Question

An article regarding interracial dating and marriage recently appeared in a newspaper. Of the 1701 randomly selected adults, 311 identified themselves as Latinos, 322 identified themselves as blacks, 252 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. Construct the 95% confidence intervals for the three Asian responses. (Round your answers to four decimal places.) welcome a white person (? ,? )/ welcome a Latino ( ?,? )/welcome a black person ( ?,? )

Answer 1

Answer:

White: (0.7397, 0.8403)

Latino: (0.6549, 0.7651)

Black: (0.6015, 0.7185)

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence interval $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

Z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

In this problem

There are 252 Asians. So $n = 252$.

Construct the 95% confidence intervals

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $z = 1.96$.

White person

Among Asians, 79% would welcome a white person into their families. This means that $\pi = 0.79$

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.79 - 1.96\sqrt{\frac{0.79*0.21}{252}} = 0.7397$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.79 + 1.96\sqrt{\frac{0.79*0.21}{252}} = 0.8403$

Latino person

71% would welcome a Latino. This means that $\pi = 0.71$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.71 - 1.96\sqrt{\frac{0.71*0.28}{252}} = 0.6549$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.71 + 1.96\sqrt{\frac{0.71*0.28}{252}} = 0.7651$

Black person

66% would welcome a black person. This means that $\pi = 0.66$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.66 - 1.96\sqrt{\frac{0.66*0.34}{252}} = 0.6015$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.66 + 1.96\sqrt{\frac{0.66*0.34}{252}} = 0.7185$