A random sample of 388 married couples found that 280 had two or more personality preferences in common. In another random sample of 562 married couples, it was found that only 36 had no preferences in common. Let p1 be the population proportion of all married couples who have two or more personality preferences in common. Let p2 be the population proportion of all married couples who have no personality preferences in common.
(a) Find a 99% confidence interval for p1 – p2. (Use 3 decimal places.)
lower limit | |||
upper limit A random sample of 388 married couples found that
280 had two or more personality preferences in common. In another
random sample of 562 married couples, it was found that only 36 had
no preferences in common. Let p1 be the
population proportion of all married couples who have two or more
personality preferences in common. Let p2 be
the population proportion of all married couples who have no
personality preferences in common.
(a) Find a 99% confidence interval for p1 – p2. (Use 3 decimal places.)
|
p1cap = X1/N1 = 280/388 = 0.7216
p1cap = X2/N2 = 36/562 = 0.0641
Here, , n1 = 388 , n2 = 562
p1cap = 0.7216 , p2cap = 0.0641
Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.7216 * (1-0.7216)/388 + 0.0641*(1-0.0641)/562)
SE = 0.025
For 0.99 CI, z-value = 2.58
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.7216 - 0.0641 - 2.58*0.025, 0.7216 - 0.0641 +
2.58*0.025)
CI = (0.593 , 0.722)
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