Lecture 8 Chi-Square STAT 3120 Statistical Methods I

Lecture 8Chi-Square

STAT 3120Statistical Methods I

STAT3120 – Chi Square

Dependent Variable

Independent (predictor) Variable

Statistical Test

Comments

Quantitative Categorical T-TEST (one, two or paired sample)

Determines if categorical variable (factor) affects dependent variable; typically used for experimental or planned change studies

Quantitative Quantitative Correlation/Regression Analysis

Test establishes a regression model; used to explain, predict or control dependent variable

Categorical Categorical Chi-Square Tests if variables are statistically independent (i.e. are they related or not?)


When presented with categorical data, one common method of analysis is the “Contingency Table” or “Cross Tab”. This is a great way to display frequencies -

For example, lets say that a firm has the following data:

120 male and 80 female employees

40 males and 10 females have been promoted


Using this data, we could create the following 2x2 matrix:

Promoted Not Promoted

Total

Male 40 80 120

Female 10 70 80

Total 50 150 200


Now, a few questions…

1)From the data, what is the probability of being promoted?2)Given that you are MALE, what is the probability of being promoted? 3)Given that you are promoted, what is the probability that you are MALE? 4)Given that you are FEMALE, what is the probability of being promoted? 5)Given that you are promoted, what is the probability that you are female?


The answers to these questions help us start to understand if promotion status and gender are related.

Specifically, we could test this relationship using a Chi-Square. This is the test used to determine if two variables are related.

The relevant hypothesis statements for a Chi-Square test are:

H0: Variable 1 and Variable 2 are NOT RelatedHa: Variable 1 and Variable 2 ARE Related

Develop the appropriate hypothesis statements and testing matrix for the gender/promotion data.


The Chi-Square Test uses the Χ2 test statistic, which has a distribution that is skewed to the right (it approaches normality as the number of obs increases). You can see an example of the distribution on pg 641.

The Χ2 test statistic calculation can be found on page 640.

The observed counts are provided in the dataset.The expected counts are the counts which would be expected if there was NO relationship between the two variables.



Total

Male 40 80 120

Female 10 70 80

Total 50 150 200

Going back to our example, the data provided is “observed”:

What would the matrix look like if there was no relationship between promotion status and gender? The resulting matrix would be “expected”…


From the data, 25% of all employees were promoted. Therefore, if gender plays no role, then we should see 25% of the males promoted (75% not promoted) and 25% of the females promoted…


Total

Male 120*.25 = 30 120*.75 = 90 120

Female 80*.25 = 20 80*.75 = 60 80

Total 50 150 200

Notice that the marginal values did not change…only the interior values changed.


Now, calculate the X2 statistic using the observed and the expected matrices:

((40-30)2/30)+((80-90)2/90)+((10-20)2/20)+((70-60)2/60) =

3.33+1.11+5+1.67 = 11.11

This is conceptually equivalent to a t-statistic or a z-score.

To determine if this is in the rejection region, we must determine the df and then use the table on page 732.

Df = (r-1)*(c-1)…

In the current example, we have two rows and two columns. So the df = 1*1 = 1.

At alpha = .05 and 1df, the critical value is 3.84…our value of 11.11 is clearly in the reject region…so what does this mean?



From the book Outliers, Malcolm Glidewell makes the point that the month in which a boy is born will determine his probability of playing in the NHL.

The months of birth for players in the NHL are on the next page…

(data taken from http://sports.espn.go.com/espn/page2/story?page=merron/081208)

January 51 February 46

March 61 April 49 May 46 June 49 July 36

August 41 September 36

October 34 November 33 December 30


Now, if there is NO relationship between birth month and playing hockey, what SHOULD the distribution of months look like?

Lets do this one in EXCEL…

Note that this is technically referred to as a “goodness of fit” test – where we are assessing if the actual distribution “fits” what would be expected.


Practice Problems for Chi-Square:

15.5515.5615.5715.58

For all of these, identify the hypothesis statements, the testing matrix, and the decision.

Categorical Example

Using credit data.

Credit

• Sample Data Set– Purchase: $: 1=$250+, 0=<$250– Age: Customer Age– Gender: male,female– Income: Low, Medium, High

What do we have?

