Statistics Essay

1147 Words Nov 21st, 2013 5 Pages
1.1.
The null hypothesis H0 is μ1=μ2=μ3 where μ1, μ2 and μ3 represent the average income for every level of education.

1.2.
The alternative hypothesis H1 is μ1 ≠ μ2 or μ2 ≠ μ3 or μ1 ≠ μ3 (at least one μ is different from another μ).

1.3.
In order to test the above pair of hypotheses, I will use ANOVA analysis that provides the F statistic significance test:
H0: FF critical H0 is rejected
The following table contains the results of the ANOVA analysis:

Anova: Single Factor SUMMARY
Groups Count Sum Average Variance
PRIMARY 115,00 3944657,40 34301,37 464689314,43
SECONDARY 137,00 5892738,36 43012,69 565872506,75
HIGHER 74,00 3556259,40 48057,56 719686158,84
…show more content…
Comparing the confidence levels of the households of secondary and higher education, it is clear that there are common points which means that their means incomes do not significantly differ.
Subject 2
2.1.
The null hypothesis H0 is μ1=μ2=μ3 where μ1, μ2 and μ3 represent the average consumption for every level of education.

2.2
The alternative hypothesis H1 is μ1 ≠ μ2 or μ2 ≠ μ3 or μ1 ≠ μ3 (at least one μ is different from another μ)

2.3.
In order to test the above pair of hypotheses, I will use ANOVA analysis that provides the F statistic significance test:

H0: FF critical H0 is rejected
The following table contains the results of the ANOVA analysis:

Anova: Single Factor SUMMARY
Groups Count Sum Average Variance
PRIMARY 115,00 3540667,04 30788,41 350039735,62
SECONDARY 137,00 5072527,76 37025,75 360411466,43
HIGHER 74,00 2993060,64 40446,77 452632644,70

ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 4679243098,80 2,00 2339621549,40 6,20 0,00 3,02
Within Groups 121962672358,68 323,00 377593412,88 Total 126641915457,47 325,00

The F statistic equals to 6,20 and is compared to F crit = 3.02. I conclude that F>F critical meaning that H0 is rejected.
2.4. In order to reach my final conclusions and following the

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