Are you likely to purchase an item promoted by a celebrity on a social media site? According to...

Question:

Are you likely to purchase an item promoted by a celebrity on a social media site? According to a survey, {eq}26 {/eq}% of social media users have made such a purchase.

Suppose that the survey had a sample size of n={eq}1000 {/eq}. Construct a {eq}95 {/eq}% confidence interval estimate for the population proportion of social media users that have purchased an item promoted by a celebrity on a social media site.

Confidence Intervalm for a Population Proportion:

An estimate about a population parameter can be made using a point value or through a confidence interval. When a confidence interval is required, it is necessary to know the point estimate, the sample size and the level of significance.

Answer and Explanation:

{eq}\begin{array}{l} {\color{Blue}{\textbf{CONFIDENCE INTERVAL FOR A POPULATION PROPORTION.}}} \\ \begin{array}{ll} n= 1000 & \text{(Sample size, Number of trials).} \\ \hat{p}= 0.26 & \text{(Sample proportion).} \end{array} \end{array} {/eq}


{eq}\begin{array}{l} \begin{array}{ll} 1-\alpha= 0.95 & \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= 0.95 \\ \displaystyle \alpha=1- 0.95 \\ \displaystyle \alpha= 0.05 \\ \displaystyle \alpha/2=\frac{ 0.05 }{ 2 } \\ \displaystyle \alpha/2= 0.0250 \\ \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( 0.0250 )\space{}\text{to the right. The cumulative area to the left is}\space{}1-\alpha/2=1- 0.0250 = \textbf{ 0.9750 }. \\ \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{} 0.9750 . \\ \begin{array}{l} ------------------------------------------ \end{array} \\ \begin{array}{ccccccccccccl} \vert{}& z & 0.00 & 0.01 & 0.02 & 0.03 & 0.04 & 0.05 & {{\color{Blue}{ 0.06 }}}& 0.07 & 0.08 & 0.09 &\vert{} \\ \vert{}& \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots &\vert{} \\ \vert{}& 1.7 & 0.9554 & 0.9564 & 0.9573 & 0.9582 & 0.9591 & 0.9599 & 0.9608 & 0.9616 & 0.9625 & 0.9633 &\vert{} \\ \vert{}& 1.8 & 0.9641 & 0.9649 & 0.9656 & 0.9664 & 0.9671 & 0.9678 & 0.9686 & 0.9693 & 0.9699 & 0.9706 &\vert{} \\ \vert{}& \color{Blue}{\textbf{ 1.9 }} & 0.9713 & 0.9719 & 0.9726 & 0.9732 & 0.9738 & 0.9744 & {{\color{Black}{ 0.9750 }}}& 0.9756 & 0.9761 & 0.9767 &\vert{} \\ \vert{}& 2.0 & 0.9772 & 0.9778 & 0.9783 & 0.9788 & 0.9793 & 0.9798 & 0.9803 & 0.9808 & 0.9812 & 0.9817 &\vert{} \\ \vert{}& 2.1 & 0.9821 & 0.9826 & 0.9830 & 0.9834 & 0.9838 & 0.9842 & 0.9846 & 0.9850 & 0.9854 & 0.9857 &\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 find}\space{} 0.9750 \space{}\text{exactly. Therefore:} \\ z_{\alpha/2}= \color{blue}{\textbf{ 1.9 }}+\color{blue}{\textbf{ 0.06 }} \\ z_{\alpha/2}= \color{blue}{\textbf{ 1.96 }} \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= 0.26 \pm 1.96 *\sqrt{\frac{ 0.26 *(1- 0.26 )}{ 1000 }} \\ \displaystyle CI= 0.26 \pm 1.96 *\sqrt{\frac{ 0.26 * 0.74 }{ 1000 }} \\ \displaystyle CI= 0.26 \pm 1.96 *\sqrt{\frac{ 0.1924 }{ 1000 }} \\ \displaystyle CI= 0.26 \pm 1.96 *\sqrt{ 0.0001924 } \\ \displaystyle CI= 0.26 \pm 1.96 * 0.0139 \\ \displaystyle CI= 0.26 \pm 0.0272 \\ \displaystyle CI=( 0.26 - 0.0272 , 0.26 + 0.0272 ) \\ \displaystyle CI=( 0.2328 , 0.2872 ) \\ \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{ 0.26 *(1- 0.26 )}{ 1000 }} \\ \displaystyle s_{\hat{p}}=\sqrt{\frac{ 0.26 * 0.74 }{ 1000 }} \\ \displaystyle s_{\hat{p}}=\sqrt{\frac{ 0.1924 }{ 1000 }} \\ \displaystyle s_{\hat{p}}=\sqrt{ 0.0001924 } \\ \displaystyle s_{\hat{p}}= 0.0139 \\ \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 = 1.96 *\sqrt{\frac{ 0.26 *(1- 0.26 )}{ 1000 }} &&&&& ME = 1.96 * 0.0139 \\ \displaystyle ME = 1.96 *\sqrt{\frac{ 0.26 * 0.74 }{ 1000 }} &&&&& ME = 0.0272 \\ \displaystyle ME = 1.96 *\sqrt{\frac{ 0.1924 }{ 1000 }} &&&&& \\ \displaystyle ME = 1.96 *\sqrt{ 0.0001924 } &&&&& \\ \displaystyle ME = 1.96 * 0.0139 &&&&& \\ \displaystyle ME= 0.0272 &&&&& \end{array} \\ \begin{array}{l} \textbf{Confidence interval}. \\ CI= 0.26 \pm 0.0272 \\ CI=( 0.26 - 0.0272 , 0.26 + 0.0272 ) \\ CI=( 0.2328 , 0.2872 ) \\ \end{array} \end{array} {/eq}


