A random sample of 362 married couples found that 290 had two or more personalit

Question

A random sample of 362 married couples found that 290 had two or more personality preferences in common. In another random sample of 550 married couples, it was found that only 38 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.)

How hard is it to reach a businessperson by phone? Let p be the proportion of calls to businesspeople for which the caller reaches the person being called on the first try.

(a) If you have no preliminary estimate for p, how many business phone calls should you include in a random sample to be 80% sure that the point estimate p will be within a distance of 0.07 from p? (Round your answer up to the nearest whole number.)
phone calls

(b) A report states that businesspeople can be reached by a single phone call approximately 13% of the time. Using this (national) estimate for p, answer part (a). (Round your answer up to the nearest whole number.)
phone calls

Explanation / Answer

a)

p1 = 290/362 =0.80, n1 = 362 , p2 = 38/550 = 0.069 , n2 = 550

99% z value = 2.576

CI = ( p1 - p2) + /- z * sqrt ( (p1 * (1 - p1) / n1 + p2 * (1 -p2) / n2)

= ( 0.80 - 0.069) + / - 2.576 * sqrt ( 0.80 * 0.2/ 362 + 0.069 * 0.931/ 550)

= (0.670 , 0.792)

lower limit = 0.670

Upper limit = 0.792

b )

If p is unknown, then p = q = 0.5
Confidence level = 80% = 0.8
alpha = 1 - 0.8 = 0.2
alpha/2 = 0.2/2 = 0.1
Z(a/2) = 1.28
Error = 0.07

Formula for error on p-hat:
E=Z(a/2)*sqrt(p*q/n)

Solving for n we get:
n = p*q*(Z/E)^2
n = (.5)(.5)(1.28/.07)^2
n = 83.59
n = 84

(b)
p = 13% = 0.13
q = 1 - p = 1 - 0.13 = 0.87
n = (.13)(.87)(1.28/.07)^2
n = 37.81
n = 38

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