STAT 3120 Statistical Methods I Lecture 8 Chi-Square 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?) 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: Promoted Not Promoted Total Male 40 80 120 Female 10 70 80 Total 50 150 200 STAT3120 – Chi Square 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? 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 ChiSquare. 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 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. STAT3120 – Chi Square Going back to our example, the data provided is “observed”: Promoted Not Promoted Total Male 40 80 120 Female 10 70 80 Total 50 150 200 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… Promoted Male Female Total Not Promoted Total 120*.25 = 30 120*.75 = 90 120 80*.25 = 20 50 80*.75 = 60 150 80 200 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)+((7060)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?pa ge=merron/081208) STAT3120 – Chi Square January February March April May June July August September October November December 51 46 61 49 46 49 36 41 36 34 33 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. 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. 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 $250 Determine ‘Scale’ • Nominal variables: – Values with no logical ordering. » Gender • Ordinal variables: – Variables have values with a logical ordering. » Income Lets Examine!? • Determine distribution of categorical values • Recognize possible associations among variables • Association ? – Two variables when one level or value of the other changes. – No changes? Distribution of the variable is the same regardless of the level of the other variable Determine Association • No Association? – Statistic professor temperament changes with golf. Great golf Sunshine Raining Bad Golf 65% 35% 65% 35% Watch Out! • Association? – Statistic professor temperament changes with golf. Great golf Sunshine Raining Bad Golf 95% 5% 30% 70% Crosstabulation Table • Table shows the number of observations for each combination of the row and the column variables Column 1 Column 1 … Column 1 Row 1 Row 2 Cell11 Cell12 … Cell1c Cell12 Cell22 … Cell2c … Row r … … … … Cellr1 Cellr2 … Cellrc - Frequency: nbr of observations falling into a category formed by row variable and column variable Percent: nbr of observations in each cell as a percentage of the total nbr of observations Row percent: nbr of observations in each cell as a percentage of the total nbr of observations in that row col percent: nbr of observations in each cell as a percentage of the total nbr of observations in that column Distributions • SAS Freq procedure – Examine distributions – Ordering values SAS Proc Freq Distributions • • • • • libname JLLP 'E:\JenniferPriestly\Chi_Square'; %let outpath=E:\JenniferPriestly\Chi_Square; %let libpath=E:\JenniferPriestly\Chi_Square; options nodate nonumber ls=95 ps=80; run; • • • • • • • • • • • • • • • • • • • • • • Proc format; • • ods rtf close; ods listing; value purfmt 1 = "$ 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; 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 graphics on; 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 graphics on; 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 Purchase Gender Purchase Frequenc y Percent Row Pct $ 100 Col Pct < $100 + Total Female 139 101 240 32.25 23.43 55.68 57.92 42.08 51.67 62.35 Male 130 61 191 30.16 14.15 44.32 68.06 31.94 48.33 37.65 Total 269 162 431 62.41 37.59 100.00 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-by2 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 Purchase Frequency Expected Cell Chi-Square Row Pct < $100 $ 100 + Total Female 139 101 240 149.79 90.209 0.7774 1.2909 57.92 42.08 Male 130 119.21 0.9769 68.06 61 71.791 1.6221 31.94 191 Total 269 162 431 Calculate cell Chi-square Results Statistic Chi-Square Likelihood Ratio Chi-Square Continuity Adj. Chi-Square Mantel-Haenszel Chi-Square Phi Coefficient Contingency Coefficient Cramer's V DF 1 1 1 1 Value 4.6672 4.6978 4.2447 4.6564 -0.1041 0.1035 -0.1041 Prob 0.0307 0.0302 0.0394 0.0309 Fisher's Exact Test Cell (1,1) Frequency (F) 139 Left-sided Pr <= F 0.0195 Right-sided Pr >= F 0.9883 Table Probability (P) Two-sided Pr <= P 0.0078 0.0355 Estimates of the Relative Risk (Row1/Row2) Type of Study Value 95% Confidence Limits Case-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 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. Gender by Purchase