{eq}\begin{array}{l} {\color{Green}{\textbf{Calculus of the confidence interval using the definition.}}} \\ \begin{array}{rcl} \displaystyle \hat{p}-z_{\alpha/2}*\sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}&\leq{p}\leq{}&\displaystyle \hat{p}+z_{\alpha/2}*\sqrt{\frac{\hat{p}*(1-\hat{p})}{n}} \\ \displaystyle 0.26 - 1.96 *\sqrt{\frac{ 0.26 *(1- 0.26 )}{ 1000 }} &\leq{}p\leq{}& \displaystyle 0.26 + 1.96 *\sqrt{\frac{ 0.26 *(1- 0.26 )}{ 1000 }} \\ \displaystyle 0.26 - 1.96 *\sqrt{\frac{ 0.26 * 0.74 }{ 1000 }} &\leq{}p\leq{}& \displaystyle 0.26 + 1.96 *\sqrt{\frac{ 0.26 * 0.74 }{ 1000 }} \\ \displaystyle 0.26 - 1.96 *\sqrt{\frac{ 0.1924 }{ 1000 }} &\leq{}p\leq{}& \displaystyle 0.26 + 1.96 *\sqrt{\frac{ 0.1924 }{ 1000 }} \\ \displaystyle 0.26 - 1.96 *\sqrt{ 0.0001924 } &\leq{}p\leq{}& \displaystyle 0.26 + 1.96 *\sqrt{ 0.0001924 } \\ \displaystyle 0.26 - 1.96 * 0.0139 &\leq{}p\leq{}& \displaystyle 0.26 + 1.96 * 0.0139 \\ \displaystyle 0.26 - 0.0272 &\leq{}p\leq{}& \displaystyle 0.26 + 0.0272 \\ \displaystyle 0.2328 &\leq{}p\leq{}& \displaystyle 0.2872 \\ \end{array} \end{array} {/eq}


Learn more about this topic:

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Finding Confidence Intervals for Proportions: Formula & Example

from Statistics 101: Principles of Statistics

Chapter 9 / Lesson 8
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