Question

one gallon of gas is put into each of 30 test cars. the resulting gas - mileage values of the sample are computed with mean of 28.5 gallons per mile, and standard deviation of 1.2 miles per gallon. what is the 95% confidence interval estimate of the mean mileage?
(28.1, 28.9)
(27.36, 29.64)
(27.3, 29.7)
(28.42, 28.58)
(28.46, 28.54)

Explanation:

Step1: Identify the formula

For a 95% confidence interval with sample size $n > 30$, the formula is $\bar{x}\pm z\frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z$ is the z - score, $s$ is the sample standard deviation and $n$ is the sample size. The z - score for a 95% confidence interval is $z = 1.96$, $\bar{x}=28.5$, $s = 1.2$ and $n = 30$.

Step2: Calculate the margin of error

The margin of error $E=z\frac{s}{\sqrt{n}}=1.96\times\frac{1.2}{\sqrt{30}}\approx1.96\times\frac{1.2}{5.477}\approx1.96\times0.219 = 0.43$.

Step3: Calculate the confidence interval

The lower limit is $\bar{x}-E=28.5 - 0.43=28.07\approx28.1$ and the upper limit is $\bar{x}+E=28.5+ 0.43=28.93\approx28.9$.

Answer:

A. $(28.1, 28.9)$