Re: Unusual formulae for confidence intervals

Just to clarify, I wrote exactly what I meant.


Ray Koopman wrote:

Stephen J. Herschkorn wrote:


One of my tutees is taking a course in statistics for public policy. I
think this is my first such client. (I have had other studetns in
psychology, sociology, and public health.)

As we were discussing confidence intervals, he pointed out two formulae
in his text and notes. I had never encountered these before.

- For a population proportion, use as the standard error 0.5 /
sqrt(n), where n is the sample size. The justification was that
since we do not know the population proportion pi, use pi = 0.5 for a
conservative interval estimate. Most formulae I have seen substitute
the sample proportion Q (my notation) to get sqrt(Q (1-Q) / n) for
the standard error. If one is going to be a stickler, one can solve the
quadratic inequality
-z <= (Q - pi) / sqrt( pi (1- pi) / n) <= z for explicit bounds on
pi. So using 0.5 seems silly to me.



That's for choosing the sample size (before the data are collected)
so that the standard error is guaranteed to be small enough.



No, they explicity said to use 0.5 / sqrt(n) for all confidence intervals for the proportion. Actually, more precisely, they cicuitously said use sqrt(P_u (1-P_u) / n), where P_u is the population proportion, and set P_u = 0.5 for a conservative estimate. I am familiar with the standard techniques for determining sample size; the book was not discussing this issue.



- For a population mean with a large sample size n, use S /
sqrt(n-1), where S is the sample standard deviation, as the standard
error with a normal distribution. I have always seen S / sqrt(n).
Personally, I always use S / sqrt(n) with the t distribution, since
computers can compute t for any degrees of freedom. (Oddly, though
the textbook discusses hypothesis testing with small samples, it does
not discuss confidence intervals with small samples.)



How did they define S^2? With n in the denominator?
Maybe the important part of that recommendation is supposed to be
the substitution of the normal for the t distribution.



S^2 was defined as usual, i.e., sum(i, (X_i - Xbar)^2) / (n-1). The part I am questioning here is using sqrt(n-1) in place ofr sqrt(n) for the denonmiator of the standard error. Again, I am very familiar with using normal in place of t for large sample sizes.



Have you seen these practices elsewhere? Are these conventions peculiar
to public policy?



Thanks for your response, RK, but you really didn't address my question.


--
Stephen J. Herschkorn sjherschko@xxxxxxxxxxxx
Math Tutor on the Internet and in Central New Jersey and Manhattan

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