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Estimating the Population Proportion Using Intervals

Consider estimating the proportion of WMU undergraduates whose   permanent residence is in Southwest Michigan. We may take a random sample of, say, n=25 students and compute the sample proportion  $\hat{p}$. If 9 out of the 25 students come from Southwest Michigan, then $\hat{p}=9/25=.36$. Because of the ``luck of the draw'', chances are the true proportion p is not exactly .36, but some number close to .36. Can we find an interval which we know contains p with say, at least 95% confidence?

Recall that the sample proportion $\hat{p}$ over different groups of 25 has a histogram that would look approximately normal with mean p and SD $\sqrt{p(1-p)}/\sqrt{n}$. This means that there is 95% likelihood that $\hat{p}$ falls within $2\sqrt{p(1-p)}/\sqrt{n}$ of p. Of course, this also means there is 95% likelihood that p falls within $2\sqrt{p(1-p)/n}$ of $\hat{p}$. In statistical shorthand,

\begin{displaymath}P( p \mbox{ is inside the interval }
\hat{p} \pm 2\sqrt{p(1-p)}/\sqrt{n}) = .95
\end{displaymath}

Recall the question posed earlier: ``Can we find an interval which we know contains p with 95% confidence?'' Answer: $\hat{p} \pm 2\sqrt{p(1-p)}/\sqrt{n}$.  In applications, we replace p inside the squareroot sign by $\hat{p}$. For our example, the 95% confidence interval for p is $.36 \pm 2 \sqrt{.36(1-.36)}/\sqrt{25}$ or $.36 \pm .19$. Furthermore, we can create an interval with any desired confidence level by replacing the critical value 2 with the appropriate standard normal percentile z.

  % latex2html id marker 4962
\fbox{ \parbox{5.5in}{
\vspace*{1ex}
{\bf Confidence...
...space is determined by $1-\alpha$\space and the normal curve.
\vspace*{1ex}
} }
   

\epsfig{file=zalpha2.ps, height=4in, width=3in, angle=-90}

The following table gives the appropriate critical value z for typical values of $1 - \alpha$.


\begin{displaymath}\begin{tabular}{c\vert ccccc}
$1-\alpha$\space & .68 & .80 &...
...ne
$z $\space & 1.00 & 1.28 & 1.64 & 2.00 & 2.58
\end{tabular}\end{displaymath}

Example:
1. Suppose that 9 out of 25 randomly selected students live in Southwest Michigan.

a. Construct a 99% confidence interval for the true proportion of WMU students who live in Southwest Michigan .
b. Construct a 75% confidence interval for the true proportion of WMU students who live in Southwest Michigan .
2. Suppose that 36 out of 100 randomly selected students live in Southwest Michigan.
a. Construct a 99% confidence interval for the true proportion of WMU students who live in Southwest Michigan .
b. Construct a 75% confidence interval for the true proportion of WMU students who live in Southwest Michigan .


next up previous contents index
Next: Sample Size for Estimating Up: Confidence Intervals Previous: Comparing Averages of Two

2003-09-08