Question

(1 point) In the following questions, use the normal distribution to find a confidence interval for a difference in proportions pu - P2 given the relevant sample results. Give the best point estimate for p. - P2, the margin of error, and the confidence interval. Assume the results come from random samples. Give your answers to 4 decimal places. 300. Use 1. A 80% interval for pı - P2 given that i = 0.15 with ni = 250 and 2 = 0.21 with n2 z* = 1.282. a. Point estimate: b. Margin of error: c. Confidence interval:( - 0999 9 -.0161 2. A 99% interval for P1 - P2 given that i = 0.69 with n = 2000 and 2 = 0.81 with n2 = 2000. Use z* = 2.576. a. Point estimate: b. Margin of error: c. Confidence interval:( 200. Use 3. A 95% interval for p - P2 given that i = 0.43 with n = 300 and 2 = 0.29 with n2 z* = 1.96. a. Point estimate: b. Margin of error: c. Confidence interval:( Note: You can earn partial credit on this problem. Preview My Answers Submit Answers Varrera

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Answer 1

1. p1 = 0.15, p2= 0.21, n1= 250, n2= 300, z = 1.282

Now,

a. point estimate = p1 - p2 = 0.15-021 = -0.06

b.

The critical value for α=0.2 is zc​=z_1−α/2​=1.282.

margin of error = 0.042

c.

The corresponding confidence interval is computed as shown below:

$\begin{array}{ccl} CI = \displaystyle \left( \hat p_1 - \hat p_2 - z_c \sqrt{\frac{\hat p_1(1-\hat p_1)}{N_1}+\frac{\hat p_2(1-\hat p_2)}{N_2}}, \hat p_1 - \hat p_2 + z_c \sqrt{(\frac{\hat p_1(1-\hat p_1)}{N_1}+\frac{\hat p_2(1-\hat p_2)}{N_2}} \right) \\ \\ = \displaystyle \left( 0.15 - 0.21 - 1.282 \times \sqrt{\frac{0.15 (1-0.15)}{250}+\frac{0.21 (1-0.21)}{300}}, 0.15 - 0.21 + 1.282 \times \sqrt{\frac{0.15 (1-0.15)}{250}+\frac{0.21 (1-0.21)}{300}} \right) \\ \\ = (-0.102, -0.018) \end{array}$

2. a. point estimate : p1 - p2 = 0.69 - 0.81 = -0.12

b.

At alpha = 0.01/2 , z = 2.576

margin of error : 0.035

c.

The critical value for α=0.01 is zc​ = 2.576. The corresponding confidence interval is computed as shown below:

$\begin{array}{ccl} CI = \displaystyle \left( \hat p_1 - \hat p_2 - z_c \sqrt{\frac{\hat p_1(1-\hat p_1)}{N_1}+\frac{\hat p_2(1-\hat p_2)}{N_2}}, \hat p_1 - \hat p_2 + z_c \sqrt{(\frac{\hat p_1(1-\hat p_1)}{N_1}+\frac{\hat p_2(1-\hat p_2)}{N_2}} \right) \\ \\ = \displaystyle \left( 0.69 - 0.81 - 2.576 \times \sqrt{\frac{0.69 (1-0.69)}{2000}+\frac{0.81 (1-0.81)}{2000}}, 0.69 - 0.81 + 2.576 \times \sqrt{\frac{0.69 (1-0.69)}{2000}+\frac{0.81 (1-0.81)}{2000}} \right) \\ \\ = (-0.155, -0.085) \end{array}$

3.

a. p1 = 0.43, n1 = 300, p2= 0.29, n2= 200, z = 1.96

p1-p2 = 0.14

b.

z = 1.96 and

margin of error:0.084

c.

The critical value for α=0.05 is z_c = 1.96. The corresponding confidence interval is computed as shown below:

$\begin{array}{ccl} CI = \displaystyle \left( \hat p_1 - \hat p_2 - z_c \sqrt{\frac{\hat p_1(1-\hat p_1)}{N_1}+\frac{\hat p_2(1-\hat p_2)}{N_2}}, \hat p_1 - \hat p_2 + z_c \sqrt{(\frac{\hat p_1(1-\hat p_1)}{N_1}+\frac{\hat p_2(1-\hat p_2)}{N_2}} \right) \\ \\ = \displaystyle \left( 0.43 - 0.29 - 1.96 \times \sqrt{\frac{0.43 (1-0.43)}{300}+\frac{0.29 (1-0.29)}{200}}, 0.43 - 0.29 + 1.96 \times \sqrt{\frac{0.43 (1-0.43)}{300}+\frac{0.29 (1-0.29)}{200}} \right) \\ \\ =; (0.056, 0.224) \end{array}$

4.

The critical value for α=0.05 is zc​=1.96. The corresponding confidence interval is computed as shown below:

CI = P1 - P2 – 2c Pi(1-P) Ni + P2(1-22) N2 -,ội – p2+zet/(P1(1 – 1) P2(1 – †2) + Ni N2 0.43 – 0.29 – 1.96 x 0.43(1 – 0.43) 300 + 0.29(1 – 0.29) 200 0.43 – 0.29 + 1.96 x 0.43(1 – 0.43) 300 + 0.29(1 - 0.29) 200 = (0.056, 0.224)

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