Question

The following data relate to a variables-sampling application: x = $ 78.50 , s = $ 13.00 , s x = $ 1.00 ,

confidence level = 90 % ( Z = 1.65 )

The achieved precision is equal to:?.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = $ 78.50

Population standard deviation = s = $ 13.00

s = 1.00

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_0.05 = 1.65

Margin of error = E = Z/2 * s

= 1.65 * 1.00

= $ 1.65

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

  ± E

$ 78.50 ± $ 1.65

( $ 76.85, $ 80.15 )