# Use the normal distribution to find a confidence interval for a proportion p given the relevant...

## Question:

Use the normal distribution to find a confidence interval for a proportion p given the relevant sample results. Give the best point estimate for p, the margin of error, and the confidence interval. Assume the results come from a random sample.

A 99% confidence interval for p given that {eq}\hat{p} {/eq} = 0.33 and n = 450.

Round answer for the best point estimate to two decimal places, and answers for the margin of error and the confidence interval to three decimal places.

## Confidence Interval for Proportions:

One way to estimate population parameters in statistics is from a point or a confidence interval. In the case of a confidence interval for the proportions, it is required to know the point estimate, the sample size and the level of confidence.

{eq}\begin{array}{l} {\color{Blue}{\textbf{CONFIDENCE INTERVAL FOR A POPULATION PROPORTION.}}} \\ \begin{array}{ll} n= 450 & \text{(Sample size, Number of trials).} \\ \hat{p}= 0.33 & \text{(Sample proportion).} \end{array} \end{array} {/eq}

Point estimation.

{eq}\hat{p}=0.33 {/eq}

{eq}\begin{array}{l} \begin{array}{ll} 1-\alpha= 0.99 & \text{(Confidence level).} \\ \end{array} \end{array} {/eq}

{eq}\begin{array}{l} {\color{Red}{\textbf{Choice of statistic distribution.}}} \\ \begin{array}{l} \text{If}\space{}n\space{}\text{is large, the distribution of}\space{}\displaystyle z=\frac{x-n\space{}p}{\sqrt{n\space{}p\space{}(1-p)}}=\frac{\hat{p}-p}{\sqrt{\frac{\hat{p}\space{}(1-\hat{p})}{n}}}\space{}\text{is approximately standard normal}. \\ \end{array} \end{array} {/eq}

{eq}\begin{array}{l} {\color{Red}{\textbf{Confidence interval for a population proportion.}}} \\ \begin{array}{l} \text{If}\space{}\hat{p}\space{}\text{is the proportion of observations in a random sample of size}\space{}n\space{}\text{that belongs to a class of interest, a}\space{}100\space{}(1-\alpha)\% \\ \text{confidence interval on the proportion}\space{}p\space{}\text{of the population that belongs to this class is:} \\ \begin{array}{ll} \text{Direct Method:} & \displaystyle CI=\hat{p}\pm{}z_{\alpha/2}*\sqrt{\frac{\hat{p}*(1-\hat{p})}{n}} \\ \text{Traditional Method:} & \displaystyle CI=\hat{p}\pm{}ME,\space{}\text{with}\space{}ME=z_{\alpha/2}*\sqrt{\frac{\hat{p}*(1-\hat{p})}{n}} \\ \text{Definition Method:} & \displaystyle \hat{p}-z_{\alpha/2}*\sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\leq{}p\leq\hat{p}+z_{\alpha/2}*\sqrt{\frac{\hat{p}*(1-\hat{p})}{n}} \\ \end{array} \\ \text{where}\space{}z_{\alpha/2}\space{}\text{is the upper}\space{}100\alpha/2\space{}\text{percentage point on the standard normal distribution}. \\ \end{array} \end{array} {/eq}

{eq}\begin{array}{l} \textbf{Calculus of}\space{}z_{\alpha/2}-\space{}\textbf{value}. \\ \begin{array}{l} \displaystyle 1-\alpha= 0.99 \\ \displaystyle \alpha=1- 0.99 \\ \displaystyle \alpha= 0.01 \\ \displaystyle \alpha/2=\frac{ 0.01 }{ 2 } \\ \displaystyle \alpha/2= 0.0050 \\ \end{array} \end{array} {/eq}

{eq}\begin{array}{l} z_{\alpha/2}-\text{value is the}\space{}z-\text{value having an area of}\space{}\alpha/2\space( 0.0050 )\space{}\text{to the right. The cumulative area to the left is}\space{}1-\alpha/2=1- 0.0050 = \textbf{ 0.9950 }. \\ \end{array} {/eq}

