A research engineer for a tire manufacturer is investigating tire life for a new rubber compound and has built 16 tires and tested them to end-of-life in a road test. The sample mean and standard deviation are 60,139.7 and 3645.94 kilometers.

Find a 95% confidence interval on mean tire life.

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.

the 1st digit is in the 100s n its 1

step-by-step explanation:

3,589÷18 = 199.389

so the first digit of the quotient is in the 100s n its 1