Suppose that a random sample of eighteen recently sold houses in a certain city has a mean sales price of , with a standard deviation of . Under the assumption that house prices are normally distributed, find a confidence interval for the mean sales price of all houses in this community. Then find the lower limit and upper limit of the confidence interval.

This question is incomplete, the complete question is;

Suppose that a random sample of eighteen recently sold houses in a certain city has a mean sales price of $280,000, with a standard deviation of $10,000. Under the assumption that house prices are normally distributed, find a 95% confidence interval for the mean sales price of all houses in this community. Then find the lower limit and upper limit of the confidence interval.

At 95%, confidence interval is ( 275026.7, 284973.3 )

Hence;

Lower limit = 275026.7

Upper limit = 284973.3

Step-by-step explanation:

 Given that;

mean x = $280,000

standard deviation σ = $10,000

sample size n = 18

degree of freedom df = n - 1 = 18 - 1 = 17

∝ = 1 - 95% = 1 - 0.95 = 0.05  

so

Critical t value = tinv( 0.05, 17 ) = 2.11

now, at 95% confidence interval for mean will be;

⇒ x ± [ Critical t value × ( σ/√n) ]

so we substitute

⇒ 280,000 ± [  2.11 × ( 10000/√18) ]

⇒ 280,000 ± [  2.11 × 2357.0226 ]

⇒ 280,000 ± 4973.3

⇒ 280000 - 4973.3, 280000 + 4973.3

At 95%, confidence interval is ( 275026.7, 284973.3 )

Hence;

Lower limit = 275026.7

Upper limit = 284973.3

160 cm

step-by-step explanation:

you have to add all the sides together