# Out of 400 people sampled, 184 preferred Candidate A. Based on this, estimate what proportion of...

## Question:

Out of 400 people sampled, 184 preferred Candidate A.

Based on this, estimate what proportion of the entire voting population (p) prefers Candidate A. Use a 99% confidence level, and give your answers as decimals, to three places.

## Confidence Interval for Proportions:

One way to estimate a population parameter, such as the mean or proportion, is through a confidence interval. To build a confidence interval, the precise estimation of the parameter and the confidence level of the interval are required.

{eq}\begin{array}{l} {\color{Blue}{\textbf{CONFIDENCE INTERVAL FOR A POPULATION PROPORTION.}}} \\ \begin{array}{ll} n= 400 & \text{(Sample size, Number of trials).} \\ x= 184 & \text{(Number of successes).} \end{array} \end{array} {/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{Sample proportion.}}} \\ \begin{array}{l} \displaystyle \hat{p}=\frac{x}{n} \\ \displaystyle \hat{p}=\frac{ 184 }{ 400 } \\ \displaystyle \hat{p}= 0.4600 \\ \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}} \\ \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 direct method.}}} \\ \begin{array}{l} \displaystyle CI=\hat{p}\pm{}z_{\alpha/2}*\sqrt{\frac{\hat{p}*(1-\hat{p})}{n}} \\ \displaystyle CI= 0.46 \pm 2.58 *\sqrt{\frac{ 0.46 *(1- 0.46 )}{ 400 }} \\ \displaystyle CI= 0.46 \pm 2.58 *\sqrt{\frac{ 0.46 * 0.54 }{ 400 }} \\ \displaystyle CI= 0.46 \pm 2.58 *\sqrt{\frac{ 0.2484 }{ 400 }} \\ \displaystyle CI= 0.46 \pm 2.58 *\sqrt{ 0.000621 } \\ \displaystyle CI= 0.46 \pm 2.58 * 0.025 \\ \displaystyle CI= 0.46 \pm 0.065 \\ \displaystyle CI=( 0.46 - 0.065 , 0.46 + 0.065 ) \\ \displaystyle CI=( 0.395 , 0.525 ) \\ \end{array} \end{array} {/eq}