{eq}\begin{array}{l} {\color{Green}{\textbf{Calculus of}}}\space{}{\color{Green}{z_{\alpha/2}}}\space{}{\color{Green}{\textbf{using the cumulative standard normal distribution table.}}} \\ \begin{array}{l} \text{We search through the probabilities to find the value that corresponds to}\space{} 0.9950 . \\ \begin{array}{l} ------------------------------------------ \end{array} \\ \begin{array}{ccccccccccccl} \vert{}& z & 0.00 & 0.01 & 0.02 & 0.03 & 0.04 & 0.05 & 0.06 & 0.07 & {{\color{Blue}{ 0.08 }}}& 0.09 &\vert{} \\ \vert{}& \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots &\vert{} \\ \vert{}& 2.3 & 0.9893 & 0.9896 & 0.9898 & 0.9901 & 0.9904 & 0.9906 & 0.9909 & 0.9911 & 0.9913 & 0.9916 &\vert{} \\ \vert{}& 2.4 & 0.9918 & 0.9920 & 0.9922 & 0.9925 & 0.9927 & 0.9929 & 0.9931 & 0.9932 & 0.9934 & 0.9936 &\vert{} \\ \vert{}& \color{Blue}{\textbf{ 2.5 }} & 0.9938 & 0.9940 & 0.9941 & 0.9943 & 0.9945 & 0.9946 & 0.9948 & 0.9949 & {{\color{Black}{ 0.9951 }}}& 0.9952 &\vert{} \\ \vert{}& 2.6 & 0.9953 & 0.9955 & 0.9956 & 0.9957 & 0.9959 & 0.9960 & 0.9961 & 0.9962 & 0.9963 & 0.9964 &\vert{} \\ \vert{}& 2.7 & 0.9965 & 0.9966 & 0.9967 & 0.9968 & 0.9969 & 0.9970 & 0.9971 & 0.9972 & 0.9973 & 0.9974 &\vert{} \\ \vert{}& \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots &\vert{} \end{array} \\ \begin{array}{l} ------------------------------------------ \end{array} \\ \begin{array}{l} \text{We do not find}\space{} 0.9950 \space{}\text{exactly; the nearest value is corresponding to}\space{} 0.9951 \space{}\text{therefore:} \\ z_{\alpha/2}= \color{blue}{\textbf{ 2.5 }}+\color{blue}{\textbf{ 0.08 }} \\ z_{\alpha/2}= \color{blue}{\textbf{ 2.58 }} \end{array} \end{array} \end{array} {/eq}

{eq}\begin{array}{l} {\color{Green}{\textbf{Calculus of the confidence interval using the traditional method.}}} \\ CI=\hat{p}\pm{}ME \\ \\ \begin{array}{l} \textbf{Margin of error}. \\ \text{There are two ways to calculate the margin of error: Directly and Using the standard error of the sample proportions}. \\ \end{array} \\ \begin{array}{l} \textbf{Standard error of the sample proportion}. \\ \displaystyle s_{\hat{p}}=\sqrt{\frac{\hat{p}*(1-\hat{p})}{n}} \\ \displaystyle s_{\hat{p}}=\sqrt{\frac{ 0.33 *(1- 0.33 )}{ 450 }} \\ \displaystyle s_{\hat{p}}=\sqrt{\frac{ 0.33 * 0.67 }{ 450 }} \\ \displaystyle s_{\hat{p}}=\sqrt{\frac{ 0.2211 }{ 450 }} \\ \displaystyle s_{\hat{p}}=\sqrt{ 0.000491333 } \\ \displaystyle s_{\hat{p}}= 0.02 \\ \end{array} \\ \begin{array}{lllllllllll} \textbf{Margin of error}. \\ \text{Directly}. &&&&& \text{Using the standard error of the sample proportions}. \\ \displaystyle ME=z_{\alpha/2}*\sqrt{\frac{\hat{p}*(1-\hat{p})}{n}} &&&&& ME=z_{\alpha/2}*s_{\hat{p}} \\ \displaystyle ME = 2.58 *\sqrt{\frac{ 0.33 *(1- 0.33 )}{ 450 }} &&&&& ME = 2.58 * 0.02 \\ \displaystyle ME = 2.58 *\sqrt{\frac{ 0.33 * 0.67 }{ 450 }} &&&&& ME = 0.052 \\ \displaystyle ME = 2.58 *\sqrt{\frac{ 0.2211 }{ 450 }} &&&&& \\ \displaystyle ME = 2.58 *\sqrt{ 0.000491333 } &&&&& \\ \displaystyle ME = 2.58 * 0.02 &&&&& \\ \displaystyle ME= 0.052 &&&&& \end{array} \\ \begin{array}{l} \textbf{Confidence interval}. \\ CI= 0.33 \pm 0.052 \\ CI=( 0.33 - 0.052 , 0.33 + 0.052 ) \\ CI=( 0.278 , 0.382 ) \\ \end{array} \end{array} {/eq}