It has long been argued that personality type is related to heart disease. Speci

Question

It has long been argued that personality type is related to heart disease. Specifically, type A people, who are competitive, driven, pressured, and impatient, are more prone to heart disease. On the other hand, type B individuals, who are less competitive and more relaxed, are less likely to have heart disease. [Friedman, M., & Rosenman, R. H. (1974). Type A behavior and your heart. New York, NY: Knopf. Suppose that an investigater would like to examine the relationship between personality type and disease. Fora random sample of individuals, personality type is assessed with a standardized test. These individuals are then examined and categorized as to whether they have a heart disorder. The observed frequencies are as follows: No Heart Disease Heart Disease Type A Type B 32 128 160 18 50 150 40 Do the data indicate that there is a relati ship between personality and disorder? Test with a-.05. Find the expected frequencies and compute the chi-square statistic. (Use two decimal places for fe and x2, three for each (fo - fo)/) Cell Type A-No Heart Disease Type A-Heart Disease Type B-No Heart Disease Type B-Heart Disease 32 18 128 22

Explanation / Answer

Given table data is as below MATRIX col1 col2 TOTALS row 1 32 18 50 row 2 128 22 150 TOTALS 160 40 N = 200 ------------------------------------------------------------------ calculation formula for E table matrix E-TABLE col1 col2 row 1 row1*col1/N row1*col2/N row 2 row2*col1/N row2*col2/N ------------------------------------------------------------------ expected frequecies calculated by applying E - table matrix formulae E-TABLE col1 col2 row 1 40 10 row 2 120 30 ------------------------------------------------------------------ calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above fo fe fo-fe (fo-fe)^2 (fo-fe)^2/fe 32 40 -8 64 1.6 18 10 8 64 6.4 128 120 8 64 0.533 22 30 -8 64 2.133 ^2 o = 10.666 ------------------------------------------------------------------ set up null vs alternative as null, Ho: no relation b/w X and Y OR X and Y are independent alternative, H1: exists a relation b/w X and Y OR X and Y are dependent level of significance, = 0.05 from standard normal table, chi square value at right tailed, ^2 /2 =3.841 since our test is right tailed,reject Ho when ^2 o > 3.841 we use test statistic ^2 o = (Oi-Ei)^2/Ei from the table , ^2 o = 10.666 critical value the value of |^2 | at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 2 - 1 ) = 1 * 1 = 1 is 3.841 we got | ^2| =10.666 & | ^2 | =3.841 make decision hence value of | ^2 o | > | ^2 | and here we reject Ho ^2 p_value =0.001 ANSWERS --------------- null, Ho: no relation b/w X and Y OR X and Y are independent alternative, H1: exists a relation b/w X and Y OR X and Y are dependent test statistic: 10.666 =10.66 critical value: 3.841 p-value:0.001 decision: reject Ho

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