Question

STATS List below are the ages (years) of randomly selected race car drivers. Construct a 95% confidence interval estimate of the mean age of all race car drivers: ages 32, 40, 27, 36, 29, 28

Answer 1

A 95% confidence interval estimate of the mean age of all race car drivers is [26.65 years, 37.35 years].

Explanation:

We are given below the ages (in years) of randomly selected race car drivers;

Ages: 32, 40, 27, 36, 29, 28.

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 mean age =
`(\sum X)/(n)` =
`(32+40+27+36+29+28)/(6)` = 32 years

s = sample standard deviation =
`\sqrt{(\sum (X-\bar X)^(2) )/(n-1) }` = 5.1 years

n = sample of car drivers = 6

`\mu` = population mean age of all race car drivers

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean,
`\mu` is ;

P(-2.571 <
`t_5` < 2.571) = 0.95 {As the critical value of t at 5 degrees of

freedom are -2.571 & 2.571 with P = 2.5%}

P(-2.571 <
`(\bar X-\mu)/((s)/(√(n) ) )` < 2.571) = 0.95

P(
`-2.571 * {(s)/(√(n) ) }` <
`{\bar X-\mu}` <
`2.571 * {(s)/(√(n) ) }` ) = 0.95

P(
`\bar X-2.571 * {(s)/(√(n) ) }` <
`\mu` <
`\bar X+2.571 * {(s)/(√(n) ) }` ) = 0.95

95% confidence interval for
`\mu` = [
`\bar X-2.571 * {(s)/(√(n) ) }` ,
`\bar X+2.571 * {(s)/(√(n) ) }` ]

= [
`32-2.571 * {(5.1)/(√(6) ) }` ,
`32+2.571 * {(5.1)/(√(6) ) }` ]

= [26.65, 37.35]

Therefore, a 95% confidence interval estimate of the mean age of all race car drivers is [26.65 years, 37.35 years].