Question

Determine the margin of error for a confidence interval to estimate the population proportion for the following confidence levels with a sample proportion equal to 0.36 and n=125.

a. 90​%

            b. 95​%

            c. 98​%

a. The margin of error for a confidence interval to estimate the population proportion for the 90% confidence level is _

b. The margin of error for a confidence interval to estimate the population proportion for the 95% confidence level is _

c. The margin of error for a confidence interval to estimate the population proportion for the 98% confidence level is _

Answer 1

Solution :

n = 125

= 0.360

1 - = 1 - 0.360 =0.640

a ) At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z/2 = Z0.05 = 1.645

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.645 * (((0.360 * 0.640) / 125)

= 0.071

Margin of error = 0.071

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

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.960 * (((0.360 * 0.640) / 125)

= 0.084

Margin of error =0.084

c) At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z/2 = Z0.01 = 2.330

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 2.330* (((0.360 * 0.640) / 125)

= 0.101

Margin of error =0.101