Mathematics, 21.12.2019 03:31, salihanegawo

An advocacy organization surveys 1060 citizens of country a and 171 of them reported being born in another country. similarly, 226 out of 1216 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.
a. what is the difference in the sample proportions, p_1 - p_2. of foreign born residents from both countries? assume p_1 is the proportion of foreign boom residents from country a and p_2 is the proportion of foreign boom residents from country b.
b. what is the pooled proportion of foreign boom in both countries combined?
c. what is the standard error of the difference in part a?

Answers: 1

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Answers

answered: Jasten

a) $\hat p_1 -\hat p_2 = 0.1613-0.1859=-0.0246$

b) $\hat p =\frac{171 +226}{1060 +1216}=0.1744$

c) $SE =\sqrt{\frac{0.1613 (1-0.1613)}{1060} +\frac{0.1859 (1-0.1859)}{1216}}=0.0159$

Step-by-step explanation:

Previous concepts and some notation

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

p_1 represent the true proportion of people reported being born in another country in country A

X_1 =171 represent the number of people reported being born in another country in country A

n_1 =1060 represent the sample in country A

n_2 =1216 represent the sample in country B

X_2 =226 represent the number of people reported being born in another country in country B

p_2 represent the true proportion of people reported being born in another country in country B

$\hat p_1=\frac{171}{1060}=0.1613$ represent the estimated proportion of people reported being born in another country in country A

$\hat p_2=\frac{226}{1216}=0.1859$ represent the estimated proportion of people reported being born in another country in country B

a. What is the difference in the sample proportions, p_1 - p_2. of foreign born residents from both countries? Assume p_1 is the proportion of foreign boom residents from country A and p_2 is the proportion of foreign boom residents from country B.

$\hat p_1 -\hat p_2 = 0.1613-0.1859=-0.0246$

b. What is the pooled proportion of foreign boom in both countries combined?

The pooled proportion is given by this formula:

$\hat p =\frac{X_1 +X_2}{n_1 +n_2}$

And if we replace we got:

$\hat p =\frac{171 +226}{1060 +1216}=0.1744$

c. What is the standard error of the difference in part a?

The standard error for a difference of proportions is given by:

$SE =\sqrt{\frac{\hat p_1 (1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

And if we replace we got:

$SE =\sqrt{\frac{0.1613 (1-0.1613)}{1060} +\frac{0.1859 (1-0.1859)}{1216}}=0.0159$

answered: Guest

the answer is the third / last answer

step-by-step explanation:

answered: Guest

answer: i wpould go with c

step-by-step explanation:

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An advocacy organization surveys 1060 citizens of country a and 171 of them reported being born in a...