Question

In a random sample of 100 measurements from a population with known standard deviation 200, the average was found to be 50. A 95% confidence interval for the true mean is

Answer 1

Solution :

Given that,

Point estimate = sample mean =     = 50

Population standard deviation =    =200

Sample size n =100

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 * ( 200 / 100 )

E= 39.2
At 95% confidence interval estimate of the population mean
is,

- E < < + E

50 - 39.2 <   < 50+ 39.2

10.8<   < 89.2

( 10.8, 89.2)