Respuesta :
Answer:
A 95% confidence interval estimate of the mean age of all race car drivers is [26.65 years, 37.35 years].
Step-by-step 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. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean age = [tex]\frac{\sum X}{n}[/tex] = [tex]\frac{32+40+27+36+29+28}{6}[/tex] = 32 years
s = sample standard deviation = [tex]\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }[/tex] = 5.1 years
n = sample of car drivers = 6
[tex]\mu[/tex] = 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, [tex]\mu[/tex] is ;
P(-2.571 < [tex]t_5[/tex] < 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 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.571) = 0.95
P( [tex]-2.571 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.571 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
P( [tex]\bar X-2.571 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.571 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.571 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.571 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]32-2.571 \times {\frac{5.1}{\sqrt{6} } }[/tex] , [tex]32+2.571 \times {\frac{5.1}{\sqrt{6} } }[/tex] ]
= [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].
