An article regarding interracial dating and marriage recentl

kuvitia9f 2021-12-17 Answered
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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Expert Answer

MoxboasteBots5h
Answered 2021-12-18 Author has 3742 answers
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,
\(\displaystyle{C}{I}=\hat{{{p}}}\pm{z}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}\)
\(\displaystyle={0.71}\pm{1.96}\sqrt{{{\frac{{{0.71}{\left({1}-{0.71}\right)}}}{{{255}}}}}}\)
\(\displaystyle={0.71}\pm{0.0557}\)
\(\displaystyle={\left({0.6543},\ {0.7657}\right)}\)
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limacarp4
Answered 2021-12-19 Author has 2882 answers
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
\(\displaystyle\hat{{{p}}}={0.66}\)
\(\displaystyle{S}.{E}=\sqrt{{{\frac{{{\left({0.66}\right)}{\left({0.34}\right)}}}{{{251}}}}}}={0.0299}\)
\(\displaystyle{95}\%{C}{I}={\left({0.66}-{\left({1.9695}\right)}{\left({0.03}\right)},\ {0.66}+{\left({1.96}\right)}{\left({0.031}\right)}\right)}\)
\(\displaystyle={\left({0.6011},\ {0.7189}\right)}\)
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nick1337
Answered 2021-12-28 Author has 9467 answers

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:
\(\text{Sample Mean}\pm\text{(Margin of Error)}\)
\(0.66\pm0.0586=(0.6014,\ 0.7186)\)

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