Question

A tire manufacturer wants to estimate the average number of miles that may be driven in a tire of a certain type before the tire wears out. Assume the population is normally distributed. A random sample of tires is chosen and are driven until they wear out and the number of thousands of miles is recorded, find the 97% confidence interval using the sample data.

Answer 1

Answer:

97% confidence interval for the average number of miles that may be driven is [26.78 miles, 33.72 miles].

Step-by-step explanation:

We are given that a random sample of tires is chosen and are driven until they wear out and the number of thousands of miles is recorded;

32, 33, 28, 37, 29, 30, 22, 35, 23, 28, 30, 36.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                                P.Q.  =  $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~ $t_n_-_1$

where, $\bar X$ = sample average number of miles = $\frac{\sum X}{n}$ = 30.25

            s  = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 4.71

            n = sample of tires = 12

            $\mu$ = population average number of miles

Here for constructing a 97% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.

So, 97% confidence interval for the population mean, $\mu$ is ;

P(-2.55 < $t_1_1$ < 2.55) = 0.97  {As the critical value of t at 11 degrees of

                                              freedom are -2.55 & 2.55 with P = 1.5%}  

P(-2.55 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.55) = 0.97

P( $-2.55 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.55 \times {\frac{s}{\sqrt{n} } }$ ) = 0.97

P( $\bar X-2.55 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.55 \times {\frac{s}{\sqrt{n} } }$ ) = 0.97

97% confidence interval for $\mu$ = [ $\bar X-2.55 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.55 \times {\frac{s}{\sqrt{n} } }$ ]

                                        = [ $30.25-2.55 \times {\frac{4.71}{\sqrt{12} } }$ , $30.25+2.55 \times {\frac{4.71}{\sqrt{12} } }$ ]

                                        = [26.78 miles, 33.72 miles]

Therefore, 97% confidence interval for the average number of miles that may be driven is [26.78 miles, 33.72 miles].