Question

statistics:

You construct a confidence interval of the form
0.24947≤p≤0.42978
from a sample of size 102 with 34 successes using the "plus 4" procedure.
Give the confidence level.

Answer 1

Sample size: \(n=102\)

Within this sample, 34 successes gives you an estimated proportion of \(\hat{p}=\dfrac{34}{102}=0.33\).

The \((1-\alpha)\times100\%\) confidence interval (that's a confidence level of \(1-\alpha\)) for a proportion has the form
\[\left(\hat{p}-Z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}},~\hat{p}+Z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right)\]
So you have two equations:
\[\begin{cases}
\hat{p}-Z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}=0.24947\\\\\\
\hat{p}+Z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}=0.42978
\end{cases}\]
We want to find \(1-\alpha\). Plugging in the numbers we know, you have
\[\begin{cases}
0.33-Z_{\alpha/2}\sqrt{\dfrac{0.33(1-0.33)}{102}}=0.24947\\\\\\
0.33+Z_{\alpha/2}\sqrt{\dfrac{0.33(1-0.33)}{102}}=0.42978
\end{cases}\]
Confidence intervals are symmetric by nature, so you really only have to focus on one of the equations and solve for \(Z_{\alpha/2}\).
\[\begin{align*}0.33-Z_{\alpha/2}\sqrt{\dfrac{0.33(1-0.33)}{102}}&=0.24947\\\\
0.08053&=Z_{\alpha/2}\cdot0.046558\\\\
Z_{\alpha/2}&=1.72967
\end{align*}\]
Checking a \(z\) table, you have \(P(Z>1.72967)=0.0418=\dfrac{\alpha}{2}\).

\[\frac{\alpha}{2}=0.0418~~\iff~~\alpha=0.0836~~\iff~~1-\alpha=0.9164\]