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Testing the Difference of Two Averages

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


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!   
Associated Topics:
High School Statistics

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