A car manufacturing company is interested in studying the fuel efficiency of their latest model. They randomly select 50 cars from their production line and measure their average fuel consumption (in miles per gallon). The sample mean is found to be 30 mpg with a standard deviation of 4 mpg.
a) Construct a 95% confidence interval for the population mean fuel consumption.
b) Interpret the confidence interval in the context of the problem.
c) If the company wants to estimate the population mean fuel consumption with a margin of error of 1 mpg, what sample size would be required?
a) To construct a 95% confidence interval for the population mean fuel consumption, we can use the formula:
CI = X̄ ± Z * (σ/√n)
where X̄ is the sample mean, Z is the critical value for a 95% confidence level (which is approximately 1.96), σ is the population standard deviation, and n is the sample size.
Plugging in the values, we get:
CI = 30 ± 1.96 * (4/√50)
CI = 30 ± 1.96 * 0.5657
CI ≈ 30 ± 1.11
CI ≈ (28.89, 31.11)
b) The 95% confidence interval (28.89, 31.11) means that we are 95% confident that the true population mean fuel consumption falls within this range. In other words, if we were to repeat this study multiple times and construct 95% confidence intervals each time, approximately 95% of those intervals would contain the true population mean.
c) To estimate the required sample size, we can use the formula:
n = (Z^2 * σ^2) / E^2
where Z is the critical value for the desired confidence level, σ is the estimated population standard deviation, and E is the desired margin of error.
Plugging in the values, we get:
n = (1.96^2 * 4^2) / 1^2
n = (3.8416 * 16) / 1
n ≈ 61.47
Therefore, a sample size of at least 62 cars would be required to estimate the population mean fuel consumption with a margin of error of 1 mpg.
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