I'm trying to understand confidence intervals but having some trouble. I've been doing some exercises I found online and I'm stuck on this question:

I have been given a 95% confidence interval for a population proportion: (0.35, 0.40), a sample size of 200, and I need to find a 99% confidence interval. The methods I would go to first involve using standard deviation, which I don't have.

How can I approach this question without knowing variance? The whole quiz is about normal distributions, to give it some context.

This isn't really homework but I'm tagging it this way because I'm looking for a similar outcome. The quiz has answers at the end; what I want is to be able to solve it myself.

  • $\begingroup$ Maybe you could start by writing down the expressions that gave the 0.35 and 0.40. $\endgroup$ – mark999 Jun 16 '11 at 8:03
  • $\begingroup$ This is all the information in the question. By writing these expressions I can get out a sample variance, I think, but I don't really know what to do with it. $\endgroup$ – Student Jun 16 '11 at 8:21

Assuming normality, the mean of the distribution of the estimator of the proportion is 0.375 because confidence intervals from normal distributions are typically given as symmetric intervals around the mean (note: technically, this isn't required, but everybody always does it like that).

Next, you know the distance from the mean to the edges of the confidence interval (0.025 here) are q * SD where SD is the standard deviation of your estimator, and q is the 1-alpha/2 standard normal quantile (note in this case: 95% CI => alpha=0.05 => q = 0.975 quantile ~ 1.96).

So you can easily find the SD from that.

Finally, using the mean and SD, create the 99% CI. If you don't know how to do this part, I believe you need more reading.

  • $\begingroup$ By the way: a 99% CI is not going to be more narrow than a 95% one. You want to be more certain that the truth is in there, so you want to consider a broader range. $\endgroup$ – Nick Sabbe Jun 16 '11 at 11:13

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