Question

1: point estimate

2: margin of error

3: lower and upper endpoint (Confidence interval)

ave & Exit Certify Lesson: 9.4 Comparing Two Population Pr... stion 4 of 10, Step 1 of 3 6/30 Correct nsider two independent random samples with the following results: ni 160 n2 = 95 84 72 Use this data to find the 90% confidence interval for the true difference between the population proportions. Step 1 of 3: Find the point estimate that should be used in constructing the confidence interval. Round your answer to three decimal Answer How to enter your answer 2020 Hawkes Learning

Answer 1

Solution:

Given

n1= 160

n2= 95

X1 = 84

X2 = 72

X1 Pi 84 160 = 0.525 n1

91 = 1 P1 = 0.475

X2 P2 = 72 95 0.758 n2

92 = 1-2 = 0.242

. Level of significance

a = 0.10

1) To find the point estimate for the difference between the population proportion is

$Point \ estimate = \hat p_1- \hat p_2$

Point estimate = 0.525 - 0.758

Point estimate = -0.233

2) To find 90% margin of error for the difference between the population proportion is

$M.E= Z_{\alpha/2}* \sqrt \frac{\hat p_1*(1-\hat p_1)}{n_1}+\frac{\hat p_2*(1-\hat p2)}{n_2}$at

a = 0.10

$Z_{\alpha/2}=Z_{0.05}=1.64$. From Z table

$M.E= 1.64*\sqrt \frac{0.525*0.475}{160}+\frac{0.758*0.242}{95}$

$M.E= 1.64*\sqrt \frac{0.249375}{160}+\frac{0.183436}{95}$

$M.E= 1.64*\sqrt 0.0015585+0.0019309$

$M.E= 1.64*\sqrt 0.0034894$

M.E = 0.0968766

Margin of error =M.E= 0.0968766

3) The 90% confidence interval for the difference between the population proportion is

$Point \ estimate \pm M.E$

$(-0.233\pm 0.0968766)$

(- 0.329876, - 0.136123)

( - 0.33, 0.136)

The 90% confidence interval for the difference between the population proportion is ( -0.33, 0.136)

Lower end point = -0.33

Upper end point = - 0.136