Getting ready for the final, don't understand how to answer these

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The discussion centers on calculating the necessary sample size to achieve a specific maximum error in confidence intervals for one mean or proportion. The formula provided is \(E=z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}\), where \(E\) represents the maximum error, \(\sigma\) is the standard deviation, and \(n\) is the sample size. To reduce the maximum error \(E\) to one-fifth of its original size, the sample size \(n\) must increase by a factor of \(25\), resulting in a required sample size of \(5000\) when starting from an initial sample size of \(200\).

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chriskeller1
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Please please urgent help needed with questions like these

How do I predict how big the sample needs to be?

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Regarding confidence intervals for one mean or proportion, we are given that the maximum error $E$ of the estimate for $\mu$ is:

$$E=z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}$$

So, by what factor must $n$ increase so that $E$ is one-fifth as large?
 
MarkFL said:
Regarding confidence intervals for one mean or proportion, we are given that the maximum error $E$ of the estimate for $\mu$ is:

$$E=z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}$$

So, by what factor must $n$ increase so that $E$ is one-fifth as large?
n must be 5000 right?
 
chriskeller1 said:
n must be 5000 right?

Yes, because $n$ is under a radical, it must increase by a factor of $5^2=25$ in order for $E$ to be 1/5 as large. and so we find:

$$200\cdot25=5000$$
 

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