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 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/
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