- 73 Views
- Uploaded on
- Presentation posted in: General

STAT 3120 Statistical Methods I

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

STAT 3120Statistical Methods I

Lecture 8

Chi-Square

STAT3120 – Chi Square

STAT3120 – Chi Square

- 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

STAT3120 – Chi Square

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

STAT3120 – Chi Square

- Now, a few questions…
- From the data, what is the probability of being promoted?
- Given that you are MALE, what is the probability of being promoted?
- Given that you are promoted, what is the probability that you are MALE?
- Given that you are FEMALE, what is the probability of being promoted?
- Given that you are promoted, what is the probability that you are female?

STAT3120 – Chi Square

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 Related

Ha: Variable 1 and Variable 2 ARE Related

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

STAT3120 – Chi Square

The Chi-Square Test uses the Χ2test 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 Χ2test 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.

STAT3120 – Chi Square

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”…

STAT3120 – Chi Square

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…

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

STAT3120 – Chi Square

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.

STAT3120 – Chi Square

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?

STAT3120 – Chi Square

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)

STAT3120 – Chi Square

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.

STAT3120 – Chi Square

Practice Problems for Chi-Square:

15.55

15.56

15.57

15.58

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

Categorical Example

Using credit data.

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

- Predictors
- Gender
- Income
- Age

- GT $250
- LT $250

- Nominal variables:
- Values with no logical ordering.
- Gender

- Values with no logical ordering.

- Variables have values with a logical ordering.
- Income

- 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

- No Association?
- Statistic professor temperament changes with golf.
Great golf Bad Golf

Sunshine

Raining

- Statistic professor temperament changes with golf.

- Association?
- Statistic professor temperament changes with golf.
Great golf Bad Golf

Sunshine

Raining

- Statistic professor temperament changes with golf.

- 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

- SAS Freq procedure
- Examine distributions
- Ordering values

- 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;
- Procformat;
- value purfmt 1 = "$ 100 +"
- 0 = "< $100"
- ;
- Run;
- ods graphics on;
- ods listing close;
- ods Rtf path="&outpath"
- style=journal
- file='freq.rtf';
- procfreq data=JLLP.Online;
- tables purchase gender income
- gender*purchase income*purchase /
- plots(only)=(freqplot);
- format purchase purfmt.;
- run;
- ods select histogram probplot;
- procunivariate 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;

- 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;
- procformat;
- value incfmt 1='Low Income'
- 2='Medium Income'
- 3='High Income';
- run;
- ods graphics on;
- ods rtf path="&outpath"
- style=statistical
- file='freq2.rtf';
- procfreq data=JLLP.Online_inc;
- tables IncLevel*Purchase;
- format IncLevel incfmt. Purchase purfmt.;
- title1 'Change Variable IncLevel to Correct Income';
- run;
- ods rtf close;

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

- ods graphics on;
- ods rtf path="&outpath"
- style=statistical
- file='freq3.rtf';
- procfreq 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;

- 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.

- Observed frequencies≠expected frequencies

- There is an association between Gender and Purchase
- Probability of purchasing items more than $100 is the same for both sexes.

- 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

- 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

- (row total * column total) / sample size

- 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.

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

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

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

- .75 / .25 = 3

- 9 / 3 = 3
Odds ratio of Group B to Group A is 3 times .

- 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

- Determine association between Gender and purchase.
- Generate expected cell frequencies and the cell’s contribution to the total chi-square statistic

Calculate cell Chi-square

P-value is 0.0307<.05 , reject the Null hypothesis

Appendix 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 $100

Odds ratio (OR-1)*100, (0.6458-1)*100=-35.42%, males have a 35.42% lower odds than females.