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

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

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. =
`(\bar X-\mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

where,
`\bar X` = sample average number of miles =
`(\sum X)/(n)` = 30.25

s = sample standard deviation =
`\sqrt{(\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 <
`(\bar X-\mu)/((s)/(√(n) ) )` < 2.55) = 0.97

P(
`-2.55 * {(s)/(√(n) ) }` <
`{\bar X-\mu}` <
`2.55 * {(s)/(√(n) ) }` ) = 0.97

P(
`\bar X-2.55 * {(s)/(√(n) ) }` <
`\mu` <
`\bar X+2.55 * {(s)/(√(n) ) }` ) = 0.97

97% confidence interval for
`\mu` = [
`\bar X-2.55 * {(s)/(√(n) ) }` ,
`\bar X+2.55 * {(s)/(√(n) ) }` ]

= [
`30.25-2.55 * {(4.71)/(√(12) ) }` ,
`30.25+2.55 * {(4.71)/(√(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].