Question

In: Statistics and Probability

An advocacy organization surveys 1050 citizens of country A and 177 of them reported being born...

An advocacy organization surveys 1050 citizens of country A and 177 of them reported being born in another country.​ Similarly, 163 out of 1191 citizens of country B reported being​ foreign-born. The researchers want to test if the proportions of foreign born are the same in country B as in country A. Complete parts a through e below.

a) what is the difference in the sample proportions phat1-phat2 of foreign born residents from both countries?

b) what is the pooled proportion of foreign born countries combined?

c) what is the SE of differences in part b

d)what is the value of the z-statistic

e) What do you conclude? P value and conclusion

Solutions

Expert Solution

Solution:

Here, we have to use two sample z test for difference in two population proportions.

a) what is the difference in the sample proportions phat1-phat2 of foreign born residents from both countries?

We are given

X1 = 177

N1 = 1050

X2 = 163

N2 = 1191

P̂1 = X1/N1 = 177/1050 = 0.168571429

P̂2 = X2/N2 = 163/1191 = 0.136859782

P̂1 - P̂2 = 0.168571429 - 0.136859782 = 0.031711647

Difference in the sample proportions = 0.031711647

b) what is the pooled proportion of foreign born countries combined?

Pooled proportion is given as below:

P = (X1+X2)/(N1+N2)

P = (177 + 163) / (1050 + 1191) = 0.1517

Pooled Proportion = 0.1517

c) what is the SE of differences in part b

Standard error is given as below:

SE = sqrt(P*(1 – P)*((1/N1) + (1/N2)))

SE = sqrt(0.1517*(1 - 0.1517)*((1/1050)+(1/1191)))

SE = 0.015186

d)what is the value of the z-statistic

A Z-test statistic for this test is given as below:

Z = (P̂1 – P̂2) / sqrt(P*(1 – P)*((1/N1) + (1/N2)))

Z = (0.168571429 - 0.136859782)/ sqrt(0.1517*(1 - 0.1517)*((1/1050)+(1/1191)))

Z = 0.031711647 / 0.015186

Z = 2.0881

Test statistic = 2.0881

e) What do you conclude? P value and conclusion

From above test statistic value, we have

P-value = 0.0368

(by using z-table)

P-value < α = 0.05

So, we reject the null hypothesis at 5% level of significance

There is sufficient evidence to conclude that proportions of foreign born are not same in country B and country A.


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An advocacy organization surveyed 960 Canadians and 192 of them reported being born in another country. Similarly, 170 out of 1250 Americans reported being foreign-born. Based on the results, construct a 95% confidence interval for the difference in proportions of Canadians and Americans who were born in foreign countries. Note z ∗ = 1.96 .
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