Question

17 randomly selected cars are driven for a tank and is found that the sample mean fuel consumption is 9.4 l/100k with a sample standard deviation of s = .437. suppose the manufactuer wishes to construct a confidence interval for mu the true mean fuel consumption for the model under these consitions

what is the margin of error of a 95% confidence interval for mu

what is the margin of error of a 90% confidence interval for mu

Answer 1

Given n = 17 ,x bar = 9.4 and standard deviation(s) = 0.437

Margin of error for confidence interval is

$Z^{*}\frac{s}{\sqrt{n}}$ Z^* = Z score for given confidence level

So, Margin of error of a 95% confidence interval for mu is

= 0.2077. Z* = 1.96 for 95% confidence level

0.437 1.969 17

And, Margin of error of a 90% confidence interval for mu is = 0.1738 Z* = 1.64 for 90% confidence level.

160.437 17