Question

7. 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 pwith 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% = ?

Answer 1

solution:

Given that,

= 0.90

1 - = 1 - 0.90 = 0.10

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 = Z_0.025 = 1.96

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

= (1.96 * 0.05)² * 0.90 * 0.10

= 138.2976

Sample size = 138

(b)at 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= (1.645 * 0.05)² * 0.90 * 0.10

= 97.41

Sample size = 97