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Data Analysis Tools - Part Two

For the second assignment of the Data Analysis Tools course, we’ve been asked to run a Chi Square Independence Test. 

By using the Gapminder dataset, I would like to understand if there is an association between income group (explanatory variable) and life expectancy group (response variable).

Ho - There is no association between income group and life expectancy.

Ha - There is an association between income group and life expectancy.

Then we run our Chi Square test in SAS, and we have the below results:

Based on the above we are able to see, the probability is <.0001 therefore we are able to reject the Ho and accept the Ha. The results show an association between  income group and life expectancy group, or in chi square terms, the explanatory variable show a dependency with the response variable.

Finally, we want to understand which of the possible combinations of probabilities show an actual significance. For this, we run a sequential analysis comparing the 10 possible pairs. Also, we will use the Bonferroni adjustment to check if the results are significant enough to reject the Ho.

Bonferroni adjustment = Probability 0.05 / # of comparisons

In this case, we have the below:

B. Adjustment = 0.05 / 10 = 0.005 - Then if the results show a probability value lower to 0.005 we’ll be able to reject the Ho.

Possible combinations:

SAS code used:

Summary results table:

As we are able to see almost all the combinations showed a significant value, except the combination of Short life vs V Short Life, and Long Life vs Average Life.

More to come! :)

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Data Analysis Tools Assignment 1

INTRODUCTION

The aim of my ANOVA statistical test was to look at correlation between the quantity of cigarettes smoked and marital status. based on NESARC data set.

PYTHON SCRIPT

import numpy import pandas import statsmodels.formula.api as smf import statsmodels.stats.multicomp as multi

data = pandas.read_csv('dataset.csv', low_memory=False)

data['S3AQ3B1'] = data['S3AQ3B1'].convert_objects(convert_numeric=True) data['S3AQ3C1'] = data['S3AQ3C1'].convert_objects(convert_numeric=True) data['CHECK321'] = data['CHECK321'].convert_objects(convert_numeric=True)

sub1=data[(data['CHECK321']==1)]

sub1['S3AQ3B1']=sub1['S3AQ3B1'].replace(9, numpy.nan) sub1['S3AQ3C1']=sub1['S3AQ3C1'].replace(99, numpy.nan)

recode1 = {1: 30, 2: 22, 3: 14, 4: 5, 5: 2.5, 6: 1} sub1['USFREQMO']= sub1['S3AQ3B1'].map(recode1)

sub1['USFREQMO']= sub1['USFREQMO'].convert_objects(convert_numeric=True)

sub1['NUMCIGMO_EST']=sub1['USFREQMO'] * sub1['S3AQ3C1']

sub1['NUMCIGMO_EST']= sub1['NUMCIGMO_EST'].convert_objects(convert_numeric=True)

ct1 = sub1.groupby('NUMCIGMO_EST').size() print (ct1)

sub2 = sub1[['NUMCIGMO_EST', 'MARITAL']].dropna()

model1 = smf.ols(formula='NUMCIGMO_EST ~ C(MARITAL)', data=sub2).fit() print (model1.summary())

print ('means for numcigmo_est by marital status') m1= sub2.groupby('MARITAL').mean() print (m1)

print ('standard deviations for numcigmo_est by marital status') sd1 = sub2.groupby('MARITAL').std() print (sd1)

mc1 = multi.MultiComparison(sub2['NUMCIGMO_EST'], sub2['MARITAL']) res1 = mc1.tukeyhsd() print(res1.summary())

DESCRIPTION

As well as in the video lessons, I have calculated the number of cigarettes smoked per month and have compared it to marital status. The results of ANOVA are below:

