Question
In: Math
The American Bankers Association reported that, in a sample of 125 consumer purchases in France, 62...
The American Bankers Association reported that, in a sample of 125 consumer purchases in France, 62 were made with cash, compared with 21 in a sample of 40 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
Solution :
Given that,
= x1 / n1 = 62 / 125 = 0.496
1-
= 0.504
= x2 / n2 = 21 / 40 = 0.525
1 -
= 0.475
At 99% confidence level the z is ,
Z
/2
= Z 0.005 = 2.576
99% confidence interval for p1 - p2 is
,
(
-
) 
Z/2 *
[(1-
) / n1 +
(1 -
) / n2]
( 0.496 - 0.525 )
2.576 *
[(0.496 * 0.504) / 125 + (0.525 * 0.475) / 40]
-0.2630 < p1 - p2 < 0.2050
-0.2630 to 0.2050
milcah answered 9 months agoRelated Solutions
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The American Bankers Association reported that, in a sample of
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