Question

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.

Answer 1

Answer:

$60139.7-2.131\frac{3645.94}{\sqrt{16}}=58197.33$    

$60139.7+2.131\frac{3645.94}{\sqrt{16}}=62082.07$    

So on this case the 95% confidence interval would be given by (58197.33;62082.07)    

Step-by-step explanation:

Previous notation

$\bar X= 60139.7$ represent the sample mean

$\mu$ population mean (variable of interest)

s=3645.94 represent the sample standard deviation

n=16 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

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

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=16-1=15$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,15)".And we see that $t_{\alpha/2}=2.131$

Now we have everything in order to replace into formula (1):

$60139.7-2.131\frac{3645.94}{\sqrt{16}}=58197.33$    

$60139.7+2.131\frac{3645.94}{\sqrt{16}}=62082.07$    

So on this case the 95% confidence interval would be given by (58197.33;62082.07)    

Answer 2

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.