Answer: For 96% confidence interval for the population mean of miles driven :
Lower bound = 10,841 miles
Upper bound= 14,949 miles
Step-by-step explanation:
Here, population standard deviation is unknown and sample size is small , So the formula is used to find the confidence interval for [tex]\mu[/tex] is given by :-
[tex]\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}[/tex]
, where n = sample size , = sample mean , t*= two tailed critical value s= sample standard deviation, .
Given, [tex]\overline{x}[/tex] =12,895 miles , s=3,801 miles, n=15 , degree of freedom = 14 [∵df=n-1]
For 96% confidence level , [tex]\alpha=0.04[/tex]
By t-distribution table ,
t-value(two tailed) for [tex]\alpha/2=0.02[/tex] and df =14 is t*=2.2638
Now , the 96% confidence interval for the population mean of miles driven will be :
[tex]12895\pm (2.2638)\dfrac{3801 }{\sqrt{15}}\\\\=12895\pm (2.0930)(981.413)\approx12895\pm 2054=(12895- 2054,\ 12895+2054)\\\\=(10,841,\ 14,949)[/tex]
Hence, For 96% confidence interval for the population mean of miles driven :
lower bound = 10,841 miles
upper bound= 14,949 miles