OLS Regression Results                             ============================================================================== Dep. Variable:           NUMCIGMO_EST   R-squared:                       0.022 Model:                            OLS   Adj. R-squared:                  0.022 Method:                 Least Squares   F-statistic:                     44.42 Date:                Mon, 12 Nov 2018   Prob (F-statistic):           1.75e-45 Time:                        01:34:48   Log-Likelihood:                -70633. No. Observations:                9804   AIC:                         1.413e+05 Df Residuals:                    9798   BIC:                         1.413e+05 Df Model:                           5                                         Covariance Type:            nonrobust                                         ===================================================================================                      coef    std err          t      P>|t|      [0.025      0.975] ----------------------------------------------------------------------------------- Intercept         429.0034      5.275     81.323      0.000     418.663     439.344 C(MARITAL)[T.2]   -28.4852     15.966     -1.784      0.074     -59.783       2.812 C(MARITAL)[T.3]    43.2257     14.581      2.965      0.003      14.645      71.807 C(MARITAL)[T.4]    57.8594      9.314      6.212      0.000      39.602      76.116 C(MARITAL)[T.5]    19.3024     15.787      1.223      0.221     -11.644      50.249 C(MARITAL)[T.6]   -78.9413      8.216     -9.608      0.000     -95.047     -62.836 ============================================================================== Omnibus:                     2361.523   Durbin-Watson:                   1.940 Prob(Omnibus):                  0.000   Jarque-Bera (JB):             7343.082 Skew:                           1.231   Prob(JB):                         0.00 Kurtosis:                       6.452   Cond. No.                         5.87 ==============================================================================

As wee see, the p-value is sufficient enough to reject the null hypothesis and assume that the number of cigarettes smoked per month depends on marital status.

The mean and standard deviation results are presented below:

NUMCIGMO_EST MARITAL               1          429.003411 2          400.518201 3          472.229094 4          486.862778 5          448.305846 6          350.062103 standard deviations for numcigmo_est by marital status         NUMCIGMO_EST MARITAL               1          321.646121 2          306.128890 3          341.175384 4          360.143462 5          335.367772 6          304.339486

groups descriptions:

1 - married

2 - unofficially married

3 - widowed

4 - divorced

5 - separated

6 - never married

The outcome of post hoc test revealed several “thrilling” correlations:

Multiple Comparison of Means - Tukey HSD,FWER=0.05 ================================================== group1 group2  meandiff   lower     upper   reject --------------------------------------------------  1      2     -28.4852  -73.9932  17.0228  False  1      3     43.2257    1.6677   84.7837   True  1      4     57.8594   31.3127    84.406   True  1      5     19.3024   -25.6949  64.2998  False  1      6     -78.9413 -102.3592  -55.5234  True  2      3     71.7109   13.8672   129.5546  True  2      4     86.3446   38.1413   134.5478  True  2      5     47.7876   -12.5745  108.1498 False  2      6     -50.4561  -97.0096  -3.9026   True  3      4     14.6337   -29.8595  59.1269  False  3      5     -23.9232  -81.3661  33.5196  False  3      6     -122.167 -164.8673  -79.4667  True  4      5     -38.5569  -86.2784   9.1645  False  4      6    -136.8007 -165.1021 -108.4992  True  5      6     -98.2437 -144.2981  -52.1893  True --------------------------------------------------

As we can see from the table, married people smoke significantly more in comparison to widowed, divorced and never married. Unofficially married smoke more than the enumerated groups as well. Uncovering that widowed smoke more than never married people was surprising for me. Other conclusions can be made based on the table.

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My DOST Coursera Grant

So, I was admitted to the DOST Coursera Grant and it was really nice to see a lot of interesting topics. I first enrolled in data analytics to try it out. It appears I need to publish a blog for the assignment so I made this blog. A lot of first times today.

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SAS vs Python vs R

I’ve already invest a lot of time learning R. I’ve spent a little time with Python though not too much yet on statistics and data mining/analysis. I’m not looking forward to having to learn a new tool. Would be great if there was an R track to mirror the SAS and Python learning options.

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Data Analysis Tools - Part One

For the first assignment of the Data Analysis Tools course, we’ve been asked to run an analysis of variance.

By using the Gapminder dataset, I would like to understand if there is an association between continent (explanatory variable) and life expectancy (response variable).

Ho - There is no association between continent and life expectancy.

Ha - There is an association between the continent and life expectancy.

Then we run our ANOVA analysis in SAS, and we have the below results:

Based on the results above, we can see there is a F Value of 57.10, and probability of <.0001, therefore we can reject the Ho and accept the Ha that there is an association between continent and life expectancy.

When running an Analysis of Variance (ANOVA), the results tells if there is a difference in means. However, it doesn’t pinpoint which means are different. Duncan’s Multiple Range test (DMRT) is a post hoc test to measure specific differences between pairs of means.

Here is the code to run Duncan’s test:

Duncan’s results:

Based on the above, we can see that there is a significance different between Africa and Europe.

More to come! :)