#chisquare

[ramputras]

🧠 Chi-Square Realness: Alcohol vs. Marital Status 🍷💔💍

Using data from the NESARC survey, we dug into how your relationship status might influence how much you drink. The stats? Unfiltered and telling.

📊 Key Insight:

There’s a significant association between marital status and alcohol consumption (Chi² = 392.18, p < .001).

Married individuals top the charts across all drinking levels.

Divorced/Separated drink moderately — but steadily.

Never Married lean toward lighter drinking, but mid-range isn't rare.

Widowed appear quietly in higher consumption brackets. (Coping, perhaps?)

⚖️ Pairwise Comparisons:

We broke it down further with post-hoc testing:

Married vs Never Married → Big difference (p < .001)

Married vs Divorced/Cohabiting → Still distinct (p < .01)

Never Married vs Separated → Subtle, yet significant (p ≈ .015)

Your marital status might be quietly shaping your drinking habits — whether you're toasting to love, numbing the breakup, or just navigating life solo. In case you want to try the code yourself ````````````````````````

import pandas as pd import numpy as np import scipy.stats as stats import seaborn as sns import matplotlib.pyplot as plt

Load the dataset

data = pd.read_csv('nesarc.csv', low_memory=False)

Convert variables to numeric

data['S2AQ8A'] = pd.to_numeric(data['S2AQ8A'], errors='coerce') # Alcohol frequency data['MARITAL'] = pd.to_numeric(data['MARITAL'], errors='coerce') # Marital status

Subset data (assuming sub1 already exists — else, create one)

sub1 = data.copy() sub2 = sub1.copy()

Replace invalid or missing codes with NaN

sub2['S2AQ8A'].replace([99, np.nan], np.nan, inplace=True) sub2['MARITAL'].replace(9, np.nan, inplace=True)

Filter only valid values

sub2 = sub2[(sub2['S2AQ8A'].between(1, 10)) & (sub2['MARITAL'].between(1, 6))]

Create a label-friendly MARITALSTATUS column (you can use actual labels instead of scores)

marital_map = { 1: 'Married', 2: 'Cohabiting', 3: 'Widowed', 4: 'Divorced', 5: 'Separated', 6: 'Never Married' } sub2['MARITALSTATUS'] = sub2['MARITAL'].map(marital_map)

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🔍 Main Chi-Square Test

-------------------------

ct_main = pd.crosstab(sub2['S2AQ8A'], sub2['MARITALSTATUS']) print("Main Contingency Table:\n", ct_main)

print("\nColumn Percentages:\n", ct_main / ct_main.sum(axis=0))

chi2, p, dof, expected = stats.chi2_contingency(ct_main) print("\nChi-square test results:") print(f"Chi2 Value: {chi2:.2f}") print(f"p-value: {p:.4f}") print(f"Degrees of Freedom: {dof}") print("Expected Counts:\n", pd.DataFrame(expected, index=ct_main.index, columns=ct_main.columns))

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📊 Visualization

-------------------------

plt.figure(figsize=(12, 6)) sns.countplot(data=sub2, x='MARITALSTATUS', hue='S2AQ8A', palette='Set2') plt.title("Drinking Frequency by Marital Status") plt.xlabel("Marital Status") plt.ylabel("Count of Respondents") plt.legend(title="Drinking Frequency Code", bbox_to_anchor=(1.05, 1), loc='upper left') plt.tight_layout() plt.show()

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🔁 Optional: Pairwise Comparisons

-------------------------

def chi_square_pairwise(df, group_col, compare_col, groups): sub_df = df[df[group_col].isin(groups)] ct = pd.crosstab(sub_df[compare_col], sub_df[group_col]) chi2, p, dof, expected = stats.chi2_contingency(ct) print(f"\nChi-square for {groups[0]} vs {groups[1]}") print(f"Chi2: {chi2:.2f}, p-value: {p:.4f}, DOF: {dof}") print("Contingency Table:\n", ct)

