Question

- P2 given the relevant sample results. Assume the results come Use the normal distribution to find a confidence interval for a difference in proportions p from random samples. A 95% confidence interval for pı - P2 given that Pı = 0.20 with nį = 60 and P2 = 0.40 with n2 = 120 Give the best estimate for p - p2, the margin of error, and the confidence interval. Round your answer for the best estimate to two decimal places and round your answers for the margin of error and the confidence interval to three decimal places. Best estimate: i Margin of error: i Confidence interval:

Answer 1

Solution:

We are given

N1 = 60

N2 = 120

P1 = 0.2

P2 = 0.4

Best estimate for difference in two proportions is given as below:

Best Estimate: P1 - P2 = 0.2 - 0.4 = -0.20

Best estimate = -0.20

We are given

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

The margin of error is given as below:

Margin of error = Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Margin of error = 1.96*sqrt[(0.2*(1 – 0.2)/60) + (0.4*(1 – 0.4)/120)]

Margin of error = 1.96* 0.0683

Margin of error = 0.133868

Margin of error = 0.134

Confidence interval for difference between two population proportions:

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Where, P1 and P2 are sample proportions for first and second groups respectively.

Confidence interval = -0.20 ± 0.133868

Lower limit = -0.20 - 0.133868 = -0.334

Upper limit = -0.20 + 0.133868 = -0.066

Confidence interval = -0.334 to -0.066

-0.334 < (p1 - p2) < -0.066