Question

Determine the margin of error for a 99​% confidence interval to estimate the population proportion with a sample proportion equal to 0.90 for the following sample sizes. a. nequals100             b. nequals180             c. nequals260 LOADING... Click the icon to view a portion of the Cumulative Probabilities for the Standard Normal Distribution table. a. The margin of error for a 99​% confidence interval to estimate the population proportion with a sample proportion equal to 0.90 and sample size nequals100 is nothing.

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = 0.90

1 - = 1 - 0.90 = 0.10

a) n = 100

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z_0.005 = 2.576

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

E = 2.576 (((0.90 * 0.10) / 100)

E = 0.077

b) n = 180

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

E = 2.576 (((0.90 * 0.10) / 180)

E = 0.058

c) n = 260

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

E = 2.576 (((0.90 * 0.10) / 260)

E = 0.048