Respuesta :
Answer:
95% confidence interval for the mean monthly rent is [433.31 , 599.69].
Step-by-step explanation:
We are given a random sample of 10 apartments advertised in the local newspaper with their rental rates. Here are the rents (in dollars per month);
610, 615, 485, 555, 570, 565, 300, 370, 660, 435
So, the pivotal quantity for 95% confidence interval for the mean monthly rent 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 = Sum of all values ÷ Total no. of observations
= 516.5
s = sample standard deviation = [tex]\sqrt{\frac{\sum (X- \bar X)^{2} }{n-1} }[/tex] = 116.3
n = sample of apartments = 10
[tex]\mu[/tex] = population mean monthly rent
So, 95% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-2.262 < [tex]t_9[/tex] < 2.262) = 0.95
P(-2.262 < [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.262) = 0.95
P( [tex]-2.262 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X - \mu}[/tex] < [tex]2.262 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
P( [tex]\bar X-2.262 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X +2.262 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.262 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X +2.262 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]516.5-2.262 \times {\frac{116.3}{\sqrt{10} } }[/tex] , [tex]516.5+2.262 \times {\frac{116.3}{\sqrt{10} } }[/tex] ]
= [433.31 , 599.69]
Therefore, 95% confidence interval for the mean monthly rent for unfurnished one-bedroom apartments available for rent in this community is [433.31 , 599.69].
