Respuesta :
Answer:
80% confidence interval for the mean waste recycled per person per day for the population of New York is [2.495 , 3.305].
Step-by-step explanation:
We are given that a random sample of 11 residents of the state of New York, the mean waste recycled per person per day was 2.9 pounds with a standard deviation of 0.98 pounds.
Firstly, the pivotal quantity for 80% confidence interval for the true mean waste recycled per person per day for the population of New York 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 waste recycled per person per day = 2.9 pounds
s = sample standard deviation = 0.98 pounds
n = sample of residents = 11
[tex]\mu[/tex] = true mean waste recycled per person per day
Here for constructing 80% confidence interval we have used t statistics because we don't know about population standard deviation.
So, 80% confidence interval for the true mean, [tex]\mu[/tex] is ;
P(-1.372 < [tex]t_1_0[/tex] < 1.372) = 0.80 {As the critical value of t at 10 degree of
freedom are -1.372 & 1.372 with P = 10%}
P(-1.372 < [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 1.372) = 0.80
P( [tex]-1.372 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X - \mu}[/tex] < [tex]1.372 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.80
P( [tex]-\bar X -1.372 \times {\frac{s}{\sqrt{n} }[/tex] < - [tex]\mu[/tex] < [tex]-\bar X +1.372 \times {\frac{s}{\sqrt{n} }[/tex] ) = 0.80
P( [tex]\bar X -1.372 \times {\frac{s}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < [tex]\bar X +1.372 \times {\frac{s}{\sqrt{n} }[/tex] ) = 0.80
80% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X -1.372 \times {\frac{s}{\sqrt{n} }[/tex] , [tex]\bar X +1.372 \times {\frac{s}{\sqrt{n} }[/tex] ]
= [ [tex]2.9 -1.372 \times {\frac{0.98}{\sqrt{11} }[/tex] , [tex]2.9 -1.372 \times {\frac{0.98}{\sqrt{11} }[/tex] ]
= [2.495 , 3.305]
Therefore, 80% confidence interval for the mean waste recycled per person per day for the population of New York is [2.495 , 3.305].
