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### Testing the Difference of Two Averages

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Date: 05/27/98 at 23:26:39
From: David Delight
Subject: Statistics Problem

Dear Dr. Math,

This is a tough one. I need to understand how to compare these two
averages using the proper steps to do so. If you can help I would
greatly appreciated it.

Thank you.

The average age at death of women in the United States is 73.2 years.
A researcher is interested in determining if the average age of men
at death is the same as that of women. A sample of 25 deaths of men
shows an average age of 68.4 years and a standard deviation of 15
years. Test to determine if the average age at death of men is the
same as that for women. Use a significance level of .0l. What is
your conclusion?
```

```
Date: 05/28/98 at 01:53:57
From: Doctor Pat
Subject: Re: Statistics Problem

David,

There are several possible approaches to this problem, all
statistically appropriate. The use of "significance level" indicates
that you probably want to use a one sample T test. I hope that sounds
familiar to you.

Usually we begin by stating what is called the null hypothesis
(usually something we want to prove is false). In this case, we assume
that there is NO difference between the true age at death of men and
women. The assumption being that any difference would be explainable
due to "chance" variation in the sample of 25 we used. A different
sample would have a different mean and it probably would be different
than this one. So we want to find out how unusual would it be to get a
sample this far from the TRUE AVERAGE (true according to our null
hypothesis) just by chance.

To answer the "how unusual is it" question, we use the standard error
of the mean. You probably already know this is:

standard error = s/sqrt(n)

where s is the sample standard deviation of 15 and n is the sample
size of 25. This gives standard error = 3.

Again, I take a moment to tell you something you probably already
know. We know that the distribution of averages from samples of size
25 will be normally distributed with an average best expressed by the
average of the sample, x-bar and a standard deviation given by the
standard error. We can then use z scores to compute the probability
that a sample is this far from the expected. In this problem:

z = (x-bar - Mu)/(standard error)

where x-bar is the average for the sample, 68.4, Mu is the predicted
true average from our null hypothesis, the women's death rate of 73.2,
and the standard error is 3, the value we computed above. This gives
us z = -1.6. For a significance test we want to compare this to the
t-critical value from a .01 two-sided t-table of values with 24
degrees of freedom. My table gives 2.797 for a critical value, and
since the absolute value of our z score is less than the critical
value, we can NOT reject the null hypothesis. If you have software or
a calculator that will compute the p-value, you should get p = .123,
indicating that if the true average death age for men was really
the same as the women, we might get an average this far away on
one side or the other, in samples of size 25, in about 12 percent of
the random samples.

To test if men's death age is really lower than women's, you would do
a one-sided test of significance, and so you would use the critical
value for a .01 one-sided t-test, which is the same as a .02 two-sided
t-test. This is a slightly lower number, but we are still comparing
the same z-score, so there is not much else to do but compare the new
critical value.

You could also approach these two problems by building a two-sided (or
one-sided) 99 percent confidence interval for the men's average death
age. If this includes the women's average in the interval, we would
conclude no significant difference. I think that should happen in this
case. If you want to understand the interrelationship between
significance tests and confidence intervals you should do that
approach and compare the values you use along the way with the numbers
we have used here.

Hope that helps.

-Doctor Pat,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Statistics

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