Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoccupied seats per flight over the past year. To accomplish this, the records of 200 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. Construct a 92\% confidence interval for the population average number of unoccupied seats per flight.

UkusakazaL 2021-08-09
Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoccupied seats per flight over the past year. To accomplish this, the records of 200 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.6 seats and the sample standard deviation is 4.3 seats. \(\displaystyle{s}_{{{x}}}=?,\ {n}=?,\ {n}-{1}=?\). Construct a \(\displaystyle{92}\%\) confidence interval for the population average number of unoccupied seats per flight. (State the confidence interval. (Round your answers to two decimal places.)

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Answered 2021-08-16 Author has 28674 answers

Step 1
\(\displaystyle\overline{{{x}}}={11.6},\ {s}={4.3},\ {n}={200}\)
\(\displaystyle{s}_{{{x}}}={\frac{{{s}}}{{\sqrt{{{n}}}}}}={\frac{{{4.3}}}{{\sqrt{{{200}}}}}}={0.3041}\)
\(\displaystyle{n}={200}\)
\(\displaystyle{n}-{1}={199}\)
Step 2
\(\displaystyle{92}\%{C}.{I}.=\overline{{{x}}}\pm{T}{I}{N}{V}{\left({0.08},\ {199}\right)}{\left({s}_{{\overline{{{x}}}}}\right)}\)
\(\displaystyle={11.6}\pm{1.76}{\left({0.3041}\right)}\)
\(\displaystyle={11.6}\pm{0.54}\)
\(\displaystyle={\left({11.06},\ {12.14}\right)}\)

\(\displaystyle{T}{I}{N}{V}{\left(\alpha,\ {n}-{1}\right)}\) is a Excel function.

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