In this situation, there are different formula for the mean and standard error, but the same logic and procedure for constructing confidence intervals remains the same.
Another example shows how to apply these ideas to an estimation problem.
mean() = p
and SE() =
The shape will be approximately normal for sufficiently large samples. A general rule of thumb is that if np > 5 and n(1-p) > 5, then the distribution will be approximately normal. The techniques of this section should not be used to construct confidence intervals if the sample size is too small.
Our sample proportion is 18/1000 = .018. We need to check that np is at least 5.
1000 * .018 = 18 > 5(This will hold whenever there are at least 5 successes and failures in the sample.)
We may conclude that the sampling distribution for the sample proportion will be approximately normal, by the central limit theorem.
We estimate the SE to by replacing p by in the SE formula .
sqrt( (.018)(.982)/1000 ) = .00420From our large sample size, we may use a reliability coefficient from the normal distribution.
.018 +/- (1.96)(.00420)or
.018 +/- .008
Bret Larget, firstname.lastname@example.org