Question

In analyzing the city and highway fuel efficiencies of many cars and trucks, the mean difference in fuel efficiencies for the 601 vehicles was 7.38 mpg with a standard
deviation of 2.51 mpg. Find a 95% confidence interval for this difference and interpret it in context.
The 95% confidence interval for the difference in fuel efficiencies is....?
(Round to two decimal places as needed.)

Answer 1

Answer:

$7.38-1.96\frac{2.51}{\sqrt{601}}=7.18$    

$7.38+1.96\frac{2.51}{\sqrt{601}}=7.58$    

For this case the confidence interval is given by :

$7.18 \leq \mu \leq 7.58$

Step-by-step explanation:

Data provided

$\bar X=7.38$ represent the sample mean for the fuel efficiencies

$\mu$ population mean

s=2.51 represent the sample standard deviation

n=601 represent the sample size  

Confidence interval

The formula for the confidence interval of the true mean is given by:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

The degrees of freedom for this case is given by:

$df=n-1=601-1=600$

The Confidence level for this case is 0.95 or 95%, and the significance level $\alpha=0.05$ and $\alpha/2 =0.025$ and the critical value is given by $t_{\alpha/2}=1.96$

Replcing into the formula for the confidence interval is given by:

$7.38-1.96\frac{2.51}{\sqrt{601}}=7.18$    

$7.38+1.96\frac{2.51}{\sqrt{601}}=7.58$    

For this case the confidence interval is given by :

$7.18 \leq \mu \leq 7.58$