Question

The sample size required to estimate, at 95% confidence, a population mean with a maximum allowable margin of error of +/- 1.5 when the population standard deviation is 5 is ____ observations.

The sample size required to estimate, at 90% confidence, a population proportion with a maximum allowable margin of error of +/- 3.00 percentage points is ____ observations.

Answer 1

Solution :

Given that,

standard deviation = = 5

margin of error = E = 1.5

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

= ((1.96 * 5) / 1.5)²

= 42.68 = 43

Sample size = 43

Given that,

= 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E = 3% = 0.03

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 / 0.03)² * 0.5 * 0.5

= 751.67 = 752

sample size = 752