Question

An international polling agency estimates that 36 percent of adults from Country X were first married between the ages of 18 and 32, and 26 percent of adults from Country Y were first married between the ages of 18 and 32. Based on the estimates, which of the following is closest to the probability that the difference in proportions between a random sample of 60 adults from Country X and a random sample of 50 adults from Country Y (Country X minus Country Y) who were first married between the ages of 18 and 32 is greater than 0.15?
(A) 0.1398
(B) 0.2843
(C) 0.4315
(D) 0.5685
(E) 0.7157

Answer 1

Answer:

(B) 0.2843

Step-by-step explanation:

Let d be the difference in proportions from Country X and Country Y who were first married between the ages of 18 and 32.

Then hypotheses are

$H_{0}$: d=0.15

$H_{a}$: d<0.15

Test statistic can be found using the equation

$z=\frac{p1-p2-0.15}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}$ where

• p1 is the sample proportion of Country X (0.36)

• p2 is the sample proportion of Country Y (0.26)

• p is the pool proportion of p1 and p2 ($\frac{60*0.36+50*0.26}{50+60}=0.31$)

• n1 is the sample size of adults from Country X (60)

• n2 is the sample size of adults from Country Y (50)

Then $z=\frac{0.36-0.26-0.15}{\sqrt{{0.31*0.69*(\frac{1}{60} +\frac{1}{50}) }}}$ ≈ 0.5646

p-value of test statistic is ≈ 0.2843

p-value states the probability that the difference in proportions between a random sample of 60 adults from Country X and a random sample of 50 adults from Country Y who were first married between the ages of 18 and 32 is at least 0.15