Question

# Use the normal distribution to find a confidence interval for a difference in proportions p1-p2 given...

Use the normal distribution to find a confidence interval for a difference in proportions

p1-p2

given the relevant sample results. Assume the results come from random samples.

A 99% confidence interval for

p1-p2

given that

p^1=0.76

with

n1=590

and

p^2=0.65

with

n2=340

Give the best estimate for

p1-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 :

Margin of error :

Confidence interval :

For the 99% Confidence interval

= 0.76 and 1 - = 0.24, n1 = 590

= 0.65 and 1 - = 0.35, n2 = 340

The Zcritical (2 tail) for = 0.01, is 2.576

(a) The Best Estimate for p1 - p2 = (- ) = 0.76 – 0.65 = 0.11

(b) The Confidence Interval is given by (- ) ME, where

ME = 0.081

(c)

The Lower Limit = 0.11 - 0.081 = 0.029 (Rounding to 3 decimal places)

The Upper Limit = 0.11 + 0.081 = 0.191 (Rounding to 3 decimal places)

The 99% Confidence Interval is 0.029 < p1 - p2 < 0.191

#### Earn Coins

Coins can be redeemed for fabulous gifts.