# A survey of several 10 to 11 year olds recorded the following amounts spent on a trip to the mall:$19.17,$21.18,$20.38,$25.08Construct the 90\%90\% confidence interval for the average amount spent by 10 to 11 year olds on a trip to the mall. Assume the population is approximately normal.

A survey of several 10 to 11 year olds recorded the following amounts spent on a trip to the mall:
$19.17,$21.18,$20.38,$25.08
Construct the $$\displaystyle{90}\%$$ confidence interval for the average amount spent by 10 to 11 year olds on a trip to the mall. Assume the population is approximately normal.
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Step 4 of 4 : Construct the $$\displaystyle{90}\%$$ confidence interval. Round your answer to two decimal places.

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Step 1
The mean and standard deviation are obtained as follows:
$$\begin{array}{|c|c|} \hline &x&x^{2} \\ \hline & 19.17&367.4889 \\ \hline & 21.18&448.5924\\ \hline &20.38&415.3444\\ \hline &25.08&629.0064\\ \hline SUM&85.81&1860.432\\ \hline \end{array}$$
Mean $$\displaystyle={\left(\overline{{{x}}}\right)}={\frac{{\sum{x}}}{{{n}}}}={\frac{{{85.81}}}{{{4}}}}={21.45}$$
Standard Deviation $$\displaystyle=\sqrt{{{\frac{{\sum{x}^{{{2}}}-{\frac{{{\left(\sum{x}\right)}}}{{{n}}}}}}{{{n}-{1}}}}}}$$
Standard Deviation $$\displaystyle=\sqrt{{{\frac{{{1860.432}-{\frac{{{85.81}^{{{2}}}}}{{{4}}}}}}{{{3}}}}}}$$
Standard Deviation $$\displaystyle={2.55}$$
Step 2
The $$\displaystyle{90}\%$$ confidence interval is obtained as follows:
Confidence Interval $$\displaystyle={\left[\overline{{{x}}}\pm{t}_{{{\left({n}-{1},{\frac{{\alpha}}{{{2}}}}\right)}}}\times{\frac{{{s}}}{{\sqrt{{{n}}}}}}\right]}$$
Confidence Intarval $$\displaystyle={\left[{21.45}\pm{t}_{{{\left({3},{\frac{{{0.10}}}{{{2}}}}\right)}}}\times{\frac{{{2.55}}}{{\sqrt{{{4}}}}}}\right]}$$
Confidence Interval $$\displaystyle={\left[{21.45}\pm{2.353}\times{\frac{{{2.55}}}{{\sqrt{{{4}}}}}}\right]}$$
Comfidence Interval $$\displaystyle={\left[{18.45},{24.45}\right]}$$
Lower Limit $$\displaystyle={18.45}$$
Upper Limit $$\displaystyle={24.45}$$