From: phunt@interpac.net
Subject: Re: Comparing Mean
Date: Sat, 24 Apr 1999 13:56:46 GMT
Newsgroups: sci.math
To: pjlee@cyberway.com.sg
Keywords: DeMoivre-Laplace Limit Theorem

In article <7frke9$as2$1@nnrp1.dejanews.com>,
  pjlee@cyberway.com.sg wrote:
>
> If I remember it correctly in school, I am supposed to find the
> z-score with the null hypothesis as p-hat = 0.5? Let me illustrate
> with an example below:-
>
> Consider
> Yes     275
> No      290
>
> I want to know whether the 290 who indicated no is statistically
> significantly higher than the number who indicated yes.
>

I think you're saying that you expected your survey to produce
replies with probability 1/2, therefore you'll want to compare
your results with a normal distribution having meanN = 1/2.

The DeMoivre-Laplace Limit Theorem enables you to treat your
binomial random sample as if it had been drawn from a normal
distribution, because your sample size is large, and both p
and q are not very small.

For your sample:

    meanS = .4867

    varS  = 141.15 = n*p*q

    stdS  = 11.88 = sqrt(n*p*q)

    numS  = 565

For a normal distribution:

    meanN = .5

    stdN  = 11.88 (assumed)

    numN  = 565 (assumed)

The test:

    Ho: meanS = meanN
    H1: neanS =/= meanN

    alpha = .05

    critical value: z(0.025) = 1.96

    sample size: numS = numN = 565

A pooled estimate of p is (.4867 + .5)/2 = .4933.  then

z = (meanS - meanN)/[ sqrt(p*q*(1/numS + 1/numN)) ]

z = .447

Because (-1.96 < z=.447 < 1.96) accept Ho with confidence of
.95.  It appears that you may be able to accept Ho with even
greater confidence, but I will leave that test up to you.  In
any case, you will want to check my arithmetic before making
any decisions.

Good luck,

/ph

- - - - - - -

In article <7frke9$as2$1@nnrp1.dejanews.com>,
  pjlee@cyberway.com.sg wrote:
[original (see above) and followup messages quoted -- djr]

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From: phunt@interpac.net
Subject: Re: Comparing Mean
Date: Sun, 25 Apr 1999 16:42:12 GMT
Newsgroups: sci.math
To: pjlee@cyberway.com.sg

Hi,

In my last reply to you I assumed you wanted some assurance that
your binomial sample was consistent with a normal distribution
having a mean of 1/2.

Suppose instead that you suspect your sample was not drawn from
a normal with mean 1/2. You may indeed test for the probability
that your sample reflects a population having a true population
mean, call it meanP, that excludes 1/2 as an upper limit.

Again, as previously, the method used here depends on the truth
of the DeMoivre-Laplace Limit Theorem.

Let's say you want determine meanP within .01.  In this case,
,4767 < meanS < .4967, hence 1/2 would be excluded as possible
meanP, if meanS, the sample mean, happens to accurately reflect
meanP.  On the other hand, suppose meanP = 1/2, then the lower
limit for meanP - .01 = .5 - .01 = .49, hence meanS = .4867 is
excluded.

Using this approach you may say with some confidence that the
ratio of yes to no answers you obtained is significant.

In other words, if S is the number of successes in n trials,
then for confidence of .95, we are saying:

   [ -.01 <= (S/n) - p <= .01 ] = .95

where p is the true population proportion of successes.

To accomplish the above we want to make n so large that:

  PHI[.01 * sqrt(n/pq)] - PHI[-.01 * sqrt(n/pq)] >= .95.

  PHI[1.96] - PHI[-1.96] = .95

Therefore:

  .01 * sqrt(n/pq) >= 1.96

  sqrt(n/pq) >= 196

  n = p*q*(196)^2 = 38416*p*q

We don't know the true value of p yet, but we do know that
it is close to 1/2, and beside that we know for a fact that
p*q can never be greater than (1/2)^2 = 1/4.  Let p = 1/2.

  n = 38416*(1/4) = 9604

By increasing your sample size to 9604 you may say with .95
confidence that the yes/no proportion found in your initial
subsample truly is significant.

Good luck,

/ph

- - - - - -

In article <7frke9$as2$1@nnrp1.dejanews.com>,
  pjlee@cyberway.com.sg wrote:
[original quoted again -- djr]

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