A major airline wants to estimate its average number of unoccupied seats per flight over the past year. To accomplish this, the records of 225 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. The results show the sample mean is 11.6 and the sample standard deviation is 4.1. Which one of the following represents a 90% confidence interval for u, the mean number of unoccupied seats per flight during the past year? 11.6 - 2.069(4.1)/15 < µ < 11.6 + 2.069(4.1)/15 11.6 - 1.96(4.1)/15 < µ < 11.6 + 1.96(4.1)/15 11.6 - 1.645(4.1)/15 < µ < 11.6 + 1.645(4.1)/15 11.6 - 2.75(4.1)/15 < µ < 11.6 + 2.75(4.1)/15 11.6 - 1.645(4.1)/225 < H < 11.6 + 1.645(4.1)/225
A major airline wants to estimate its average number of unoccupied seats per flight over the past year. To accomplish this, the records of 225 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. The results show the sample mean is 11.6 and the sample standard deviation is 4.1. Which one of the following represents a 90% confidence interval for u, the mean number of unoccupied seats per flight during the past year? 11.6 - 2.069(4.1)/15 < µ < 11.6 + 2.069(4.1)/15 11.6 - 1.96(4.1)/15 < µ < 11.6 + 1.96(4.1)/15 11.6 - 1.645(4.1)/15 < µ < 11.6 + 1.645(4.1)/15 11.6 - 2.75(4.1)/15 < µ < 11.6 + 2.75(4.1)/15 11.6 - 1.645(4.1)/225 < H < 11.6 + 1.645(4.1)/225
MATLAB: An Introduction with Applications
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Transcribed Image Text:A major airline wants to estimate its average number of unoccupied seats per flight over the past year. To accomplish
this, the records of 225 flights are randomly selected and the number of unoccupied seats is noted for each of the
sampled flights. The results show the sample mean is 11.6 and the sample standard deviation is 4.1. Which one of the
following represents a 90% confidence interval for u, the mean number of unoccupied seats per flight during the past
year?
11.6 - 2.069(4.1)/15 < µ < 11.6 + 2.069(4.1)/15
11.6 - 1.96(4.1)/15 < µ < 11.6 + 1.96(4.1)/15
11.6 - 1.645(4.1)/15 < µ < 11.6 + 1.645(4.1)/15
11.6 - 2.75(4.1)/15 < µ < 11.6 + 2.75(4.1)/15
11.6 - 1.645(4.1)/225 < H < 11.6 + 1.645(4.1)/225
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