1. Confidence Interval

Missed class and have no idea how to start.

Question is Construct a 90% confidence interval for the proportion of smokers who quit smoking by using either the patch or e-cigarettes within 6 months.
There was a sample size of 584 and just 38 of the smokers given the e-cigarette or the patch.

2. Re: Confidence Interval

I don't think the question is written correctly. Are you saying that there are 584 people who quit smoking, and 38 of those used the e-cigarette or patch?

3. Re: Confidence Interval

yes i believe thats what they mean. Yeah it isnt written very well.

4. Re: Confidence Interval

Could this be the equation?

38/584=0.07
0.07 ± (1.645) √(0.07)(0.93)/584
=0.053,0.087

5. Re: Confidence Interval

Ok so we can see that \displaystyle \begin{align*} \mu = \hat{p} = \frac{38}{584} = \frac{19}{292} \end{align*} and \displaystyle \begin{align*} \sigma = \sqrt{\frac{\hat{p} \left( 1 - \hat{p} \right) }{n}} = \sqrt{\frac{\frac{19}{292} \cdot \frac{273}{292}}{584}} = \frac{\sqrt{757\,302}}{85\,264} \end{align*}. Since we want a 90% Confidence Interval, we want the endpoints which have 90% of the area between them, so we want \displaystyle \begin{align*} z_{0.05} = -1.645 \end{align*} and \displaystyle \begin{align*} z_{0.95} = 1.645 \end{align*}. So using the standardisation formula

\displaystyle \begin{align*} z &= \frac{x - \mu}{\sigma} \\ -1.645 &= \frac{x - \frac{19}{292}}{\frac{\sqrt{757\,302}}{85\,264}} \\ -1.645 \cdot \frac{\sqrt{757\,302}}{85\,264} &= x - \frac{19}{292} \\ x &= \frac{19}{292} - 1.645 \cdot \frac{\sqrt{757\,302}}{85\,264} \\ x &\approx 0.0483 \end{align*}

Now do the same with the other z value.

6. Re: Confidence Interval

Originally Posted by Simplelife1107
Could this be the equation?

38/584=0.07
0.07 ± (1.645) √(0.07)(0.93)/584
=0.053,0.087
You appear to have a LOT of roundoff error. You should only ever make approximations where exact values are not possible!