/* Population size: 29.477.792
/* Parametros:
N P Q P*Q Z_95% Z2_95% E_ 5% E2_5%
29477792 0,5 0,5 0,25 1,9600 3,8416 0,05 0,0025
/* Formula : n = N*(p*q)*Z2 / (p*q)*Z2 + (N-1)*E2 */
How can I do a program to calculate the sample size, according to this formula ???
data want; n=29477792; p= 0.5; q= 0.5; pq = p*q; z_95 = 1.96; z2_95 = 3.8416; e_5 = .05; e2_5 = 0.0025; nwant = n*pq*z2_95/pq*z2_95 +(n-1)*e2_5; run;
But I think you need to verify your "formula".
Note I use . instead of , as that is the NLS setting for my system. You should be able to reverse things easy enough.
data want; n=29477792; p= 0.5; q= 0.5; pq = p*q; z_95 = 1.96; z2_95 = 3.8416; e_5 = .05; e2_5 = 0.0025; nwant = n*pq*z2_95/pq*z2_95 +(n-1)*e2_5; run;
But I think you need to verify your "formula".
Note I use . instead of , as that is the NLS setting for my system. You should be able to reverse things easy enough.
Hi, Thank you so much!
It's worked.
data want;
n=10000;
p=0.5;
q=0.5;
pq = p*q;
z_95 = 1.96;
z2_95 = 3.8416;
e_5 = 0.05;
e2_5 = 0.0025;
nwant = (n*pq*z2_95) / ((pq*z2_95) + ((n-1)*e2_5));
run;
Yea I agree I think you need to verify your formula. I think what you are trying to do is:
n = N*X / (X + N – 1),
where,
X = Z(α/2)^2 *p*(1-p) / MOE^2,
and Zα/2 is the critical value of the Normal distribution at α/2 (e.g. for a confidence level of 95%, α is 0.05 and the critical value is 1.96), MOE is the margin of error, p is the sample proportion, and N is the population size. Note that a Finite Population Correction has been applied to the sample size formula.
I think it would look something like this
n=(N*((Z^2 *p*(1-p)) / MOE^2)) / (((Z^2 *p*(1-p)) / MOE^2)+N+1)
Thank you. Thank you.
You really helped me.
you're so right. that's the 'exact format' of formula.
before,I guess that I didn't express right.
sorry about my english 😉
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