Run pairwise tests

pairwise_groups = ('Married', 'Never Married'), ('Married', 'Divorced'), ('Married', 'Cohabiting'), ('Never Married', 'Separated')

for g1, g2 in pairwise_groups: chi_square_pairwise(sub2, 'MARITALSTATUS', 'S2AQ8A', [g1, g2])

````````````````````````

[ramputras]

🧠 Chi-Square Realness: Alcohol vs. Marital Status 🍷💔💍

Using data from the NESARC survey, we dug into how your relationship status might influence how much you drink. The stats? Unfiltered and telling.

📊 Key Insight:

There’s a significant association between marital status and alcohol consumption (Chi² = 392.18, p < .001).

Married individuals top the charts across all drinking levels.

Divorced/Separated drink moderately — but steadily.

Never Married lean toward lighter drinking, but mid-range isn't rare.

Widowed appear quietly in higher consumption brackets. (Coping, perhaps?)

⚖️ Pairwise Comparisons:

We broke it down further with post-hoc testing:

Married vs Never Married → Big difference (p < .001)

Married vs Divorced/Cohabiting → Still distinct (p < .01)

Never Married vs Separated → Subtle, yet significant (p ≈ .015)

Your marital status might be quietly shaping your drinking habits — whether you're toasting to love, numbing the breakup, or just navigating life solo. In case you want to try the code yourself ````````````````````````

import pandas as pd import numpy as np import scipy.stats as stats import seaborn as sns import matplotlib.pyplot as plt

Load the dataset

data = pd.read_csv('nesarc.csv', low_memory=False)

Convert variables to numeric

data['S2AQ8A'] = pd.to_numeric(data['S2AQ8A'], errors='coerce') # Alcohol frequency data['MARITAL'] = pd.to_numeric(data['MARITAL'], errors='coerce') # Marital status

Subset data (assuming sub1 already exists — else, create one)

sub1 = data.copy() sub2 = sub1.copy()

Replace invalid or missing codes with NaN

sub2['S2AQ8A'].replace([99, np.nan], np.nan, inplace=True) sub2['MARITAL'].replace(9, np.nan, inplace=True)

Filter only valid values

sub2 = sub2[(sub2['S2AQ8A'].between(1, 10)) & (sub2['MARITAL'].between(1, 6))]

Create a label-friendly MARITALSTATUS column (you can use actual labels instead of scores)

marital_map = { 1: 'Married', 2: 'Cohabiting', 3: 'Widowed', 4: 'Divorced', 5: 'Separated', 6: 'Never Married' } sub2['MARITALSTATUS'] = sub2['MARITAL'].map(marital_map)

-------------------------

🔍 Main Chi-Square Test

-------------------------

ct_main = pd.crosstab(sub2['S2AQ8A'], sub2['MARITALSTATUS']) print("Main Contingency Table:\n", ct_main)

print("\nColumn Percentages:\n", ct_main / ct_main.sum(axis=0))

chi2, p, dof, expected = stats.chi2_contingency(ct_main) print("\nChi-square test results:") print(f"Chi2 Value: {chi2:.2f}") print(f"p-value: {p:.4f}") print(f"Degrees of Freedom: {dof}") print("Expected Counts:\n", pd.DataFrame(expected, index=ct_main.index, columns=ct_main.columns))

-------------------------

📊 Visualization

-------------------------

plt.figure(figsize=(12, 6)) sns.countplot(data=sub2, x='MARITALSTATUS', hue='S2AQ8A', palette='Set2') plt.title("Drinking Frequency by Marital Status") plt.xlabel("Marital Status") plt.ylabel("Count of Respondents") plt.legend(title="Drinking Frequency Code", bbox_to_anchor=(1.05, 1), loc='upper left') plt.tight_layout() plt.show()

-------------------------

🔁 Optional: Pairwise Comparisons

-------------------------

def chi_square_pairwise(df, group_col, compare_col, groups): sub_df = df[df[group_col].isin(groups)] ct = pd.crosstab(sub_df[compare_col], sub_df[group_col]) chi2, p, dof, expected = stats.chi2_contingency(ct) print(f"\nChi-square for {groups[0]} vs {groups[1]}") print(f"Chi2: {chi2:.2f}, p-value: {p:.4f}, DOF: {dof}") print("Contingency Table:\n", ct)

Run pairwise tests

pairwise_groups = ('Married', 'Never Married'), ('Married', 'Divorced'), ('Married', 'Cohabiting'), ('Never Married', 'Separated')

for g1, g2 in pairwise_groups: chi_square_pairwise(sub2, 'MARITALSTATUS', 'S2AQ8A', [g1, g2])

