Statistics questions

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people who black out at G forces greater than 6. In an earlier study the population proportion was estimated at 0.45. How large of a sample would be required to estimate people who black out at 6 or more Gs at the 90% confidence level with error 0.03

Mar 4th, 2015

Here we first find the critical value of Z for a 90% confidence interval (0.05 to 0.95) which is the value at p = 0.95 of the unit normal cdf:

z_crit = N.inv ( p = 0.95, 0, 1 ) = 1.65

Now, generally, the required sample size n would be given by: 

n = [ (z_crit x s.d.) / s.e.m ]^2 = [1.65sd/0.03]^2 = 3006sd^2

So in order to calculate the required n, you would need to know the standard deviation of your sample, which isn't included in your question.

But you have a special case here, where you're measuring a 'known' proportion (45% in this case). That means we can calculate:

n = (z_crit^2 × P(1 – P)) / error^2 where P  is the proportion you're trying to estimate (0.45):

n = ( 1.65^2 x 0.45(0.55) ) / 0.03^2

= 744

Hope this helped!


Mar 4th, 2015

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