# An article regarding interracial dating and marriage recentl

An article regarding interracial dating and marriage recently appeared in a newspaper. Of the 1713 randomly selected adults, 301 identified themselves as Latinos, 325 identified themselves as blacks, 255 identified themselves as Asians, and 778 identified themselves as whites. Among Asians, 79% would welcome a white person into their families, 71% would welcome a Latino, and 66% would welcome a black person.
If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)
Construct the 95% confidence intervals for the three Asian responses. (Round your answers to four decimal places.)
Construct the 95% confidence intervals for the three Asian responses. (Round your answers to four decimal places.)
a) Welcome a white peron.
b) welcome a black person.
c) welcome a latino

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Step 1
From the Standard Normal table, the z value for 95% confidence level is 1.96.
The 95% confidence intervals for the Asian responses "Welcome a white peron" is,
$$\displaystyle{C}{I}=\hat{{{p}}}\pm{z}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}$$
$$\displaystyle={0.79}\pm{1.96}\sqrt{{{\frac{{{0.79}{\left({1}-{0.79}\right)}}}{{{255}}}}}}$$
$$\displaystyle={0.79}\pm{0.0499}$$
$$\displaystyle={\left({0.7401},\ {0.8399}\right)}$$
Step 2
The 95% confidence intervals for the Asian responses "welcome a black person" is,
$$\displaystyle{C}{I}=\hat{{{p}}}\pm{z}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}$$
$$\displaystyle={0.66}\pm{1.96}\sqrt{{{\frac{{{0.66}{\left({1}-{0.66}\right)}}}{{{255}}}}}}$$
$$\displaystyle={0.66}\pm{0.0581}$$
$$\displaystyle={\left({0.6019},\ {0.7181}\right)}$$
The 95% confidence intervals for the Asian responses "welcome a latino" is,
$$\displaystyle{C}{I}=\hat{{{p}}}\pm{z}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}$$
$$\displaystyle={0.71}\pm{1.96}\sqrt{{{\frac{{{0.71}{\left({1}-{0.71}\right)}}}{{{255}}}}}}$$
$$\displaystyle={0.71}\pm{0.0557}$$
$$\displaystyle={\left({0.6543},\ {0.7657}\right)}$$
###### Not exactly what you’re looking for?
limacarp4
a) Welcome a white person
$$\displaystyle\hat{{{p}}}={0.79}$$
$$\displaystyle{S}.{E}=\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}=\sqrt{{{\frac{{{\left({0.79}\right)}{\left({0.21}\right)}}}{{{251}}}}}}={0.02571}$$
$$\displaystyle{95}\%{C}{I}={\left(\hat{{{p}}}-{t}_{{{\frac{{\alpha}}{{{2}}}}}}\times{S}.{E},\ \hat{{{p}}}+{t}_{{{\frac{{\alpha}}{{{2}}}}}}\times{S}.{E}\right)}$$
$$\displaystyle{t}_{{{\frac{{\alpha}}{{{2}}}}}}$$ for $$\displaystyle{d}{f}={n}-{1}={250}$$
$$\displaystyle{t}_{{{0.025}}},\ {250}={1.9695}$$
$$\displaystyle\therefore{95}\%{C}{I}={\left({0.79}-{\left({1.9695}\right)}{\left({0.02571}\right)},\ {0.79}+{\left({1.9695}\right)}{\left({0.0257}\right)}\right)}$$
$$\displaystyle={\left({0.7394},\ {0.8406}\right)}$$
b) Welcom a latino
$$\displaystyle\hat{{{p}}}={0.71}$$
$$\displaystyle{S}.{E}=\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}=\sqrt{{{\frac{{{\left({0.71}\right)}{\left({0.29}\right)}}}{{{251}}}}}}={0.02864}$$
$$\displaystyle{95}\%{C}{I}={\left({0.71}-{\left({1.9695}\right)}{\left({0.02864}\right)},\ {0.71}+{\left({1.9695}\right)}{\left({0.02864}\right)}\right)}$$
$$\displaystyle={\left({0.6536},\ {0.7664}\right)}$$
c) Welcome a black
$$\displaystyle\hat{{{p}}}={0.66}$$
$$\displaystyle{S}.{E}=\sqrt{{{\frac{{{\left({0.66}\right)}{\left({0.34}\right)}}}{{{251}}}}}}={0.0299}$$
$$\displaystyle{95}\%{C}{I}={\left({0.66}-{\left({1.9695}\right)}{\left({0.03}\right)},\ {0.66}+{\left({1.96}\right)}{\left({0.031}\right)}\right)}$$
$$\displaystyle={\left({0.6011},\ {0.7189}\right)}$$
nick1337

Step 1
White
Sample size, $$n=251$$
We use normal approximation, for this we check that both np and $$n(1-p)>5$$
Since $$n\times p=198.29>5$$ and $$n\times(1-p)=52.71>5$$, we can take binomial random variable as normally distributed, with mean $$=p=0.79$$ and std error $$=\sqrt{\frac{p}\times(1-p)}{n}=0.025709$$
For constructing Confidence interval,
Margin of Error
95% confidence interval is given by:
$$\text{Sample Mean}\pm\text{(Margin of Error)}$$
$$0.79\pm0.0504=(0.7369,\ 0.8404)$$
Step 2
Latino
Sample size $$n=251$$
Sample proportion, $$p=0.71$$
We use normal approximation, for this we check that both np and $$n(1-p)>5$$
Since $$n\times p=178.21>5$$ and $$n\times(1-p)=72.79>5$$, we can take binomial random variable as normally distributed, with mean $$=p=0.71$$ and std error $$=\sqrt{\frac{p\times(1-p)}{n}}=0.02864$$
For constructing Confidence interval,
Margin of Error $$(ME)=z\times SD=0.0561$$
95% confidence interval is given by:
$$\text{Sample Mean}\pm\text{(Margin of Error)}$$
$$0.71\pm0.0561=(0.6359,\ 0.7661)$$
Step 3
Black
Sample size n=251
Sample proportion, p=0.66
We use normal approximation, for this we check that both np and $$n(1-p)>5$$
Since  and $$​ n\times(1-p)=85.34>5$$, we can take binomial random variable as normally distributed, with mean $$=p=0.66$$ and std error $$=\sqrt{\frac{p\times(1-p)}{n}}=0.0299$$
For constructing Confidence interval,
Margin of Error  $$(ME)=z\times SD=0.0586$$
95% confidence interval is given by:
$$\text{Sample Mean}\pm\text{(Margin of Error)}$$
$$0.66\pm0.0586=(0.6014,\ 0.7186)$$