Question
In: Math
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 9 months agoRelated Solutions
The American Bankers Association reported that, in a sample of 150 consumer purchases in France, 60...
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decimal places.)
The 95 percent confidence interval is from ______ to _______
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