Question

The lengths of text messages are normally distributed with a population standard deviation of 5 characters and an unknown population mean. If a random sample of 21 text messages is taken and results in a sample mean of 28 characters, find a 99% confidence interval for the population mean. Round your answers to two decimal places.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 28

Population standard deviation =    = 5
Sample size = n =21

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576   ( Using z table )

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

= 2.576* (5 / 21)

= 2.81

At 99% confidence interval estimate of the population mean is,

- E < < + E

28-2.81 < < 28+2.81

25.19< < 30.81

(25.19 , 30.81)