Date: 04/12/97 at 15:49:05 From: Anonymous Subject: P-Value Calculation? Can you please tell me how I would calculate a p value. I have made a contingency table of two populations, diseased and controls, that do or do not have a given mutation. I was able to determine the chi-square value, but I do not know how to determine the p-value, or if I can even do that. Any ideas? I do not know the presumed variance. Thanks, Dr. Cindy McGrath --a resident physician who cannot remember how to do this.
Date: 04/12/97 at 22:27:36 From: Doctor Steven Subject: Re: P-Value Calculation? I'm assuming you are trying to test a hypothesis. A p-value is the probability that your hypothesis is true and you get results that are equal to or farther away from your results. Depending on your level of significance you might find that you should either reject or fail to reject your original hypothesis. (Many testings use a level of significance of .05, but each test might need a more stringent level of significance.) If your p-value is smaller than the level of significance, you should reject your original hypothesis. If it is larger than your level of significance you should fail to reject your hypothesis. Here's an example using normal distributions from a Statistics text I used in college: Test the hypothesis that the average content of containers of a particular lubricant is 10 liters if the contents of a random sample of 10 containers are 10.2, 9.7, 10.1, 10.3, 10.1, 9.8, 9.9, 10.4, 10.3, and 9.8 liters. Use a .01 level of significance and assume the distribution of contents is normal. Setup: Our hypothesis is that the content of a container is 10 liters. u = 10. Our alternative is that the contents of a container is not 10 liters (u does not equal 10). X = 10.06 S = .2459 a = .01 n = 10. Scratch: T = (X - u)/(S/Sqrt(n)) = (10.06 - 10)/(.2459/3.1623) = .06/.0777 = .7717. Since our alternative is two-sided we check for values both above and below our hypothesis so we need the probability that |T| >= .7717, and we use the Student t-distribution with 9 degrees of freedom. p = P(|T| >= .7717) = 2*P(T >= .7717) = somewhere in the range (.4, .6) (sorry, the table in the book is not very complete). So our p-value is in the range (.4, .6). Results and Conclusions: Our p-value in the range of (.4, .6) is not significant at the .01 level of significance. Based on this information we fail to reject our hypothesis. We conclude that there is not enough evidence to state the average contents of the containers is anything but 10 liters. Hope this helps. -Doctor Steven, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 04/14/97 at 16:04:09 From: Anonymous Subject: Re: P-Value Calculation? My sister placed her inquiry with my email address. I passed your response to her. She asked me to express her gratitude. She was preparing all weekend for an hour's presentation to be given on Monday and could not find the information she needed from her old college stat text. You saved her! She was very impressed by your Web site and service. So thank you very much for your time and attention.
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