Large Sample Estimation of a Population Proportion

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Mar 12th, 2015

a) The best point estimate is the sample proportion, which is:

42/900 = 0.046666667

b) The sample size is 900, so it's certainly large enough to construct a confidence interval... it sounds as though you're being asked to quote a "rule-of-thumb" you may have been taught though, so check your notes for that.

c) Since we don't know the population standard deviation, only that for the sample, we should use a t-distribution here, with df = n-1 = 899. The critical t-stat for an 80% confidence interval will be:

t_crit = t.inv(p=0.9,df=899) = 1.282493968

The sample standard deviation is given by sqrt(p(1-p)) = sqrt(0.046666(1-0.046666)) 

0.210923894

The standard error is sample sd / sqrt(n) = 0.210923894/sqrt(900)

SEM = 0.007030796

So the 95% confidence interval for the sample mean is given by 

sample mean +/- t_crit x SEM

= 0.046666667 +/- 1.282493968 x 0.007030796

= [0.0376 to 0.0557]


Mar 12th, 2015

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