Question

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.) lower limit     upper limit

Answer 1

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)