Question

The lengths of text messages are normally distributed with a population standard deviation of 4 characters and an unknown population mean. If a random sample of 24 text messages is taken and results in a sample mean of 30 characters, find a 95% confidence interval for the population mean. Round your answers to two decimal places.

z0.10	z0.05	z0.04	z0.025	z0.01	z0.005
1.282	1.645	1.751	1.960	2.326	2.576

You may use a calculator or the common z-values above.

Select the correct answer below:

(28.66,31.34)

(28.57,31.43)

(27.90,32.10)

(28.10,31.90)

(28.40,31.60)

(28.95,31.05)

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 30

Population standard deviation =    = 4

Sample size = n =24

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960 ( Using z table )

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

= 1.960 * (4 /  24 )

E= 1.60
At 95% confidence interval estimate of the population mean
is,

- E < < + E

30 - 1.60 <   < 30 + 1.60

28.40<   < 31.60

(  28.40 , 31.60)