````````````````````````

#DataScience #MentalHealth #NESARC #Sociology #ChiSquare #AlcoholUse #Relationships #ResearchVibes

[onlinespsscom]

HOW TO RUN CHI SQUARE TEST IN SPSS

[oxybiostats]

Chi-Square Test - Practice

Question: You cross two strains of mice and want to test whether the proportion of offspring with the genotypes AA, Aa, and aa fit the expected proportions under Mendelian inheritance. Results = AA = 20, Aa = 55, aa = 25. What are the expected frequencies? Test this hypothesis with a X2 test.

Answer: https://drive.google.com/file/d/0B3QbRK1QvL4sTk9tdXRmdjhQQjA/view?usp=sharing

[oxybiostats]

Poison Dart Frog Mate Choice

For those of you who were curious in class today how to calculate the opposite results for the poison dart frog example, here is an update. I added 3 new slides to the lecture (slides 14-16) to help clarify.

Imagine your red and green female frogs both choose 75 red male frogs and 25 green male frogs. If you write up a contingency table, you’ll see that your column/row totals now change. As a result, your expected frequencies change as well. In fact now your observed and expected frequencies are exactly the same. So, when you go to calculate your X2 value, your observed and expected frequencies completely cancel out. You end up with X2 = 0, P-value = 1, and your results are not significant.

I hope that clarifies the question!

[giuliojcg]

Running a Chi-Square Test of Independence

 Model Interpretation for Chi-Square Tests

When analysing the association between degree of urbanisation in a country (categorical response) and income level (categorical explanatory), there is sufficient evidence to infer that the proportions of people living in urban areas with income level above median (67.21%) differ significantly from those with income level below the median average (32.79%).

Model Interpretation for post hoc Chi-Square Test results

There is strong evidence to conclude that urbanisation depends strongly on income level (collapsed into 3 groups). 

On the other hand, using Post Hoc comparisons for each pair of income level (3 categories) and degree of urbanisation (2 categories), we found sufficient evidence to infer that the level of urbanisation increases as income levels rise up for all income categories (Low, Middle and High income countries).

LIBNAME mydata "/courses/d1406ae5ba27fe300 " access=readonly; Data Analysis2; Set mydata.gapminder; Length Income_Level2 $13;

