Question

A travel survey reported that Americans spend a mean of $160 a day with a standard deviation of 12.25 a day when they sent on vacation. The sample size was 36. Find the 95% confidence interval for the population mean.

Answer 1

Solution :

Given that,

Point estimate = sample mean =     = 160

Population standard deviation =    =12.25

Sample size n =36

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E =   Z/2    * ( /n)
= 1.96* ( 12.25 / 36)

= 4.00
At 95% confidence interval population mean
is,

- E < < + E

160  - 4 <   < 160  + 4

156<   <164

(156 , 164)