Question

A large advertising firm specializes in creating television commercials for children’s products. The firm wants to design a study to investigate factors that may affect the lengths of time a commercial is able to hold a child’s attention. A preliminary study determines that two factors that may be important are the age of the child and the type of product being advertised. The firm wants to determine whether there were large differences in the mean length of time that the commercial is able to hold the child’s attention depending on these two factors. If there proves to be a difference, the firm would then attempt to determine new types of commercials depending on the product and targeted age group. Three age groups are used: A1: 5-6 years, A2: 7-8 years, and A3: 9-10 years. The types of products selected are P1: Breakfast cereals and P2: Video games. The data are below:

	A1	A2	A3
P1	19	19	37
	36	35	6
	40	22	28
	30	28	4
	4	1	32
	10	27	16
	30	27	8
	5	16	41
	24	3	29
	21	18	18
P2	39	30	51
	18	47	52
	32	6	43
	22	27	48
	16	44	39
	2	26	33
	36	33	56
	43	48	43
	7	23	40
	16	21	51a.

a. Create a two-way ANOVA table in Excel.

b. Summarize your findings.

Answer 1

Here we have data:

	A1	A2	A3
P1	19	19	37
P1	36	35	6
P1	40	22	28
P1	30	28	4
P1	4	1	32
P1	10	27	16
P1	30	27	8
P1	5	16	41
P1	24	3	29
P1	21	18	18
P2	39	30	51
P2	18	47	52
P2	32	6	43
P2	22	27	48
P2	16	44	39
P2	2	26	33
P2	36	33	56
P2	43	48	43
P2	7	23	40
P2	16	21	51

Excel output;

Anova: Two-Factor With Replication
SUMMARY	A1	A2	A3	Total
P1
Count	10	10	10	30
Sum	219	196	219	634
Average	21.9	19.6	21.9	21.13333
Variance	157.6556	117.8222	177.6556	141.8437
P2
Count	10	10	10	30
Sum	231	305	456	992
Average	23.1	30.5	45.6	33.06667
Variance	191.8778	171.8333	51.15556	219.4437
Total
Count	20	20	20
Sum	450	501	675
Average	22.5	25.05	33.75
Variance	165.9474	168.4711	256.1974
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Sample	2136.067	1	2136.067	14.76544	0.000323	4.019541
Columns	1391.7	2	695.85	4.810023	0.011956	3.168246
Interaction	1273.633	2	636.8167	4.401959	0.016943	3.168246
Within	7812	54	144.6667
Total	12613.4	59

Here we have sufficient evidence to reject the null hypotheses because F-observed value (14.76544) is grater than F-critical value (4.019541) so, it is in the rejection region.

Conclusion: we can say that difference in the mean value.