Question

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

$26,327.16

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

$8,100

Complete parts​ (a) through​ (c) below.

a. Determine a​ 95% confidence interval for the mean​ cost,

muμ​,

of all recent weddings in this country.

The​ 95% confidence interval is from ?????

​(Round to the nearest cent as​ needed.)

Answer 1

Solution :

Given that,

= 26327.16

s = 8100

n = 20

Degrees of freedom = df = n - 1 = 20 - 1 = 19

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,19 =2.093

Margin of error = E = t_/2,df * (s /n)

= 2.093 * (8100/ 20) = 3791

The 95% confidence interval estimate of the population mean is,

- E < < + E

26327.16 - 3791< < 26327.16 + 3791

22536.16 < <301180.16

(25, 26 )