Question

120. 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 role="math" localid="1650173946070" $X$and $P^{\text{'}}$, 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 number of black adults who would welcome a white individual into their family is a random variable.

$0.822\le p\le 0.898\text{and}EBM-0.038\text{.}$

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

c. The population proportion has a 95 percent confidence interval is

localid="1650174575459" $0.822\le p\le 0.898andEBM-0.038$

Step-by-step solution

Introduction

The given data is about a Washington post about interracial dating and marriage

The objective is to find the random variable, the distribution kind, and the confidence level

Step 1 

a) Of the $1709$ adults that were randomly chosen, $323$ identified as black.

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

As a result, the point estimate for the genuine population proportion is role="math" localid="1650175047198" $\hat{p}=86\%=0.86$.

The random variable $X$represents the number of black adults who would welcome a white individual into their family.

As a result, $n=32andx=332\times 0.86=285.52$.

Step 2

Given $\hat{p}=0.86andn=323$, the distribution we should apply to estimate a proportion is

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

Step 3

$\hat{p}$is the proportion of observers in a random sample of size n that belong to a class of interest; a $100(1-\alpha )\%$confidence interval on the proportion $p$of the population that belongs to this class is approximately $100(1-\alpha )\%$.

$\overrightarrow{p}{z}_2\sqrt{\frac{\hat{p}(1-\overrightarrow{p})}{n}}\le p\le \hat{p}+z_2\sqrt{\frac{\hat{p}(1-\overrightarrow{p})}{n}}$

where $z\alpha$is the standard normal distribution's upper$\frac{\alpha }{2}$percent age point. Two-sided confidence interval for $95 percent $.

$\frac{a}{2}=\frac{1-0.95}{2}=0.025$

and

$z_{\frac{3}{2}}=1.96\text{.}$

From the Equations ,$95\%$the two-sided CI for the proportion of population is

$0.86-1.96\sqrt{\frac{0.86(1-0.86)}{323}}\le p\le 0.86+1.96\sqrt{\frac{0.86(1-0.86)}{323}}0.86-0.038\le p\le 0.86+0.038$

- A $95\%$CI for $p$is

localid="1650177091618" $0.822\le p\le 0.898$

- The graph for this problem is

The bound of error is determined as

$EBM=0.038$