Question

1 1 CUINE IVI Calculator X + -s/35882/assignments/4122612?module_item_id=13064730 NP3 - Your answer is partially correct. Use the normal distribution to find a confidence interval for a difference in proportions P - P2 given the relevant sample results. Assume the results come from random samples. A 90% confidence interval for Pi-P2 given that = 0.73 with n = 520 and 2 = 0.68 with n2 = 240 Give the best estimate for Pi-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: 1.645 Margin of error: i 1.645 Confidence interval: -0.009 0.109 e Textbook and Media Hint Assistance Used If we have large samples of sizes and n2 from two different groups, we can construct a confidence interval for Pi-P2. the difference in proportions between the two groups, using 4/12/2020

Answer 1

a) The point estimate for the difference in proportion is computed here as :

$\hat p_1 - \hat p_2 = 0.73 -0.68 = 0.05$

Therefore 0.05 is the best estimate value here.

b) From standard normal tables, we have:
P(-1.645 < Z < 1.645) = 0.9

The pooled proportion here is computed as:
$P = \frac{p_1 n_1 + p_2 n_2}{n_1 + n_2} = \frac{0.73*520 + 0.68*240}{520 + 240} = 0.7142$

The standard error now is computed here as:

$SE = \sqrt{P(1-P)(\frac{1}{n_1} + \frac{1}{n_2})} = \sqrt{0.7142(1 - 0.7142)(\frac{1}{520} + \frac{1}{240})} = 0.0353$Now the margin of error here is computed as:

MOE = z*SE = 1.645*0.0353 = 0.0580

Therefore 0.0580 is the required margin of error here.

The confidence interval here is thus obtained as:

p1 + p2 +- MOE

0.05 +- 0.0580

(-0.0080, 0.1080)

This is the required 90% confidence interval here.