Question

In the planning stage, a sample proportion is estimated as pˆp^ = 72/80 = 0.90. Use this information to compute the minimum sample size n required to estimate p with 95% confidence if the desired margin of error E = 0.05. What happens to n if you decide to estimate p with 90% confidence? (You may find it useful to reference the z table. Round intermediate calculations to at least 4 decimal places and "z" value to 3 decimal places. Round up your answers to the nearest whole number.)  

Answer 1

Solution :

Given that,

= 0.90

1 - = 1 - 0.90 = 0.1

margin of error = E = 0.05

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 ( see the 0.025 value in standard normal (z) table corresponding z value is 1.96 )

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

= (1.96 / 0.05)2 * 0.90 * 0.1

= 138.29

Sample size = 138

b.

At 90% confidence level z

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.65 ( Using z table ( see the 0.05 value in standard normal (z) table corresponding z value is 1.65 )   

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

= (1.65 / 0.05)2 * 0.90 * 0.10

=98.01

Sample size = 98