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> wrote:

>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. > >Good ol' > >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? >> >>a. 31.410 >>b. 30.144 >>c. 10.851 >>d. 10.117 >> >>Any help with this would be greatly appreciated. >> >>Best wishes, >> >>Matt