Label incomeperperson='Income per Capita' HIVrate='Estimated HIV Prevalence %' Internetuserate='Internet users (per 100 people)' breastcancerper100TH='Breast cancer new cases per 100,000 female' lifeexpectancy='Life expectancy at birth (years)' polityscore='Democracy score (Polity)' Income_Level='Income per Capita by Category' breastcancerper100th_G= 'Breast cancer new cases per 100,000 female by Group'  lifeexpectancy_G= 'Life expectancy interval'  polityscore_G='Polity Score by Group'  internetuserate_G='Internet users (per 100 people)by Group'; /*Data Management for incomeperperson*/ if incomeperperson >= 0 and incomeperperson <= 560 then Income_Level = "US$ 0-560       "; else if incomeperperson > 560 and incomeperperson <= 1850 then Income_Level ="US$ 560-1,850"; else if incomeperperson > 1850 and incomeperperson <= 5000 then Income_Level = "US$ 1,850-5,000"; else if incomeperperson > 5000 and incomeperperson <= 14000 then Income_Level = "US$ 5,000-14,000"; else if incomeperperson > 14000 and incomeperperson then Income_Level = ">US$ 14,000"; else Income_Level='NA'; /*Data Management for lifeexpectancy*/ if lifeexpectancy >= 0 and lifeexpectancy <= 68 then lifeexpectancy_G = " 0-68" ; else if lifeexpectancy > 68 and lifeexpectancy <= 74 then lifeexpectancy_G ="68-74" ; else if lifeexpectancy > 74 then lifeexpectancy_G = "> 76" ; else if lifeexpectancy = . then lifeexpectancy_G = "NA" ; /*Data Management for polityscore*/ if polityscore >= -10 and polityscore <= -4 then polityscore_G = "-10 and -4" ; else if polityscore > -4 and polityscore <= 4 then polityscore_G ="-4 and 4" ; else if polityscore > 4 and polityscore <= 7 then polityscore_G ="4 and 7" ; else if polityscore > 7 and polityscore <= 9 then polityscore_G ="7 and 9" ; else if polityscore > 9 then polityscore_G = "> 9" ; else if polityscore = . then polityscore_G = "NA"; /*Data Management for internetuserrate*/ if internetuserate >= 0 and internetuserate <= 7.5 then internetuserate_G = "0 - 7.5" ; else if internetuserate > 7.5 and internetuserate <= 20 then internetuserate_G ="7.5-20"; else if internetuserate > 20 and internetuserate <= 41 then internetuserate_G = "20-41" ; else if internetuserate > 41 and internetuserate <= 66 then internetuserate_G = "41-66" ; else if internetuserate > 66 then internetuserate_G = ">66" ; else if internetuserate = . then internetuserate_G = "NA"; /*Data Management for breastcancerper100th*/ if breastcancerper100th >= 0 and breastcancerper100th <= 19 then breastcancerper100th_G = " 0 and 19" ; else if breastcancerper100th > 19 and breastcancerper100th <= 26 then breastcancerper100th_G = "19 and 26" ; else if breastcancerper100th > 26 and breastcancerper100th <= 34 then breastcancerper100th_G = "26 and 34" ; else if breastcancerper100th > 34 and breastcancerper100th <= 53 then breastcancerper100th_G = "34 and 53" ; else if breastcancerper100th > 53 then breastcancerper100th_G = ">53" ; else if breastcancerper100th = . then breastcancerper100th_G = "NA"; /*Urban Rate Groups*/ If urbanrate >= 50 then urban=1; else if urbanrate <50 then urban=0; else urban=99;

/*Income per person median*/ If incomeperperson >= 2549.5584738 then Median_income='Above Median Income'; else if incomeperperson <2549.5584738 then Median_income='Below Median Income'; else Median_Income='No Data';

/*Income per person The Middle*/ If incomeperperson <= 1295 then Income_Level2='Low Income'; else if incomeperperson > 1295 and incomeperperson <= 5900 then Income_Level2='Middle Income'; else if incomeperperson > 5900 then Income_Level2='High Income'; else Income_Level2='No Data';

keep country incomeperperson breastcancerper100TH HIVrate Internetuserate lifeexpectancy polityscore urbanrate Income_Level lifeexpectancy_G polityscore_G internetuserate_G breastcancerper100th_G urban Median_Income Income_Level2;

Proc Sort;by country;Run; Proc Freq Data=analysis2;Tables Income_Level2;Run; Proc Freq Data=analysis2;Tables urban*Median_income/ChisQ;Run;

Proc Freq Data=analysis2;Tables urban*Income_Level2/ChisQ;Run;

Data COMPARISON1; Set Analysis2; If Income_Level2='Low Income' or Income_Level2='Middle Income'; Proc Sort; by country; Proc Freq; Tables urban*Income_Level2/ChisQ; Run; Data COMPARISON2; Set Analysis2; If Income_Level2='Low Income' or Income_Level2='High Income'; Proc Sort; by country; Proc Freq; Tables urban*Income_Level2/ChisQ; Run; Data COMPARISON3; Set Analysis2; If Income_Level2='Middle Income' or Income_Level2='High Income'; Proc Sort; by country; Proc Freq; Tables urban*Income_Level2/ChisQ; Run;

[omgblairzy-blog]

Data Tools - Chi Square

1. Introduction

In the previous assignment (ANOVA), I discovered there was a lack of statistical significance in the association between female employment and urbanisation (Gapminder). This prompted an adjustment to my research question by elevating my third variable ‘internetuserate’ in place of ‘urbanrate’.

