What is the formula for determining the length of a confidence interval?

[mathmari]

Hey!

A research institute wants to establish a confidence interval for the quota of working people in a city. Let $\hat{p}_n $ be the estimated quota, based on a sample of size $n$. It is assumed that $n> 30$.
How can one determine the length of the confidence interval?

Generally this is equal to $L = 2 \cdot \frac{\sigma \cdot z}{\sqrt{n}}$, right?

Do you have to use the $\hat{p}_n$ in the formula in this case? But how?

(Wondering)

 

[I like Serena]

Hey mathmari! (Wave)

The distribution of a proportion is a special case.
It has a z-distribution with an estimated standard deviation $\sigma_{\hat{p}_n} = \sqrt{\hat{p}_n(1-\hat{p}_n)}$.
And we should have $n \hat{p}_n > 10$ and $n (1 − \hat{p}_n) > 10$ to ensure it is sufficiently reliable.

 

[mathmari]

The distribution of a proportion is a special case.
It has a z-distribution with an estimated standard deviation $\sigma_{\hat{p}_n} = \sqrt{\hat{p}_n(1-\hat{p}_n)}$.
And we should have $n \hat{p}_n > 10$ and $n (1 − \hat{p}_n) > 10$ to ensure it is sufficiently reliable.

Ah ok!

So, the length is equal to $$L = 2 \cdot \frac{\sigma_{\hat{p}_n} \cdot z}{\sqrt{n}}= 2 \cdot \frac{ \sqrt{\hat{p}_n(1-\hat{p}_n)} \cdot z}{\sqrt{n}}$$ where $z$ depends on the confidence level $1-\alpha$, right? (Wondering)
Suppose that $1-\alpha=0.95$. I want to determine $n$ so that the length of confidence interval is not bigger that $L_{\text{max}}=0.03$.

We have that $z =1.96$.

Then $$L_{\text{max}}=0.03 \Rightarrow 2 \cdot \frac{ \sqrt{\hat{p}_n(1-\hat{p}_n)} \cdot 1.96}{\sqrt{n}}\leq 0.03\Rightarrow \frac{ \sqrt{\hat{p}_n(1-\hat{p}_n)} }{\sqrt{n}}\leq \frac{3}{392} \Rightarrow \frac{ \hat{p}_n(1-\hat{p}_n) }{n}\leq \frac{9}{153664}\Rightarrow n\geq \frac{153664\hat{p}_n(1-\hat{p}_n)}{9}$$ right? We cannot continue from here, can we? (Wondering)

 

[I like Serena]

Indeed. (Nod)
Btw, do we have $L_{\text{max}}=0.04$ or $L_{\text{max}}=0.03$?

 

[mathmari]

Indeed. (Nod)
Btw, do we have $L_{\text{max}}=0.04$ or $L_{\text{max}}=0.03$?

Oh, it is $L_{\text{max}}=0.03$. (Tmi)