how to get Margin of Error reduced to 1/4 current level

J

Jaskaran

Junior Member
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May 5, 2006
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Hi, I'd appreciate if I could know how to do this.
We have calculated a confidence interval based on a sample size n=100. Now we want to get a better estimate with a margin of error that is only one-fourth as large. How large does the new sample need to be?.
(from the book)..Because the standard error declines only with the square root of the sample size, to cut he standard error (and thus the ME) in half, we must quadruple the sample size...

So the new sample size is 800?.

Also,
ASuppose that a manufacturer is testing one of its machines to make sure that the machine is producing more than 97% good parts (H0=.97 and Ha=P>.97). The test results in a P-value of .122. Unknown to the manufacturer, the machine is actually producing 99% good parts. What probably happens as a result of the testing?

I'm not sure what they're looking for as an answer.
 
I'm interested in the size of your population. I could make a difference.

\(\displaystyle \L\;4*\sqrt{n}\;=\;sqrt{16*n}\)
 
But that would give me a sample size of 40, wouldn't the sample size be significantly larger for a lower margin of error?
 
..is 1600, however, the square root of that is 40. Am I missing something?
 
\(\displaystyle \sqrt{100} = 10\)
\(\displaystyle \sqrt{1600} = 40\)

\(\displaystyle \frac{1}{40} = \frac{(\frac{1}{10})}{4}\)
 
I'm not sure I follow, and I've been following...where do all these numbers come from?
 
It's my Bertrand Russel complex. I'm trying to communicate without words. Oops. I guess I talked.

\(\displaystyle ME\;=\;\frac{something}{\sqrt{n}}\)

\(\displaystyle \frac{ME}{4}\;=\;\frac{something}{\sqrt{16*n}}\)

You have n = 100.
 
Similarily, the margin of error could be \(\displaystyle ME ={1.96}\;sqrt{(.5)(.5)/100}\)

ME=.098
But reducing it to 1/4 as large, would mean solving for ME/4, at .0245

So..
\(\displaystyle .0245 ={1.96}\;sqrt{(.5)(.5)/n}\) => \(\displaystyle \;sqrt{n}={1.96}\;sqrt{(.5)(.5)/.0245}\) =>\(\displaystyle \;sqrt{n} = 40\)

Since it solves for the square root of 40, I square it at the end? So the sample size required to reduce margin of error to 1/4 at n=100 would provide, is 1600.
 
In you example, \(\displaystyle \L\;1.96*\sqrt{(0.5)(0.5)}\;=\;something\), and that is exactly what I said.

I think you have it.

I am not encouraged by your algebra skills. Have you considered brushing up on that? It may help you see things more clearly.
 
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