1) There are four Populations in this problem they are a)all 10 year old ii) all 30 year old iii) all 50 year old iv) all 70 year old
Hypothesis
H0:
H1 : Atleast one mean is different.
One-way ANOVA: 10 year-olds, 30 year-olds, 50year-olds, 70year-olds
Method
Null hypothesis | All means are equal |
Alternative hypothesis | Not all means are equal |
Significance level | α = 0.05 |
Equal variances were assumed for the analysis.
Factor Information
Factor | Levels | Values |
Factor | 4 | 10 year-olds, 30 year-olds, 50year-olds, 70year-olds |
Analysis of Variance
Source | DF | Adj SS | Adj MS | F-Value | P-Value |
Factor | 3 | 61.00 | 20.333 | 5.60 | 0.003 |
Error | 36 | 130.60 | 3.628 | ||
Total | 39 | 191.60 |
since p-value is less than 0.05 we can conclude that there is a significant difference exists between the means of the population.
Tukey Pairwise Comparisons
Grouping Information Using the Tukey Method and 95% Confidence
Factor | N | Mean | Grouping | |
10 year-olds | 10 | 7.800 | A | |
30 year-olds | 10 | 6.500 | A | B |
50year-olds | 10 | 5.700 | A | B |
70year-olds | 10 | 4.400 | B |
Means that do not share a letter are significantly different.
Tukey Simultaneous 95% CIs
2)
Hypothesis:
for factor A
H0: all the music types are equal
H1 : Atleast one music is different.
For factor B
H0: all the genders have equal interest in music
H1 : Genders have not equal interest in music
General Linear Model: Response versus Music type, Gender
Method
Factor coding | (-1, 0, +1) |
Factor Information
Factor | Type | Levels | Values |
Music type | Fixed | 4 | 1, 2, 3, 4 |
Gender | Fixed | 2 | 1, 2 |
Analysis of Variance
Source | DF | Adj SS | Adj MS | F-Value | P-Value |
Music type | 3 | 64.20 | 21.400 | 11.43 | 0.000 |
Gender | 1 | 33.80 | 33.800 | 18.05 | 0.000 |
Music type*Gender | 3 | 23.40 | 7.800 | 4.17 | 0.009 |
Error | 72 | 134.80 | 1.872 | ||
Total | 79 | 256.20 |
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