Question

An article regarding interracial dating and marriage recently appeared in the Washington Post. Of the 1,709 randomly selected adults, 315 identified themselves as Latinos, 323 identified themselves as blacks, 254 identified themselves as Asians, and 779 identified themselves as whites. In this survey, 86% of blacks said that they would welcome a white person into their families. Among Asians, 77% would welcome a white person into their families, 71% would welcome a Latino, and 66% would welcome a black person. Construct a 95% confidence interval for the percent of all black adults who would welcome a white person into their families:

Answer 1

Answer:

The 95% confidence interval for the percent of all black adults who would welcome a white person into their families is (0.8222, 0.8978).

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 level of $1-\alpha$, 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 $1 - \frac{\alpha}{2}$.

For this problem, we have that:

323 blacks, 86% of blacks said that they would welcome a white person into their families. This means that $n = 323, p = 0.86$

95% confidence level

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

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.86 - 1.96\sqrt{\frac{0.86*0.14}{323}} = 0.8222$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.86 + 1.96\sqrt{\frac{0.86*0.14}{323}} = 0.8978$

The 95% confidence interval for the percent of all black adults who would welcome a white person into their families is (0.8222, 0.8978).