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Happiness Data part 4

Data set 1. Happiness Data.   [Info]

1. My data set is a sample because these questions are applicable for the entire population of the world. However, this is just a sample of people in the world to understand basic information about them. If it were a population, the researcher would have to survey every person in thw world. This sample can be used to infer the results for a survey of the entire population.

Result 1: One sample T hypothesis test   [Info]
One sample T hypothesis test:
μ : Mean of variable
H0 : μ = 46
HA : μ > 46

Hypothesis test results:
VariableSample MeanStd. Err.DFT-StatP-value
Age46.8141031.26441411550.643857530.2603

- variable: Age of participants

- null hypothesis: The mean equals 46

- alternative hypothesis:The mean is greater than 46

- t- statistic = 0.64385753

- Conclusion: The t statistic of.0.64385753 is smaller than the critical value of t, 1.960.

Therefore, you fail to reject the null hypothesis that the sample mean is equal to 46.

2. Two-way table hypothesis test

Result 2: Contingency table (with data)   [Info]
Contingency table results:
Rows: Children
Columns: Age
 10 to 20 20 to 30 30 to 40 40 to 50 50 to 60 60 to 70 70 to 80 80 to 90 Total 0 3 16 9 5 9 3 0 0 45 1 0 4 5 5 5 1 0 0 20 2 0 1 9 13 10 9 3 1 46 3 0 1 4 5 4 7 1 3 25 4 0 0 3 3 0 3 3 0 12 5 0 0 0 0 1 3 0 0 4 6 0 0 0 1 2 0 0 0 3 7 0 0 1 0 0 0 0 0 1 Total 3 22 31 32 31 26 7 4 156

Chi-Square test:
StatisticDFValueP-value
Chi-square4990.1106730.0003
Warning: over 20% of cells have an expected count less than 5.
Chi-Square suspect.

- variables: Age of participants and Number of children that participants have.

- null hypothesis: Age and number of children are independent of one another.

- alternative hypothesis: Age and number of children are not independent of one another.

- Chi squared statistic = 90.110673

- Conclusion: The p value of .0003 is less than the significance level of .05. Therefore, you reject the null hypothesis that age and number of children are independent of one another.

3. One-way ANOVA

Result 3: One Way ANOVA   [Info]
Analysis of Variance results:
Data stored in separate columns.

Column statistics
ColumnnMeanStd. Dev.Std. Error
Age15646.81410315.7925281.2644141
Children1561.79487181.55633130.12460623
Education15613.3589743.09411240.24772726
Income15616.7692315.25951810.42109846

ANOVA table
SourceDFSSMSF-StatP-value
Columns3172312.1757437.39794.81023<0.0001
Error62044804.63572.26554
Total623217116.81

- variables: Age, number of children, educational level, and income level.

- null hypothesis:The sample means of each set of data are all equal.

- alternative hypothesis: The sample means of each data set are different.

- F statistic = 794.81023

- Conclusion: The p value of less than .0001 is less than the significance level of .05. Therefore, you reject the null hypothesis that The sample means of each set of data are all equal.

Extra Credit:

1.

Result 4: One sample Z hypothesis test   [Info]
One sample Z hypothesis test:
μ : Mean of variable
H0 : μ = 46
HA : μ > 46
Standard deviation not specified.

Hypothesis test results:
VariablenSample MeanStd. Err.Z-StatP-value
Age15646.8141031.26441410.643857530.2598

- variable: Age

- null hypothesis: the sample mean is equal to 46

- alternative hypothesis: The sample mean is greater than 46

- Z statistic = 0.64385753

- Conclusion: You would fail to reject the null hypothesis because the p value of 0.2598 is greater than the significance level of .05.

2.

Result 5: Contingency table (with data)   [Info]
Contingency table results:
Rows: Income
Columns: Education
 0 6 7 9 10 11 12 13 14 15 16 17 18 19 20 Total 0 to 5 0 0 0 1 1 0 2 1 1 0 1 0 0 0 0 7 5 to 10 2 1 0 3 0 1 3 1 1 0 0 0 0 0 0 12 10 to 15 0 0 1 3 0 1 7 2 4 0 1 0 0 1 0 20 15 to 20 0 1 0 0 1 5 25 3 6 1 9 5 2 0 0 58 20 to 25 0 0 0 0 2 1 14 4 6 1 17 1 6 0 1 53 25 to 30 0 0 0 0 0 0 0 0 1 0 2 0 1 0 2 6 Total 2 2 1 7 4 8 51 11 19 2 30 6 9 1 3 156

Chi-Square test:
StatisticDFValueP-value
Chi-square70138.84036<0.0001
Warning: over 20% of cells have an expected count less than 5.
Chi-Square suspect.

- variables: Education level and income level

- null hypothesis: Education level and income level are independent of one another

- alternative hypothesis: Education level and income level are not independent of one another

- Chi squared statistic = 138.84036

- Conclusion: The p value of less than .0001 is  less than the significance level of .05. Therefore, you reject the null hypothesis that education and income level are independent of one another.

3.

Result 6: One Way ANOVA (2)   [Info]
Analysis of Variance results:
Data stored in separate columns.

Column statistics
ColumnnMeanStd. Dev.Std. Error
Age15646.81410315.7925281.2644141
Children1561.79487181.55633130.12460623
Education15613.3589743.09411240.24772726

ANOVA table
SourceDFSSMSF-StatP-value
Columns2170544.6785272.335978.64335<0.0001
Error46540516.94287.133209
Total467211061.61

- variables: Age, number of children, and education level

- null hypothesis: The sample means of each set of data are all equal.

- alternative hypothesis:The sample means of each set of data are all different.

- F statistic = 978.64335

- Conclusion:  The p value of  less than .0001 is less than the significance level of .05. Therefore, you reject the null hypothesis that The sample means of each set of data are all equal.

4.  I believe that the data set could be largely improved by giving people more options for their happiness level besides 1,2, or 3. I believe it would have been more beneficial to have surveyed this question with a scale of 1-10 or even a scale of 1-50. This would have made it much easier to conduct more statistical tests on. Also, if the data provided the numerical value of participants income rather than a level. If this was the case, I could have conducted more statistical tests in regards to the income of the participants. In addition, more graphs could have been used in several instances to better depict my data set.