The ANOVA suggested statistical significance with a Probability F Statistic of  0.00137. The Tukey Post Hoc test highlighted three paired comparisons where we should accept statistical significance.

For this assignment, I have continued to apply a modified research question: Is there an association between the rate of female employment (response variable) and the rate of internet usage (explanatory variable). My hypothesis is that there is a positive association.

To perform the chi square test for independence, I have had to convert my quantitative research variables to categoric.

The code for this assignment follows below.

2. Code

2.1 Convert quantitative variables to categoric

#Quantile (5) split using the pandas.qcut function. (Explanatory variable) sub2['internet5'] = pandas.qcut(sub2.internetuserate_rd, 5, labels = \ ["Q1", "Q2", "Q3", "Q4", "Q5"])

''' Create two level categorical variable for female employment using 42% as the top level of the 25th percentile representing middling to high rates of female employment ''' def FEMCAT (row):    if row['femaleemployrate_rd'] < 42:        return 0    else :        return 1

sub2['FEMCAT'] = sub2.apply (lambda row: FEMCAT (row), axis = 1)

2.2 New dataframe with categorical variables of interest

# New Dataframe for explanatory categorical variable with 5 levels sub3 = sub2[['FEMCAT','internet5']].dropna()

2.3 Chi Square Test of Independence

#Contingency table of observed counts ct1 = pandas.crosstab(sub3['FEMCAT'], sub3['internet5']) print(ct1)

#Column percentages. Axis = 0 sums values in each column colsum = ct1.sum(axis = 0) colpct = ct1/colsum print(colpct)

#Request chi-square calculations which include chi-square, p value and exp print('Chi-square valuee, p value and the expected counts') cs1 = scipy.stats.chi2_contingency(ct1) print(cs1)

2.4 Post Hoc Comparison | Bonferroni Adjustment

#Bonferroni adjustment. Performed by recode pair (x 10). P value < 0.005 #Recode1 recode1 = {'Q1': 'Q1', 'Q2': 'Q2'} sub3['COMP1v1'] = sub3['internet5'].map(recode1)

#Contingency table of observed counts print('Observed counts and column percentage for recode1') ct2 = pandas.crosstab(sub3['FEMCAT'], sub3['COMP1v1']) print(ct2)

#column percentage colsum = ct2.sum(axis = 0) colpct = ct2 / colsum print(colpct)

#chi-square print('Chi-square value, p value, expected counts') cs2 = scipy.stats.chi2_contingency(ct2) print(cs2)

#Recode2 recode2 = {'Q1': 'Q1', 'Q3': 'Q3'} sub3['COMP1v2'] = sub3['internet5'].map(recode2)

#Contingency table of observed counts print('Observed counts and column percentage for recode2') ct3 = pandas.crosstab(sub3['FEMCAT'], sub3['COMP1v2']) print(ct3)

#column percentage colsum = ct3.sum(axis = 0) colpct = ct3 / colsum print(colpct)

#chi-square print('Chi-square value, p value, expected counts') cs3 = scipy.stats.chi2_contingency(ct2) print(cs3)

#Recode3 recode3 = {'Q1': 'Q1', 'Q4': 'Q4'} sub3['COMP1v3'] = sub3['internet5'].map(recode3)

#Contingency table of observed counts print('Observed counts and column percentage for recode3') ct4 = pandas.crosstab(sub3['FEMCAT'], sub3['COMP1v3']) print(ct4)

#column percentage colsum = ct4.sum(axis = 0) colpct = ct4 / colsum print(colpct)

#chi-square print('Chi-square value, p value, expected counts') cs4 = scipy.stats.chi2_contingency(ct4) print(cs4)

#Recode4 recode4 = {'Q1': 'Q1', 'Q5': 'Q5'} sub3['COMP1v4'] = sub3['internet5'].map(recode4)

#Contingency table of observed counts print('Observed counts and column percentage for recode4') ct5 = pandas.crosstab(sub3['FEMCAT'], sub3['COMP1v4']) print(ct5)

#column percentage colsum = ct5.sum(axis = 0) colpct = ct5 / colsum print(colpct)

