Runners and Heart Attacks
Date: 01/14/97 at 11:54:21 From: casey beck Subject: Statistics Hello, I have a statistics problem. I tried all last night to figure it out and I am just stumped. The problem is: The influence of running on preventing heart attacks has been studied by a local runners club. The following two-way table classifies 350 people as runners or non-runners, and whether or not they have had a heart attack. The variables are runner status (yes or no) and heart attack (yes or no). heart attack Yes No Runner yes 12 112 no 36 190 A. Of the two categorical variables, runner and heart attack status, which is the response variable and which is the explanatory variable? B. Calculate the marginal distributions of runner status and heart attack. C. Calculate the conditional distribution of heart attacks given runner status. D. Is there evidence for an association between runner status and heart attacks? Why or why not? (Give relevant percentages to support your answer.) I hope you can help me out. I greatly appreciate any time you put into it. Thanks, Casey Beck
Date: 01/29/97 at 18:24:33 From: Doctor Statman Subject: Re: Statistics Dear Casey, The response variable responds to changes in the explanatory variable. For example, running might help to explain the risk of heart attack, but heart attacks wouldn't help to explain exercise, under normal circumstances. Marginal distributions are found by adding across rows or columns, that is, by looking in the margins of the table. For example, the marginal probability that a person had a heart attack would be: (12+36)/350 = 0.137 (about 14 percent) Conditional distributions are found by considering only one row (or column) at a time. For example, the conditional probability that a person had a heart attack, given that the person is a runner, can be found by considering only the people who are runners: 12/(12+112) = 0.097 (about 10 percent) Telling you someone is a runner (a conditional distribution) reduces the probability of a heart attack from the marginal distribution value. Dr. Statman, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 02/03/97 at 07:48:54 From: Doctor Anthony Subject: Re: Statistics The contingency table is calculated as follows: Find the sums of rows and columns and then calculate the marginal probabilities assuming independence of running and heart attacks. The table is: Heart Attack Yes No Runner Yes 12 112 | 124 No 36 190 | 226 ACTUAL FREQUENCIES ----------------------- Totals 48 302 | 350 If you look at the totals in the margin of the table we see that out of 350 people, 48 had heart attacks and 302 did not. We also see that 124 were runners and 226 were not. Assuming independence of these factors, the probabilty that someone is a runner and has a heart attack would be: 48/350 * 124/350 The expected number would be 350 times this answer, giving: 48/350 * 124 = 17 Since marginal totals must agree, we can fill in the rest of the table for expected totals giving: Heart Attack Yes No Runner Yes 17 107 | 124 No 31 195 | 226 EXPECTED FREQUENCIES ---------------------- Totals 48 302 | 350 From this we see that the expected number of heart attacks for runners would be 17 (assuming independence from running) whereas in actual fact only 12 had heart attacks. So it seems that heart attacks can be avoided to some extent by running. We can do a Chi-squared test to see if our result is significant, and what the level of signifance is. I can complete the analysis, but this will require some knowledge of the Chi-squared test. Dr.Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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