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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

Can you please explain how do this both manually and using calculator

Respuesta :

Answer:

option B is correct

Step-by-step explanation:

P[tex]_{1}[/tex] = 0.36  

P[tex]_{2}[/tex] = 0.26

n[tex]_{1}[/tex]= 60

n[tex]_{2}[/tex]=50

Sampling distribution

μ =  P[tex]_{1}[/tex] - P[tex]_{2}[/tex] = 0.36 - 0.26 = 0.10

σ = [tex]\sqrt{\frac{p_{1}(1-p_{1}) }{n_{1}} + \frac{p_{2}(1-p_{2}) }{n_{2}} }[/tex] = [tex]\sqrt{\frac{0.36(1-0.36) }{60} + \frac{0.26(1-0.26) }{50} }[/tex]

= 0.0877

the Z-score =   [tex]\hat{p_{1}} - \hat{p_{2}}[/tex] = 0.15

z = [tex]\frac{0.15-0.10}{0.0877}[/tex] = 0.57

P([tex]\hat{p_{1}} - \hat{p_{2}}[/tex] > 0.15) = P(z >0.57)= 1 - P(z ≤0.57)

1-0.7157 = 0.2843

option B is correct

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