Question

Suppose that a random sample of nine recently sold houses in a certain city has a mean sales price of

$280,000

, with a standard deviation of

$12,000

. Under the assumption that house prices are normally distributed, find a

90%

confidence interval for the mean sales price of all houses in this community. Then complete the table below.

Carry your intermediate computations to at least three decimal places. Round your answers to the nearest whole number. (If necessary, consult a list of formulas.)

What is the lower limit of the confidence interval? $ What is the upper limit of the confidence interval?

Answer 1

SOLUTION:

From given data,

Suppose that a random sample of nine recently sold houses in a certain city has a mean sales price of $280,000 , with a standard deviation of $12,000  . Under the assumption that house prices are normally distributed, find a  90% confidence interval for the mean sales price of all houses in this community. Then complete the table below.

Carry your intermediate computations to at least three decimal places. Round your answers to the nearest whole number. (If necessary, consult a list of formulas.)

We have,

Sample Mean = = 280000

Standard deviation = s = 12,000

Sample size = n = 9

Degree of freedom = n-1 = 9-1 = 8

90% confidence interval.

90% = 90/100 = 0.9

= 1-0.9 = 0.1

/2 = 0.1/2 = 0.05

Critical value

t_/2,df = t_0.05,8 = 1.859

We have to calculate 90% confidence interval

The lower limit of the confidence interval is  -E

The upper limit of the confidence interval is  +E

Where,

E = t_/2,df * (s/sqrt(n))

E = 1.859* (12000 / sqrt(9))

E = 7436

Then,

The lower limit of the confidence interval is

-E = 280000- 7436 = 272564

The lower limit of the confidence interval is $ 272564

The upper limit of the confidence interval is

+E = 280000+7436 = 287436

The upper limit of the confidence interval is $ 287436