In: Statistics and Probability
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
43 | 30 | 52 |
24 | 38 | 58 |
42 | 46 | 43 |
35 | 54 | 49 |
28 | 47 | 63 |
31 | 42 | 53 |
17 | 34 | 48 |
31 | 43 | 58 |
20 | 57 | 47 |
47 | 51 | |
44 | 51 | |
54 |
a. Complete the ANOVA table using a .05 significance level (Round the SS and MS values to 1 decimal place and F value to 2 decimal places. Leave no cells blank — be certain to enter "0" wherever required. Round the df values to nearest whole number.)
b. Find the values of mean and standard deviation. (Round the mean and standard deviation values to 3 decimal places.)
c. Is there a difference in the mean attention span of the children for the various commercials?
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d.
Are there significant differences between pairs of means?
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One-way ANOVA: cloths, food, toys
Method
Null hypothesis All means are equal
Alternative hypothesis At least one mean is different
Significance level α = 0.05
Equal variances were assumed for the analysis.
Factor Information
Factor Levels Values
Factor 3 cloths, food, toys
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
Factor 2 2437 1218.52 20.81 0.000
Error 29 1698 58.57
Total 31 4136
Model Summary
S R-sq R-sq(adj) R-sq(pred)
7.65296 58.93% 56.10% 49.78%
Means
Factor N Mean StDev 95% CI
cloths 9 30.11 9.01 (24.89, 35.33)
food 12 44.67 8.11 (40.15, 49.19)
toys 11 52.09 5.72 (47.37, 56.81)
Pooled StDev = 7.65296
a)
Source DF Adj SS Adj MS F-Value P-Value
Factor 2 2437 1218.52 20.81 0.000
Error 29 1698 58.57
Total 31 413
b)
Factor N Mean StDev
cloths 9 30.11 9.01
food 12 44.67 8.11
toys 11 52.09 5.72
c)
p value =0 < alpha
we reject null hypothesis,there is sufficient evidence to support that atleast two of the mean is differ from the other
d)all differences are significant