Question

A statistician is investigating the mean time that people wait in line while checking out at a grocery store. She observed 100 customers and computed a mean time of 12.35 minutes and a standard deviation of 1.38 minutes.

A)Calculate a 90% and a 95% confidence interval for the population mean.

B)Interpret the meaning of a confidence interval.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 12.35

Population standard deviation =    = 1.38

Sample size = n =100

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E = Z/2    * ( /n)

= 1.645 * ( 1.38 /  100 )

E= 0.2270
At 90% confidence interval estimate of the population mean
is,

- E < < + E

12.35 - 0.2270 <   < 12.35 + 0.2270

12.1230 <   < 12.5770

( 12.1230 , 12.5770 )

(B)

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 * ( 1.38 /  100 )

= 0.2705
At 95% confidence interval estimate of the population mean
is,

- E < < + E

12.35  - 0.2705 <   < 12.35  +0.2705

12.0795 <   < 12.6205

( 12.0795 , 12.6205)