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.
Answer:
At 95%, confidence interval is ( 275026.7, 284973.3 )
Hence;
Lower limit = 275026.7
Upper limit = 284973.3
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