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### 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

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/
```
Associated Topics:
College Probability

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