Problem 2. Permutation Chi-square Test Typically, we test for an association bet

Question

Problem 2. Permutation Chi-square Test Typically, we test for an association between categorical variables using the chi-square test, which measures how much the counts in each cell differ from what we would expect the counts to be if the nul hypothesis were true. However, if the expected counts is less than 5 for some cells in the table, then this chi-square distribution may not be accurate Consider the following example on baldness and myocardial infarction, self-assessed bald- ness. A study was completed to learn about the relationship between male-pattern baldness and the risk of cardiovascular disease. Cases were men less than 55 years of age hospitalized for a heart attack. Cases were excluded if they had a prior history of serious heart prob- lems. Controls were men with no history of heart disease admitted to the same hospital for nonfatal, non-cardiac problems. The data is displayed in the table below: one Little Some Much xtreme 663 772 1435 isease 251 165 19550 Control 331 221 18 3 582 386 380 84 The data are from Lesko, Rosenberg, and Shapiro (1993). A case-control study of baldness in relation to myocardial infarction in men. JAMA, 269(8), 998-1003. Question: Is there an association between between baldness and risk of cardiovascular dis- ease? To access this create your own permutation distribution for chi-square and test the hypothesis of independence.

Explanation / Answer

here null hypothesis: baldness and cardiovascular disease are independent/

altetrnate hypothesis:baldness and cardiovascular disease are dependent/

applying chi sqaure test for independence:

here degree of freedom =(row-1)*(column-1) =(2-1)*(5-1)=4

for 4 degree of freedom and above test stat p value =0.0057

as p value is significantly low we reject null hypotheiss and conclude that baldness and cardiovascular disease are dependent/

Observed O None little some much extreme Total disease 251 165 195 50 2 663 control 331 221 185 34 1 772 Total 582 386 380 84 3 1435 Expected E=rowtotal*column total/grand total None little some much extreme Total disease 268.90 178.34 175.57 38.81 1.39 663 control 313.10 207.66 204.43 45.19 1.61 772 Total 582 386 380 84 3 1435 chi square =(O-E)^2/E None little some much extreme Total disease 1.1911 0.9979 2.1508 3.2265 0.2719 7.838 control 1.0229 0.8570 1.8471 2.7710 0.2335 6.731 Total 2.214 1.855 3.998 5.998 0.505 14.570

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