Question

A study of the effect of television commercials on 12-year-old children measured their attention span, in seconds. The commercials were for clothes, food, and toys.

Clothes	Food	Toys
22	49	50
20	33	53
48	38	42
35	51	55
28	47	63
31	42	53
17	34	48
31	43	58
20	57	47
	47	51
	44	51
	54

1. Complete the ANOVA table. Use 0.05 significance level. (Round the SS and MS values to 1 decimal place and F value to 2 decimal places.)

2. Find the values of mean and standard deviation. (Round the mean and standard deviation values to 3 decimal places.)

3. Is there a difference in the mean attention span of the children for the various commercials?

4. Are there significant differences between pairs of means?

Answer 1

solutions:

Here we are going to perform one way anova to check there a difference in the mean attention span of the children for the various commercials.

So the procedure for performing one way anova is given below

Conclusion :It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that clothes, food, and toys population means are not equal, at the α=0.05 significance level.

so from above we have

1)Complete the ANOVA table. Use 0.05 significance level. (Round the SS and MS values to 1 decimal place and F value to 2 decimal places.)

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	2936.0	2	1468.0	25.34	4.31169E-07	3.33
Within Groups	1679.9	29	57.9
Total	4615.9	31

2)

SUMMARY
Groups	Count	Sum	Average	Variance
Clothes	9.000	252.000	28.000	94.000
Food	12.000	539.000	44.917	55.720
Toys	11.000	571.000	51.909	31.491

3)

Is there a difference in the mean attention span of the children for the various commercials?

from anova table

Since F cal = 25.34 > F crit =3.33 It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that clothes, food, and toys population means are not equal, at the α=0.05 significance level.

4) Are there significant differences between pairs of means?

If the null hypothesis is computed, you will need to conduct a Post-Hoc test. to check there significant differences between pairs of means?

so we will perform Fisher's LSD to check Fisher's LSD.

so from above table we can the p-values for all pairs are < 0.05 level of significance so we reject null hypothesis and conclude that there significant differences between all pairs of means.