• Predictors• Gender

• Income

• Age

• Outcome• GT $250

• LT $250

Determine ‘Scale’

• Nominal variables: – Values with no logical ordering.

» Gender

• Ordinal variables:– Variables have values with a logical ordering.

» Income

Lets Examine!?

• Determine distribution of categorical values• Recognize possible associations among variables

• Association ?– Two variables when one level or value of the

other changes.– No changes? Distribution of the variable is the

same regardless of the level of the other variable

Determine Association• No Association?

– Statistic professor temperament changes with golf.

Great golf Bad Golf

Sunshine

Raining65% 35%

65% 35%

Watch Out!

• Association?– Statistic professor temperament changes with

golf. Great golf Bad Golf

Sunshine

Raining95% 5%

30% 70%

Crosstabulation Table

• Table shows the number of observations for each combination of the row and the column variables

Column 1 Column 1 … Column 1

Row 1

Row 2

…

Row r

- Frequency: nbr of observations falling into a category formed by row variable and column variable

- Percent: nbr of observations in each cell as a percentage of the total nbr of observations

- Row percent: nbr of observations in each cell as a percentage of the total nbr of observations in that row

- col percent: nbr of observations in each cell as a percentage of the total nbr of observations in that column

Cell11 Cell12 … Cell1c

Cell12 Cell22 … Cell2c

… … … …

Cellr1 Cellr2 … Cellrc

Distributions

• SAS Freq procedure– Examine distributions– Ordering values

SAS Proc Freq Distributions• libname JLLP 'E:\JenniferPriestly\Chi_Square';

• %let outpath=E:\JenniferPriestly\Chi_Square;

• %let libpath=E:\JenniferPriestly\Chi_Square;

• options nodate nonumber ls=95 ps=80;

• run;

• Proc format;

• value purfmt 1 = "$ 100 +"

• 0 = "< $100"

• ;

• Run;

• ods graphics on;

• ods listing close;

• ods Rtf path="&outpath"

• style=journal

• file='freq.rtf';

• proc freq data=JLLP.Online;

• tables purchase gender income

• gender*purchase income*purchase /

• plots(only)=(freqplot);

• format purchase purfmt.;

• run;

• ods select histogram probplot;

• proc univariate data=JLLP.Online;

• var age;

• histogram age / normal (mu=est sigma=est);

• probplot age / normal (mu=est sigma=est);

• run;

• ods rtf close;

• ods listing;

SAS Ordering Values

• Change Income• ods graphics on;

• ods listing;

• data JLLP.Online_inc;

• set JLLP.Online;

• if income='Low' then IncLevel=1;

• else if income='Medium' then IncLevel=2;

• else if income='High' then IncLevel=3;

• run;

• proc format;

• value incfmt 1='Low Income'

• 2='Medium Income'

• 3='High Income';

• run;


• ods rtf path="&outpath"

• style=statistical

• file='freq2.rtf';

• proc freq data=JLLP.Online_inc;

• tables IncLevel*Purchase;

• format IncLevel incfmt. Purchase purfmt.;

• title1 'Change Variable IncLevel to Correct Income';

• run;

• ods rtf close;

Tests for Association

• Determine– Chi-square test for association– Examine strength of the association– Calculate exact p-value– Cramer’s V

Chi-Square Test


• ods rtf path="&outpath"

• style=statistical

• file='freq3.rtf';

• proc freq data=JLLP.Online_inc;

• tables Gender*purchase

• / chisq expected cellchi2 nocol nopercent

• relrisk;

• format purchase purfmt.;

• Title1 'Association Between Gender and Purchase';

• run;

• ods rtf close;

Gender by Purchase

Table of Gender by PurchaseGender Purchase

FrequencyPercentRow PctCol Pct < $100

$ 100 + Total

Female 13932.2557.9251.67

10123.4342.0862.35

24055.68

Male 13030.1668.0648.33

6114.1531.9437.65

19144.32

Total 26962.41

16237.59

431100.00

Table of Gender by PurchaseGender Purchase

FrequencyPercentRow PctCol Pct < $100

$ 100 + Total

Female 13932.2557.9251.67

10123.4342.0862.35

24055.68

Male 13030.1668.0648.33

6114.1531.9437.65

19144.32

Total 26962.41

16237.59

431100.00

Chi-Square Test

• No association• Observed frequencies=expected frequencies

– Null Hypothesis:• No association between Gender and Purchase

• Probability of purchasing items more than $100 is the same for both sexes.

