An airline research department measures the weight of a randomly selected 400 unaccompanied...

Question:

An airline research department measures the weight of a randomly selected 400 unaccompanied luggage bags which were checked in various airports across Australia. It is found the sample mean and sample standard deviation to be 29.88 kg and 2.4 kg respectively. The lower limit of a {eq}95\% {/eq} confidence interval for the population mean weight (in kg) of unaccompanied luggage bags is estimated to be _____ (2 decimal places).

Confidence Interval of the Mean:

A confidence interval of the population mean is discriminated against if the population variance is known or if the sample is small (less than 30) or large (greater than or equal to 30). If the population variance is known or a large sample is available, the normal distribution is used. If the variance is unknown and the sample is small, the student's t distribution is used.

Answer and Explanation:

{eq}{\color{Blue}{\textbf{CONFIDENCE INTERVAL OF THE MEAN.}}} \\ \begin{array}{ll} n= 400 & \text{(Sample size).} \\ \overline{x}= 29.88 \space\text{ kg } & \text{(Sample mean).} \\ s= 2.4 \space\text{ kg } & \text{(Sample standard deviation).} \end{array} {/eq}


{eq}\begin{array}{ll} 1-\alpha= 0.95 & \text{(Confidence level).} \\ \end{array} {/eq}


{eq}{\color{Red}{\textbf{Choice of statistic distribution.}}} \\ \begin{array}{l} \text{When}\space{}n\space{}\text{is large}\space{}(n\geq30),\space{}\text{the quantity}\space{}\displaystyle z=\frac{\overline{x}-\mu}{s/\sqrt{n}}\space{}\text{has an approximate}\space{}\underline{\text{standard normal distribution}}. \\ \end{array} {/eq}


{eq}{\color{Red}{\textbf{Confidence interval of the mean.}}} \\ \begin{array}{ll} \text{Direct Method:} & \displaystyle \displaystyle CI=\overline{x}\pm{}z_{\alpha/2}*\frac{s}{\sqrt{n}} \\ \text{Traditional Method:} & \displaystyle \displaystyle CI=\overline{x}\pm{}ME,\space{}\text{with}\space{}ME=z_{\alpha/2}*\frac{s}{\sqrt{n}} \\ \text{Definition Method:} & \displaystyle \displaystyle \overline{x}-z_{\alpha/2}*\frac{s}{\sqrt{n}}\leq{}\mu\leq{\overline{x}+z_{\alpha/2}*\frac{s}{\sqrt{n}}} \\ \end{array} \\ \text{is a large sample confidence interval for}\space{}\mu,\space{}\text{with confidence level of approximately}\space{}100\space{}(1-\alpha)\%. \\ {/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}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 }. \\ {/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=\overline{x}\pm{}z_{\alpha/2}*\frac{s}{\sqrt{n}} \\ \displaystyle CI= 29.88 \pm 1.96 *\frac{ 2.4 }{ \sqrt{ 400 }} \\ \displaystyle CI= 29.88 \pm 1.96 *\frac{ 2.4 }{ 20 } \\ \displaystyle CI= 29.88 \pm 1.96 * 0.12 \\ \displaystyle CI= 29.88 \pm 0.24 \\ \displaystyle CI=( 29.88 - 0.24 , 29.88 + 0.24 ) \\ \displaystyle CI=( 29.64 , 30.12 ) \\ \end{array} \end{array} {/eq}


The lower limit of a {eq}95\% {/eq} confidence interval for the population mean weight (in kg) of unaccompanied luggage bags is estimated to be 29.64 kg.


Learn more about this topic:

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Finding Confidence Intervals with the Normal Distribution

from Statistics 101: Principles of Statistics

Chapter 9 / Lesson 3
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