Question

Determine the margin of error for a confidence interval to estimate the population mean with n=34 and σ=48 for the following confidence levels.

91%

95%

98%

Answer 1

Solution :

Given that,

= 48

n = 34

a) At 91% confidence level the z is ,

  = 1 - 91% = 1 - 0.91 = 0.09

/ 2 = 0.09 / 2 = 0.045

Z_/2 = Z_0.045 = 1.695

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

= 1.695 * ( 48/ 34)

E = 13.95

b) 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 = 1.96

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

= 1.96 * ( 48/ 34)

E = 16.13

c) At 98% confidence level the z is ,

  = 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z_/2 = Z_0.01 = 2.326

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

= 2.326 * ( 48/ 34)

E = 19.15