Answer:
[tex]7.38-1.96\frac{2.51}{\sqrt{601}}=7.18[/tex]
[tex]7.38+1.96\frac{2.51}{\sqrt{601}}=7.58[/tex]
For this case the confidence interval is given by :
[tex]7.18 \leq \mu \leq 7.58[/tex]
Step-by-step explanation:
Data provided
[tex]\bar X=7.38[/tex] represent the sample mean for the fuel efficiencies
[tex]\mu[/tex] 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:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
The degrees of freedom for this case is given by:
[tex]df=n-1=601-1=600[/tex]
The Confidence level for this case is 0.95 or 95%, and the significance level [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex] and the critical value is given by [tex]t_{\alpha/2}=1.96[/tex]
Replcing into the formula for the confidence interval is given by:
[tex]7.38-1.96\frac{2.51}{\sqrt{601}}=7.18[/tex]
[tex]7.38+1.96\frac{2.51}{\sqrt{601}}=7.58[/tex]
For this case the confidence interval is given by :
[tex]7.18 \leq \mu \leq 7.58[/tex]
