An international polling agency estimates that 36 percent of adults from Country X were first married between the ages of 18 and

Question

2 months ago

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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

Mathematics

1 answer:

Leona [12.6K] · Answer 1 · 2026-03-06T15:34:00+03:00

Answer:

(B) 0.2843

Step-by-step explanation:

Let d represent the difference in proportions between Country X and Country Y for adults who first married from ages 18 to 32.

The hypotheses can be stated as

$H_{0}$ : d=0.15

$H_{a}$ : d<0.15

The test statistic can be computed with the formula

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

p1 corresponds to the sample proportion from Country X (0.36)

p2 corresponds to the sample proportion from Country Y (0.26)

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

n1 is the sample size from Country X (60)

n2 is the sample size from Country Y (50)

Thus, $z=\frac{0.36-0.26-0.15}{\sqrt{{0.31*0.69*(\frac{1}{60} +\frac{1}{50}) }}}$ is approximately 0.5646

The p-value for this test statistic is approximately 0.2843

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