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? (27.3, 29.7) (27.36, 29.64) (28.1, 28.9) (28.42, 28.58) (28.46, 28.54)

Explanation:

Step1: Identify the formula

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

Step2: Calculate the margin of error

Given $\bar{x}=28.5$, $s = 1.2$, $n = 30$, and $z = 1.96$. 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:

C. (28.1, 28.9)