Question

A random sample of 20 recent weddings in a country yielded a mean wedding cost of

$26,395.79.

Assume that recent wedding costs in this country are normally distributed with a standard deviation of

​$8500

a. Determine a​ 95% confidence interval for the mean​ cost,μ​,of all recent weddings in this country.

B. margin of error

Answer 1

Sample size = n = 20

Sample mean = = 26395.79

Population standard deviation = = 8500

a)

We have to construct 95% confidenc interval for the population mean.

Here population standard deviation is known so we have to use one sample z-confidence interval.

z confidence interval

Here E is a margin of error

Zc = 1.96    ( Using z table)

So confidence interval is ( 26395.79 - 3725.289 , 26395.79 + 3725.289) = > ( 22670.5 , 30121.08)

b)

E is a margin of error

Zc = 1.96    ( Using z table)