Question

A research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes during a week. They found the distribution of amounts spent per week followed the normal distribution with a population standard deviation of $5. A sample of 64 steady smokers revealed that   X¯¯¯=$20X¯=$20 .

a.	What is the 95% confidence interval estimate of μμ  ? (Round your answers to 3 decimal places.)

  Confidence interval is between $  and $  .

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 20

Population standard deviation =    = 5

Sample size = n = 64

a) At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025 =} 1.96

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

= 1.96 * ( 5 /  64 )

= 1.225

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

  ± E

20 ± 1.225   

( $18.775, $21.225 )