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.

a. We are interested in finding the $95\%$confidence interval for the percent of all black adults who would welcome a white person into their families. Define the random variables $\mathrm{X}$and ${\mathrm{P}}^{\text{'}}$, in words.

b. Which distribution should you use for this problem? Explain your choice.

c. Construct a $95\%$confidence interval.

i. State the confidence interval.

ii. Sketch the graph.

iii. Calculate the error bound.

Short answer

a.

The percentage of black adults who would welcome a white family member is$P^{\text{'}}$,$X$is The percentage of black adults who would welcome a white family member into their home is a random variable.

b. Contribution is $\mathcal{N}\left(0.86,\sqrt{\frac{0.86\times 0.14}{323}}\right).$$\mathcal{N}\left(0.86,\sqrt{\frac{0.86\times 0.14}{323}}\right).$

c. Confident level is localid="1649856649792" $0.822\le \mathrm{p}\le 0.898$,

Graph is ,

Error bound is$0.038$.

Step-by-step solution

Explanation (part a)

a).

$323$of the$1709$adults chosen at random described themselves as black.

The percentage of black adults who would welcome a white person into their family is $\hat{p}$.

As a result, the true population proportion point estimate is

$\hat{\mathrm{p}}=86\%=0.86.$

Variable at random is the percentage of black adults who would accept a white family member into their home.

Therefore,

$\mathrm{n}=332$

And $\mathrm{x}=332\times 0.86$

$=285.52.$

Explanation  (part b)

b)

Given $\hat{p}=0.86$and $n=323$,

The distribution we should apply for predicting a proportion is,

$\mathcal{N}\left(0.86,\sqrt{\frac{0.86\times 0.14}{323}}\right).$

Explanation part (c)

c)

If the proportion of observations in a random sample of size$n$that belong to a class of interest is $\hat{\mathrm{p}}$,

An approximate $100(1-\mathrm{\alpha })\%$confidence interval on the proportion $p$of the population that belongs to this class is $\hat{p}$

localid="1649855567861" $\hat{\mathrm{p}}-{\mathrm{z}}_{\frac{\mathrm{\alpha }}{2}}\sqrt{\frac{\hat{\mathrm{p}}(1-\hat{\mathrm{p}})}{\mathrm{n}}}\le \mathrm{p}\le \hat{\mathrm{p}}+{\mathrm{z}}_{\frac{\mathrm{\alpha }}{2}}\sqrt{\frac{\hat{\mathrm{p}}(1-\hat{\mathrm{p}})}{\mathrm{n}}}.........1$

Where

$\frac{\mathrm{\alpha }}{2}=\frac{1-0.95}{2}=0.025$

and

localid="1649855580829" ${\mathrm{z}}_{\frac{\mathrm{\alpha }}{2}}=1.96.......2$

Therefore, from Equations $1$and $2$

$0.86-1.96\sqrt{\frac{0.86(1-0.86)}{323}}\le \mathrm{p}\le 0.86+1.96\sqrt{\frac{0.86(1-0.86)}{323}}$

A $95\%$CI is

$0.822\le \mathrm{p}\le 0.898$

Graph and error bound part(c) solution

Graph is,

$\mathrm{E}\mathrm{B}\mathrm{M}=0.038$