Question

For the specified margin of error and confidence​ level, obtain a sample size that will ensure a margin of error of at most the one specified.

margin of ERROR=0.01; confidence level= 90​%.

Find n=?

Answer 1

Solution :

Given that,

= 0.5 ( when estimate is not given than assume 0.5)

1 - = 1 - 0. 5= 0.5

margin of error = E =1 % = 0.01

At 90% confidence level z

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645} ( Using z table ( see the 0.05 value in standard normal (z) table corresponding z value is 1.645 )   

Sample size = n = (Z_/2 / E)² * * (1 - )

= (1.645 / 0.01)² * 0.5 * 0.5

=6765.06

Sample size = 6765