#chi-square print('Chi-square value, p value, expected counts') cs5 = scipy.stats.chi2_contingency(ct5) print(cs5)

#Recode5 recode5 = {'Q2': 'Q2', 'Q3': 'Q3'} sub3['COMP1v5'] = sub3['internet5'].map(recode5)

#Contingency table of observed counts print('Observed counts and column percentage for recode5') ct6 = pandas.crosstab(sub3['FEMCAT'], sub3['COMP1v5']) print(ct6)

#column percentage colsum = ct6.sum(axis = 0) colpct = ct6 / colsum print(colpct)

#chi-square print('Chi-square value, p value, expected counts') cs6 = scipy.stats.chi2_contingency(ct6) print(cs6)

#Recode6 recode6 = {'Q2': 'Q2', 'Q4': 'Q4'} sub3['COMP1v6'] = sub3['internet5'].map(recode6)

#Contingency table of observed counts print('Observed counts and column percentage for recode6') ct7 = pandas.crosstab(sub3['FEMCAT'], sub3['COMP1v6']) print(ct7)

#column percentage colsum = ct7.sum(axis = 0) colpct = ct7 / colsum print(colpct)

#chi-square print('Chi-square value, p value, expected counts') cs7 = scipy.stats.chi2_contingency(ct7) print(cs7)

#Recode7 recode7 = {'Q2': 'Q2', 'Q5': 'Q5'} sub3['COMP1v7'] = sub3['internet5'].map(recode7)

#Contingency table of observed counts print('Observed counts and column percentage for recode7') ct8 = pandas.crosstab(sub3['FEMCAT'], sub3['COMP1v7']) print(ct8)

#column percentage colsum = ct8.sum(axis = 0) colpct = ct8 / colsum print(colpct)

#chi-square print('Chi-square value, p value, expected counts') cs8 = scipy.stats.chi2_contingency(ct8) print(cs8)

#Recode8 recode8 = {'Q3': 'Q3', 'Q4': 'Q4'} sub3['COMP1v8'] = sub3['internet5'].map(recode8)

#Contingency table of observed counts print('Observed counts and column percentage for recode8') ct9 = pandas.crosstab(sub3['FEMCAT'], sub3['COMP1v8']) print(ct9)

#column percentage colsum = ct9.sum(axis = 0) colpct = ct9 / colsum print(colpct)

#chi-square print('Chi-square value, p value, expected counts') cs9 = scipy.stats.chi2_contingency(ct9) print(cs9)

#Recode9 recode9 = {'Q3': 'Q3', 'Q5': 'Q5'} sub3['COMP1v9'] = sub3['internet5'].map(recode9)

#Contingency table of observed counts print('Observed counts and column percentage for recode9') ct10 = pandas.crosstab(sub3['FEMCAT'], sub3['COMP1v9']) print(ct10)

#column percentage colsum = ct10.sum(axis = 0) colpct = ct10 / colsum print(colpct)

#chi-square print('Chi-square value, p value, expected counts') cs10 = scipy.stats.chi2_contingency(ct10) print(cs10)

#Recode10 recode10 = {'Q4': 'Q4', 'Q5': 'Q5'} sub3['COMP1v10'] = sub3['internet5'].map(recode10)

#Contingency table of observed counts print('Observed counts and column percentage for recode10') ct11 = pandas.crosstab(sub3['FEMCAT'], sub3['COMP1v10']) print(ct11)

#column percentage colsum = ct11.sum(axis = 0) colpct = ct11 / colsum print(colpct)

#chi-square print('Chi-square value, p value, expected counts') cs11 = scipy.stats.chi2_contingency(ct11) print(cs11)

2.5 Graph proportions within explanatory category

''' Graph the percent of middling/high rates of female employment ( > 42%) within each category (quantile) for rates of internet usage '''

#Set variable types. First set explanatory variable to cat and response to num sub3['internet5'] = sub3['internet5'].astype('category') sub3['FEMCAT'] = pandas.to_numeric(sub3['FEMCAT'], errors = 'coerce')

#Bivariate bar chart, internet categories on x axis, mean for female employment #(proportion of 1's on y axis) seaborn.factorplot(x = 'internet5', y = 'FEMCAT', data = sub3, kind = 'bar',\ ci = None) plt.xlabel('Rate of internet usage grouped by quantile') plt.ylabel('proportion of middling/high rate of female employment')

3. Output

3.1 Chi Square Test of Independence (prior to post hoc)

3.2 Bonferroni Adjustment (p < 0.005)

3.3 Graph of proportions within categories

4. Interpretation

Prior to performing a Bonferroni Adjustment, the chi square test for independence revealed a chi square value of 13.2 (rounded), and a p value of 0.0103 with one degrees of freedom. 

