Question

Determine the margin of error for a 95​% confidence interval to estimate the population proportion with a sample proportion equal to 0.70 for the following sample sizes. a. nequals100             b. nequals200             c. nequals250 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 95​% confidence interval to estimate the population proportion with a sample proportion equal to 0.70 and sample size nequals100 is nothing. ​(Round to three decimal places as​ needed.) b. The margin of error for a 95​% confidence interval to estimate the population proportion with a sample proportion equal to 0.70 and sample size nequals200 is nothing. ​(Round to three decimal places as​ needed.) c. The margin of error for a 95​% confidence interval to estimate the population proportion with a sample proportion equal to 0.70 and sample size nequals250 is nothing. ​(Round to three decimal places as​ needed.)

Answer 1

Solution:

The formula for margin of error for the population proportion is given as below:

Margin of error = Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

Confidence level = 95%

So, the critical value Z by using z-table is given as below:

Z = 1.96

Sample proportion = P = 0.70

Part a)

We are given

n = 100

P = 0.70

(1 – P) = 1 – 0.70 = 0.30

Z = 1.96

Margin of error = Z* sqrt(P*(1 – P)/n)

Margin of error = 1.96*sqrt(0.70*0.30/100)

Margin of error = 1.96* 0.045826

Margin of error = 0.089819

The margin of error for a 95​% confidence interval to estimate the population proportion with a sample proportion equal to 0.70 and sample size n =100 is 0.090.

Part b

We are given

n = 200

P = 0.70

(1 – P) = 1 – 0.70 = 0.30

Z = 1.96

Margin of error = Z* sqrt(P*(1 – P)/n)

Margin of error = 1.96*sqrt(0.70*0.30/200)

Margin of error = 1.96* 0.032404

Margin of error = 0.063512

The margin of error for a 95​% confidence interval to estimate the population proportion with a sample proportion equal to 0.70 and sample size n =200 is 0.064.

Part c

We are given

n = 250

P = 0.70

(1 – P) = 1 – 0.70 = 0.30

Z = 1.96

Margin of error = Z* sqrt(P*(1 – P)/n)

Margin of error = 1.96*sqrt(0.70*0.30/250)

Margin of error = 1.96* 0.028983

Margin of error = 0.056807

The margin of error for a 95​% confidence interval to estimate the population proportion with a sample proportion equal to 0.70 and sample size n =250 is 0.057.