Answer:
The 95% confidence interval for mean tire life is between 52,368.4 kilometers and 67,911 kilometers.
Step-by-step explanation:
We are in posession of the sample's standard deviation, so we use the students t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 16 - 1 = 15
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 15 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.95}{2} = 0.975. So we have T = 2.68
The margin of error is:
M = T*s = 2.1315*3645.94 = 7771.3
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 60139.7 - 7771.3 = 52,368.4.
The upper end of the interval is the sample mean added to M. So it is 60139.7 + 7771.3 = 67,911.
The 95% confidence interval for mean tire life is between 52,368.4 kilometers and 67,911 kilometers.
i hope this answer helps you!