This suggests we should reject a null hypothesis that the rate of female employment and the rate of internet usage are independent. We should therefore accept the alternate hypothesis that the rate of female employment and the rate of internet usage are dependent (statistically associated).

Since the null hypothesis was rejected, I performed a post hoc test using the Bonferroni Adjustment. The explanatory variable has been grouped into five categories. As a result, the post hoc test was performed across 10 pairwise comparisons. To reduce the risk of type one error, the p value for statistical significance was lowered to 0.005 (0.05/10).

Using the adjusted Bonferroni Adjustment post hoc comparison, none of the pairwise comparisons satisfies the threshold for statistical significance. This suggests the initial chi square test for independence was subject to overall type one error. We should therefore accept the null hypothesis that female employment and internet usage rates are independent.

[nmansoorn]

Chi Square Test

Here I am examining whether drug usage status is associated with gambling habit.

H0: Gambling is not associated with status of drug use

H1: Gambling is associated with status of drug use

Since the P value is 0.0024, we can conclude that status of drug use is associated with gambling. That is, if the person has drugs ever, or in the past 12 months, this raises the chances that they have also gambled. With this in mind, we can reject the null hypothesis saying there is no association between drug status and gambling, and accept the alternate hypothesis.

The Tables:

The Code Screenshot:

The Code Text:

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LIBNAME mydata "/courses/d1406ae5ba27fe300" access=readonly; DATA new; set mydata.nesarc_pds; LABEL DGSTATUS="Drug Use Status" S12Q1="Ever Gambled 5+ Times in Any One Year" S1Q6A="Highest Grade of Year of School Completed" SPOUSE="Respondent Has a Spouse" S3BQ1A9A="Ever Used Heroin" S3BD9Q2A="Age First Used Heroin" S3BD9Q2B="Used Heroin in Last 12 Months" S3BD9Q2C="How Often Used Heroin in Last 12 Months" S3BD9Q2E="How Often Used Heroin When Using the Most" AGEGROUP="Age First Used Heroin"; /* Just giving AGEGROUP a lable too */ IF S3BQ1A9A=9 THEN S3BQ1A9A=.; IF S12Q1=9 THEN S12Q1=.; IF S3BQ1A9A NE 9 AND S3BD9Q2B=. THEN S3BD9Q2B=11; IF S3BQ1A9A NE 9 AND S3BD9Q2A=. THEN S3BD9Q2A=99; /* Assiging 99 to avoide misrepresentation */ IF S3BD9Q2A NE 99 AND S3BD9Q2A LE 17 THEN AGEGROUP=1; /* Ages 17 and below as group 1*/ IF S3BD9Q2A NE 99 AND S3BD9Q2A GE 18 AND S3BD9Q2A LE 30 THEN AGEGROUP=2; /* Ages 18 - 29 as group 2 */ IF S3BD9Q2A NE 99 AND S3BD9Q2A GE 31 THEN AGEGROUP=3; /* Ages 30 and above as group 3 */ IF S3BQ1A9A NE 9 AND S3BD9Q2E=. THEN S3BD9Q2E=11; IF DGSTATUS LE 2; /* Retriving data only for current or ex- drug users */ IF AGE LE 45; /* Intrested in the age group of 45 and below */ PROC SORT; by IDNUM; /*PROC ANOVA; CLASS DGSTATUS; MODEL SPOUSE=DGSTATUS; MEANS DGSTATUS/DUNCAN;*/

PROC FREQ; TABLES S12Q1*DGSTATUS/CHISQ;

RUN;

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