Date: Mon, 22 May 2006 08:34:01 -0400
Reply-To: Edzard van Santen <evsanten@ACESAG.AUBURN.EDU>
Sender: "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From: Edzard van Santen <evsanten@ACESAG.AUBURN.EDU>
Subject: Re: Critical Value Question - Correction
The calculation I gave below should of course have been 19*(23.2/25)^2 = 16.36 since we are
dealing with variances. Thanks to Robert Vogel to pointing this out in a private email.
On Sun, 21 May 2006 18:45:05 -0400, Edzard van Santen <evsanten@ACESAG.AUBURN.EDU>
>One critical piece of information missing is whether you are considering a
>one- or two-tailed alternative. If you have two-tailed alternative the
>answer is none of the above. If you are considering a one-tailed alternative
>it's either(a) for the upper or (d) for lower. Given the scant information
>provided I would assume that the lower alternative is what you wanted. The
>ratio of two variances (n-1)*S is distributed as a Chisquare with n-1 df.
>Your test statistic would be 19*23.2/25 = 17.6, which is distributed as a
>CHI square with 19 df. One thing to remember is that confidence intervals
>based on Chi-square are not symmetrical.
>Snedecor, George W. and Cochran, William G. (1989), Statistical Methods,
>Eighth Edition, Iowa State University Press
>is a good source for further edification.
>On Sun, 21 May 2006 09:30:24 -0700, Matt <matthewamack@YAHOO.COM> wrote:
>>A statistician believes that the standard deviation of the weights of
>>firemen is less than 25 pounds. A sample of 20 firemen had a standard
>>deviation of 23.2 pounds. Assume that the variable is normally
>>distributed with a = 0.05. What is the critical value?
>>Any help with this would be greatly appreciated.