Problem 2 A company with a large fleet of cars wants to study the gasoline usage
ID: 3361311 • Letter: P
Question
Problem 2 A company with a large fleet of cars wants to study the gasoline usage. They check the gasoline usage for 80 company trips chosen at random, finding a mean of 24.92 mpg. Based on the past experience, they believe that the general gasoline usage roughly follows normal distribution with the standard deviation 4.63 mpg. 1. Which kind of confidence intervals is appropriate to use here, one-sample z-interval or one-sample t-interval? 2. Please find the multiplier they need when constructing a 99% C 3. Please construct a 99% C.1, for the mean of the general gasoline usage. 4. Please interpret the C.I. 5. If they want to control the margin of error to be within 0.7 mpg, at least how many trips do they have to sample if we still have the 99% confidence level? 6. If they want to control the margin of error to e wn 0.7 mpg using the same sample, then what's your confidence level? 7. If we decrease the confidence level, are we going to obtain a more precise C.I.? 8. If we sample more brands, are we going to obtain a more precise C.I?Explanation / Answer
(1) z-interval as the population standard deviation is known
(2) z(0.01/2)=2.5758
(3)(1-alpha)*100% confidence interval for population mean=sample mean±z(alpha/2)*sd/sqrt(n)
99% confidence interval for population mean=24.92±z(0.01/2)*4.63/sqrt(80)=24.92±2.5758*4.63/sqrt(80)=
=24.92±1.33=(23.59,26.25)
(4)To interpret a confidence interval remember that the sample information is random - but there is a pattern to its behavior if we look at all possible samples. Each possible sample gives us a different sample proportion and a different interval. But, even though the results vary from sample-to-sample, we are "confident" because the margin-of-error would be satisfied for 99% of all samples
(5)99% margin of error=z(alpha/2)*sd/sqrt(n)=z(0.01/2)*4.63/sqrt(n)
or, 0.7=2.5758*4.63/sqrt(n)
or sqrt(n)=2.5758*4.63/0.7=17.04
or, n=290.36 ( next integer is 291)
(6) CI=24.92±0.7=(24.22, 25.62)
(7) larger the confidence level , wider is CI
statement is False
(8) more the sample size, smaller the CI
statement is True
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