# Confidence interval

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• Dec 4th 2011, 05:10 PM
deezy
Confidence interval
Find a 95% confidence interval for the percentage of cars on a certain highway that have poorly adjusted brakes, using a random sample of 500 cars stopped at a roadblock on that highway, 87 of which had poorly adjusted brakes.

I think I need to find k, with

$\displaystyle k = \frac{cs}{\sqrt n}$

n = 500, $\displaystyle \mu = 87$.

Not sure how to find c or s.
• Dec 4th 2011, 05:21 PM
pickslides
Re: Confidence interval
Try this

$\displaystyle \frac{87}{500}\pm 1.96\times \sqrt{\frac{\frac{87}{500}\left(1-\frac{87}{500}\right)}{500}}$
• Dec 4th 2011, 06:05 PM
deezy
Re: Confidence interval
Quote:

Originally Posted by pickslides
Try this

$\displaystyle \frac{87}{500}\pm 1.96\times \sqrt{\frac{\frac{87}{500}\left(1-\frac{87}{500}\right)}{500}}$

Where did that come from? When I calculated it I got .5063 and -0.1583.

http://edugen.wileyplus.com/edugen/c...25/math086.gif
This is the answer, but I don't know how c or $\displaystyle s^2$ was calculated.

For c, I tried $\displaystyle .5(1+.95) = .975$ and looking up the value for .975 on the t distribution chart with 50 - 1 = 49 degrees of freedom. But 49 was not on the chart.

For $\displaystyle s^2$, I'm not understanding why it is $\displaystyle 87 * 413 / 500$.