Question

In the planning stage, a sample proportion is estimated as pˆ p ^ = 64/80 = 0.80. Use this information to compute the minimum sample size n required to estimate p with 99% confidence if the desired margin of error E = 0.08. What happens to n if you decide to estimate p with 95% 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,

= 0.80

1 - = 1 - 0.80 = 0.20

margin of error = E = 0.08

a ) At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576

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

= (2.576 / 0.08)2 * 0.8 * 0.8

= 165.87

n = sample size =166

b ) At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.960

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

= (1.960 / 0.08)2 * 0.8 * 0.8

= 96

n = sample size =96