Question

13. In the planning stage, a sample proportion is estimated as p^ = 36/60 = 0.60. Use this information to compute the minimum sample size n required to estimate p with 95% confidence if the desired margin of error E = 0.09. 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.)  

Confidence Level	n
95%	?
90%	?

Please do not answer if you are not sure, THANK YOU! :)

Answer 1

13)

Solution :

Given that,

= 0.60

1 - = 0.40

margin of error = E = 0.09

At 95% confidence

Z_/2 = Z_0.025 = 1.96

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.96 / 0.09)² * 0.60 * 0.40

= 114

sample size = n = 114

At 90% confidence

Z_/2 = Z_0.05 = 1.645

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.645 / 0.09)² * 0.60 * 0.40

= 81

sample size = n = 81