Respuesta :
Answer:
99% confidence interval for the true mean resale value of a 5-year-old car of this model is [$12,173.24 , $13,306.76].
Step-by-step explanation:
We are given that you manage to obtain data on 17 recently resold 5-year-old foreign sedans of the same model.
These 17 cars were resold at an average price of $ 12 comma 740 with a standard deviation of $ 800.
Firstly, the pivotal quantity for 99% confidence interval for the true 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 average price = $12,740
s = sample standard deviation = $800
n = sample of cars = 17
[tex]\mu[/tex] = true population mean
Here for constructing 99% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.
So, 99% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-2.921 < [tex]t_1_6[/tex] < 2.921) = 0.99 {As the critical value of t at 16 degree of
freedom are -2.921 & 2.921 with P = 0.5%}
P(-2.921 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.921) = 0.99
P( [tex]-2.921 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.921 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.99
P( [tex]\bar X-2.921 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.921 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.99
99% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.921 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.921 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]12,740-2.921 \times {\frac{800}{\sqrt{17} } }[/tex] , [tex]12,740+2.921 \times {\frac{800}{\sqrt{17} } }[/tex] ]
= [$12,173.24 , $13,306.76]
Therefore, 99% confidence interval for the true mean resale value of a 5-year-old car of this model is [$12,173.24 , $13,306.76].
