Large Sample Estimation of a Population Proportion

Statistics
Tutor: None Selected Time limit: 1 Day

Mar 12th, 2015

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

822/1250 = 0.6576

b) 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 = 1249. The critical t-stat for a 98% confidence interval will be:

t_crit = t.inv(p=0.99,df=1249) = 2.329337192

The sample standard deviation is given by sqrt(p(1-p)) = sqrt(0.6576(1-0.6576)) 

0.474512634

The standard error is sample sd / sqrt(n) = 0.474512634/sqrt(1250)

SEM = 0.013421244

So the 98% confidence interval for the sample mean is given by 

sample mean +/- t_crit x SEM

= 0.6576 +/- 2.329337192 x 0.013421244

= [0.6263 to 0.6889]


Mar 12th, 2015

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