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
20	33	55
24	43	53
35	34	40
35	50	44
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.)

	Source df SS MS F p Factors       Error    Total

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

	Level N Mean StDev Clothes Food Toys

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

The hypothesis of identical means can definitely REJECTED/ NOT REJECTED be.There is A DİFFERENCE /NO DİFFERENCE in the mean attention span.

Answer 1

ANSWERS:

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Ans 1) Excel Output:

ANOVA: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Clothes	9	241	26.77778	45.94444
Food	12	528	44	59.09091
Toys	11	563	51.18182	41.16364
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	3056.3	2	1528.2	31.00	6.27E-08	3.33
Within Groups	1429.2	29	49.3
Total	4485.5	31

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Ans 2)

Descriptive statistics for clothes, foods, toys :

Level	N	Mean	StDev
Clothes	9	26.778	6.778
Food	12	44.000	7.687
Toys	11	51.182	6.416

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Ans 3)

According to the descriptive statistics table,

Yes, there a difference in the mean attention span of the children for the various commercials.

Here the F-cal value is 31 which is greater than F-critical 3.33 which means our null hypothesis is rejected and there is a difference in the mean attention span.

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