The American Bankers Association reported that, in a sample of 150 consumer purchases in France, 74...

The American Bankers Association reported that, in a sample of 150 consumer purchases in France, 74 were made with cash, compared with 28 in a sample of 60 consumer purchases in the United States.

Construct a 99 percent confidence interval for the difference in proportions. (Round your intermediate value and final answers to 4 decimal places.)

The 99 percent confidence interval is from to _____ . _____

Solutions

Expert Solution

Sample proportion 1 = 74 / 150 = 0.4933

Sample proportion 2 = 28 / 60 = 0.4667

99% confidence interval for p1 - p2 is

(1 - 2) - Z/2 * sqrt[ 1 ( 1- 1) / n1 + 2 ( 1 - 2) / n2) ] < p1 - p2 <

(1 - 2) + Z/2 * sqrt[ 1 ( 1- 1) / n1 + 2 ( 1 - 2) / n2) ]

(0.4933 - 0.4667) - 2.5758 * sqrt ( 0.4933 ( 1 - 0.4933) / 150) + 0.4667 * ( 1 - 0.4667) / 60 ) < p1 - p2 <

(0.4933 - 0.4667) + 2.5758 * sqrt ( 0.4933 ( 1 - 0.4933) / 150) + 0.4667 * ( 1 - 0.4667) / 60 )

-0.1698 < p1 - p2 < 0.2230

99% CI is from -0.1698 to 0.2230

milcah answered 3 months ago

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