# Thread: Confidence Interval 90% help

1. ## Confidence Interval 90% help

Generate 50 random samples of size n=35 from a normal population with mean ( μ = 30 ) standard deviation ( σ = 3 ) ... Based on the confidence interval used i this example ( 90% ), what percent of the intervals did you expect to fail to contain μ.

I'm doing this on minitab. I've already generated my random data, and found out what the actual percentage was, but I don't know how I'm supposed to estimate..

Help?

Thanks!

2. ## Re: Confidence Interval 90% help

The average of 35 samples, from a normal distribution of mean $\displaystyle /mu$ and standard deviation $\displaystyle \sigma$ is normally distributed with mean $\displaystyle \mu$ and standard deviation $\displaystyle \sigma \sqrt{35}$. The average of 50 such samples is normally distributed with mean $\displaystyle \mu$ and standard deviation $\displaystyle \sqrt{50}(\sqrt{35}\sigma)= \sqrt{(50)(35)}\sigma$. That is the same as the distribution of (35)(50) separate samples.

3. ## Re: Confidence Interval 90% help

Originally Posted by HallsofIvy
The average of 35 samples, from a normal distribution of mean $\displaystyle /mu$ and standard deviation $\displaystyle \sigma$ is normally distributed with mean $\displaystyle \mu$ and standard deviation $\displaystyle \sigma \sqrt{35}$. The average of 50 such samples is normally distributed with mean $\displaystyle \mu$ and standard deviation $\displaystyle \sqrt{50}(\sqrt{35}\sigma)= \sqrt{(50)(35)}\sigma$. That is the same as the distribution of (35)(50) separate samples.
I appreciate the response.I don't see how this helps me with the question I asked though. Maybe I'm missing something, but the question I need answered is

" What percent of the intervals did you expect to fail to contain μ " ?

Thanks again.