Question

I. In a random sample of 1200 Canadians, 225 reported being born in another country; similarly, 210 out of 1300 Americans reported being foreign-born. Using a 99% confidence level, build a confidence interval to test the claim that the two countries have about the same proportion of foreign-born inhabitants. (a) X = (b) n = (c)2 = (d) x2 = (e) n2 = (1) P2 = SE, = V8, 162,582) = 0.0113 SE = />•(7782) = 0.0102 SE, = / SE? + SE? (8) Ộ2 - =— (h) SE, =— (i) z* = — (1) MoE=— (k) CI = ( — ,- )

Answer 1

Here,

a) x₁ = 225

b) n₁ = 1200

c) $\widehat{p}$ ₁ = ( x₁ / n₁ )

= 225 /1200

= 0.1875

d) x₂ = 210

e) n₂ = 1300

f)  $\widehat{p}$₂ = ( x₂/ n₂)

=( 210 / 1300 )

= 0.1615

g)  $\widehat{p}$₁ -  $\widehat{p}$₂ = (0.1875 - 0.1615)

= 0.026

h) SE_t = $\sqrt$ ( SE₁² + SE₂² )

= $\sqrt$ {(0.0113)² + (0.0102)² }

= 0.0151

i) From the critical z table, the critical value of z is:

z^* = $\pm$ 2.576

j) MoE = z*( SE_t)

= $\pm$ 2.576 * 0.0151

= $\pm$ 0.0388

k) CI =  ($\widehat{p}$₁ -  $\widehat{p}$₂) $\pm$ MoE

= 0.026  $\pm$ 0.0388

= (-0.012 , 0.064)