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

0.35 and n=120

a)90​%

b)95​%

c)98​%

Click the icon to view a portion of the Cumulative Probabilities for the Standard Normal Distribution table.

a. The margin of error for a confidence interval to estimate the population proportion for the 90 % confidence level is _____​(Round to three decimal places as​ needed.)

b. The margin of error for a confidence interval to estimate the population proportion for the 95 % confidence level is _____​(Round to three decimal places as​ needed.)

c. The margin of error for a confidence interval to estimate the population proportion for the 98 % confidence level is _____​(Round to three decimal places as​ needed.)

Answer 1

a) At 90% confidence interval the critical value is z_0.05 = 1.645

Margin of error = z_0.05 * sqrt(p(1 - p)/n)

                            = 1.645 * sqrt(0.35 * (1 - 0.35)/120)

                            = 0.072

The margin of error for a confidence interval to estimate the population proportion for the 90% confidence level is 0.072

b) At 95% confidence interval the critical value is z_0.025 = 1.96

Margin of error = z_0.025 * sqrt(p(1 - p)/n)

                       = 1.96 * sqrt(0.35 * (1 - 0.35)/120)

                      = 0.085

The margin of error for a confidence interval to estimate the population proportion for the 95% confidence level is 0.085

c) At 98% confidence interval the critical value is z_0.01 = 2.33

Margin of error = z_0.01 * sqrt(p(1 - p)/n)

                            = 2.33 * sqrt(0.35 * (1 - 0.35)/120)

                            = 0.101

The margin of error for a confidence interval to estimate the population proportion for the 98% confidence level is 0.101