ12+− μ7+		0.54		35

Answer 1

$(f)~SSE=sum~of~square~due~to~error\\ =(22-1)*1.33^2+(21-1)*1.7^2+(18-1)*1.5^2+(19-1)*2.03^2\\ =207.373\\ G=rand~mean=(22*7.2+21*6.2+18*6+19*5)/(22+21+18+19)\\ =6.145\\ SStr=sum~of~square~due~to~treatment\\ =22*(7.2-6.145)^2+21*(6.2-6.145)^2+18*(6-6.145)^2+19*(5-6.145)^2\\ =49.838\\ with~df=4-1=3\\ SST=sum~of~square~due~to~total=SSE+SStr=207.373+49.838\\=257.211\\ with~df=22+21+18+19-1=79\\ error~df=79-3=76\\$

Source of Variation	df	Sum of Squares	Mean Square	F
Treatment	3	49.838	49.838/3=16.613	16.613/2.729=6.09
Error	76	207.373	207.373/76=2.729	X
Total	79	257.211	X	X

(g) Critical value=F_0.05,3,76=2.72

Since F=6.09>Critical value so the data provide convincing evidence that the mean rating associated with the game description by 12- to 13-year-old boys is not the same for all four restrictive rating labels.

Parameter	difference in sample means	standard error of difference	t-test statistic	degrees of freedom	two sided p-value
μ18+− μ16+	7.2-6.2=1	0.50	1/0.5=2	41	0.0522
μ18+− μ12+	7.2-6=1.2	0.53	1.2/0.53=2.264	38	0.0294
μ18+− μ7+	7.2-5=2.2	0.52	2.2/0.52=4.231	39	0.0001
μ16+− μ12+	6.2-6=0.2	0.53	0.2/0.53=0.377	37	0.7083
μ16+− μ7+	6.2-5=1.2	0.52	1.2/0.52=2.308	38	0.0265
μ12+− μ7+	6-5=1	0.54	1/0.54=1.852	35	0.0725

Error Rate for each pair=0.05/6=0.0083

Hence μ₁₈₊ and μ₇₊ are significantly different (since p-value<0.0083) and other pairs are insignificantly different (since p-value>0.0083).