Chapter 7.2, Problem 27E

Reconsider the CI (7.10) for p, and focus on a confidence level of 95%. Show that the confidence limits agree quite well with those of the traditional interval (7.11) once two successes and two failures have been appended to the sample [i.e., (7.11) based on x + 2 S’s in n + 4 trials]. [Hint: 1.96 ≈ 2. Note: Agresti and Coull showed that this adjustment of the traditional interval also has an actual confidence level close to the nominal level.]

To determine

Show that the 95% large sample confidence interval of population proportion $\frac{\widehat{p}+\frac{z_{\frac{\alpha }{2}}^2}{2n}}{1+\frac{z_{\frac{\alpha }{2}}^2}{n}}\pm z_{\frac{\alpha }{2}}\times \frac{\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}+\frac{z_{\frac{\alpha }{2}}^2}{4n^2}}}{1+\frac{z_{\frac{\alpha }{2}}^2}{n}}$ agrees quite well with traditional interval $\widehat{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}$ in case of $x+2$ successes in $n+4$ trails.

Answer to Problem 27E

The 95% large sample confidence interval of population proportion $\frac{\widehat{p}+\frac{z_{\frac{\alpha }{2}}^2}{2n}}{1+\frac{z_{\frac{\alpha }{2}}^2}{n}}\pm z_{\frac{\alpha }{2}}\times \frac{\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}+\frac{z_{\frac{\alpha }{2}}^2}{4n^2}}}{1+\frac{z_{\frac{\alpha }{2}}^2}{n}}$ agrees quite well with traditional interval $\widehat{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}$ in case of $x+2$ successes in $n+4$ trails.

Explanation of Solution

Calculation:

Critical value:

For 95% confidence level

$\begin{array}{c}1-\alpha =1-0.95\\ \alpha =0.05\\ \frac{\alpha }{2}=\frac{0.05}{2}\\ =0.025\end{array}$

Hence, cumulative area to the left is,

$\begin{array}{c}\text{Area to the left}=1-\text{Area to the right}\\ =1-0.025\\ =0.975\end{array}$

From Table A3 of the standard normal distribution in Appendix, the critical value is 1.96.

Here the critical value 1.96 is approximately equal to 2.

Thus, the critical value $z_{0.025}\simeq 2$

The 95% large sample confidence interval of population proportion is $CI=\frac{\widehat{p}+\frac{4}{2n}}{1+\frac{4}{n}}\pm 2\times \frac{\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}+\frac{4}{4n^2}}}{1+\frac{4}{n}}$.

Point estimate of population proportion:

The point estimate of population proportion is obtained as follows:

$\begin{array}{c}\widehat{p}=\frac{\text{Lower bound}+\text{Upper bound}}{2}\\ =\frac{\frac{\widehat{p}+\frac{4}{2n}}{1+\frac{4}{n}}-2\times \frac{\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}+\frac{4}{4n^2}}}{1+\frac{4}{n}}+\frac{\widehat{p}+\frac{4}{2n}}{1+\frac{4}{n}}+2\times \frac{\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}+\frac{4}{4n^2}}}{1+\frac{4}{n}}}{2}\\ =\frac{\widehat{p}+\frac{4}{2n}}{1+\frac{4}{n}}\end{array}$

The general formula for the estimate of population proportion is $\widehat{p}=\frac{x}{n}$.

Now, the point estimate reduces as follows:

$\begin{array}{c}\widehat{p}=\frac{\widehat{p}+\frac{4}{2n}}{1+\frac{4}{n}}\\ =\frac{\frac{x}{n}+\frac{4}{2n}}{\frac{n+4}{n}}\\ =\frac{x+2}{n+4}\end{array}$

Thus, the point estimate of the large sample population proportion at 95% confidence level is $\widehat{p}=\frac{x+2}{n+4}$.

Margin of error for confidence interval:

The margin of error for each confidence interval $\left(E\right)$ is obtained as follows:

$\begin{array}{c}E=\frac{\text{Upper bound}-\text{Lower bound}}{2}\\ =\frac{\left(\frac{\widehat{p}+\frac{4}{2n}}{1+\frac{4}{n}}+2\times \frac{\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}+\frac{4}{4n^2}}}{1+\frac{4}{n}}\right)-\left(\frac{\widehat{p}+\frac{4}{2n}}{1+\frac{4}{n}}-2\times \frac{\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}+\frac{4}{4n^2}}}{1+\frac{4}{n}}\right)}{2}\\ =2\times \frac{\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}+\frac{4}{4n^2}}}{1+\frac{4}{n}}\\ \simeq z_{\frac{\alpha }{2}}\times \sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n+4}}\end{array}$

Thus, the margin of error for each confidence interval $\left(E\right)$ is $z_{\frac{\alpha }{2}}\times \sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n+4}}$.

Here, the point estimate is $\widehat{p}=\frac{x+2}{n+4}$ and the margin of error is $z_{\frac{\alpha }{2}}\times \sqrt{\frac{p\left(1-p\right)}{n+4}}$

Therefore, the large sample interval of population proportion is,

$CI=\left(\frac{x+2}{n+4}\right)\pm z_{\frac{\alpha }{2}}\times \sqrt{\frac{\frac{x+2}{n+4}\left(1-\frac{x+2}{n+4}\right)}{n+4}}$

Substituting $x^{*}=\frac{x+2}{n+4}$ and $n^{*}=n+4$, the large sample confidence interval is,

$CI={\widehat{p}}^{*}\pm z_{\frac{\alpha }{2}}\times \sqrt{\frac{{\widehat{p}}^{*}\left(1-{\widehat{p}}^{*}\right)}{n^{*}}}$

Thus, the95% large sample confidence interval of population proportion $\frac{\widehat{p}+\frac{z_{\frac{\alpha }{2}}^2}{2n}}{1+\frac{z_{\frac{\alpha }{2}}^2}{n}}\pm z_{\frac{\alpha }{2}}\times \frac{\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}+\frac{z_{\frac{\alpha }{2}}^2}{4n^2}}}{1+\frac{z_{\frac{\alpha }{2}}^2}{n}}$ agrees quite well with traditional interval $\widehat{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}$ in case of $x+2$ successes in $n+4$ trails.