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))
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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