Use the normal distribution to find a confidence interval for a difference in proportions
given the relevant sample results. Assume the results come from random samples.
A 99% confidence interval for
Give the best estimate for
, 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 :
Enter your answer; Best estimate
Margin of error :
Enter your answer; Margin of error
Confidence interval :
Enter your answer; Confidence interval, value 1
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
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
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