Rather than relying on estimates from computerized flight models, an airline com
ID: 3181894 • Letter: R
Question
Rather than relying on estimates from computerized flight models, an airline company wants to use actual flight time data to develop a 99% confidence interval to estimate the true average flight time from Chicago to Baltimore. After collecting 17 actual flight observations, the sample mean flight time was found to be 85 minutes. The population standard deviation is known to be 6 minutes. The distribution of flight times follows a normal distribution.
a. Compute the margin of error for the 99% confidence interval.
Round your critical value to three decimal places. Do not round any other intermediate calculations. Round your final answer to three decimal places.
Margin of error =---------- minutes
b. Using your answer to Part a, compute the 99% confidence interval for the true average flight time.
Round your final answers to three decimal places.
Lower confidence limit =--------- minutes. Upper confidence limit =----------- minutes.
c. Suppose that a 90% confidence interval is developed instead. Assuming everything else remains the same, will this 90% confidence interval be narrower or wider than the original 99% confidence interval?
*Narrower *WiderExplanation / Answer
Here it is given that distribution is normal with sd know so we will use normal distribution
z value for 99% CI is 2.58
a. Hence margin of error E=z*sd/sqrt(n)=2.58*6/sqrt(17)=3.75
b. CI=mean+/-E=85+/-3.75=(81.25,88.75)
c. Keeping other things same for 90% z value will be 1.645 hence E=2.39
Hence CI=mean+/-E=(82.61,87.39) so we can see that will 90% it becomes Narrower
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