• Association• Observed frequencies≠expected frequencies

– Alternative Hypothesis:• There is an association between Gender and Purchase

• Probability of purchasing items more than $100 is the same for both sexes.

Pearson Chi-square Test

• Commonly used test to determine whether there is association between 2 categorical values

• Test measure the difference between the observed cell frequencies and the cell frequencies that are expected if there is no association between the variables

• Significant test statistic, strong evidence an association exists

Frequencies Calculation

• Expected frequencies are calculated by:» (row total * column total) / sample size

No association between Row and Column variable the expected percentage in any R*C will be equal to the percentage in that cell rows (R/T) times the percentage in the cell column (C/T). The expected percentage times the total sample size.

Expected count=(R/T)*(C/T)*T=(R*C)/T

Chi-square tests

• Measures of association

– P-value tests only indicates how confident you can be that the null hypothesis if no association exists.

– Cramer’s V statistics: measures association between two nominal variables. Range from -1 to 1 for a 2-by-2 table. 0 to 1 for larger tables. Values further from 0 indicate the presence of a relativity strong association.

– Odds Ratios indicates how much more likely, with the respect to odds a certain event occurs in one group relative to its occurrence in another group.

Odds Ratio

Probability of odds of an outcome

No Yes Total

Group A 20 60 80

Group B 10 90 100

Total 30 150 180

Prob of Yes outcome in Group B = 90/100 (.90)

Prob of a No Outcome in Group B = 10/100 (.10)

Odds Ratio

• Odds of outcome in Group B» .90 / .10 = 9

• Odds of outcome in Group A» .75 / .25 = 3

• Odds Ratio of Group B to Group A» 9 / 3 = 3

Odds ratio of Group B to Group A is 3 times .

Properties of the Odds Ratio, B to A

• Odds ratio shows strength of association.– If odds ration is 1 then there is no association– If odds ratio is greater than 1then Grp B is

more likely to have the outcome.– If odds ratio is less than 1 then Grp A is more

likely to have the outcome

Example

• Determine association between Gender and purchase.

• Generate expected cell frequencies and the cell’s contribution to the total chi-square statistic

Results

Table of Gender by Purchase

Gender PurchaseFrequencyExpectedCell Chi-SquareRow Pct < $100 $ 100 + TotalFemale 139

149.790.777457.92

10190.2091.290942.08

240

Male 130119.210.976968.06

6171.7911.622131.94

191

Total 269 162 431

Calculate cell Chi-square

Results

Statistic DF Value ProbChi-Square 1 4.6672 0.0307

Likelihood Ratio Chi-Square 1 4.6978 0.0302

Continuity Adj. Chi-Square 1 4.2447 0.0394

Mantel-Haenszel Chi-Square 1 4.6564 0.0309

Phi Coefficient -0.1041

Contingency Coefficient 0.1035

Cramer's V -0.1041

Fisher's Exact TestCell (1,1) Frequency (F) 139

Left-sided Pr <= F 0.0195

Right-sided Pr >= F 0.9883

Table Probability (P) 0.0078

Two-sided Pr <= P 0.0355

Estimates of the Relative Risk (Row1/Row2)

Type of Study Value 95% Confidence LimitsCase-Control (Odds Ratio) 0.6458 0.4339 0.9612

Cohort (Col1 Risk) 0.8509 0.7360 0.9839

Cohort (Col2 Risk) 1.3177 1.0214 1.7000

P-value is 0.0307<.05 , reject the Null hypothesisAppendix A.5: .05<p-value<.025

Cramer’s V indicates association is relatively weak. Relative Risk at 95% CI that Males in the right column (+100) compared to Females has value of .6458. Males has a 65% odds of purchasing more then $100Odds ratio (OR-1)*100, (0.6458-1)*100=-35.42%, males have a 35.42% lower odds than females.

Gender by Purchase

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Lecture 8 Chi-Square STAT 3120 Statistical Methods I