Question

Consider two independent random samples with the following results:

n1=123pˆ1=0.48   n2=367pˆ2=0.63

Use this data to find the 80% confidence interval for the true difference between the population proportions.

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Step 1 of 3 :  

Find the point estimate that should be used in constructing the confidence interval.

Step 2 of 3: Find the value of the margin of error. Round your answer to six decimal places.

Step 3 of 3: Construct the 80% confidence interval. Round your answers to six decimal places.

Answer 1

Given :

Consider two independent random samples with the following results:

n1 = 123,   ${\hat{p}}$1 = 0.48   

n2 = 367,   ${\hat{p}}$2 = 0.63

1)) Point estimate = ${\hat{p}}$ 1 - ${\hat{p}}$ 2 = 0.48 - 0.63 = -0.15

2) Margin of error :

Margin of error = 2212x Ô C1-2, P2C1-Ⓡ2 + n2 =to 28* 10.48(1-0.48), 0.63(1-0.63 + 123 367 0.066071 .: E=0.066071

3) 80% confidence interval :

Significance level = $\alpha$ = 1-0.80 = 0.20

At 0.2 significance level the critical value of Z is

2x12 = 1.28 ::80./. CI is CT-CP-P2) + 2412 x Pic - Po) P261-82) ni t 02 = -0.15 + 1.28X 0.48(1-0.48) 0.63(1-0.63) + 123 367 = -0.15 + 0.066071 =(-0 (-0.216071,-0.083929)

2) The margin of error = E = 0.066071

3) 80% confidence interval is (-0.216071, -0.083929)