91Ó°ÊÓ

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 $\mathrm{x}=332×0.86$

$=285.52.$

b)

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

The distribution we should apply for predicting a proportion is,

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

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

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

Where

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

and

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

Therefore, from Equations $1$and $2$

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

A $95\%$CI is

$0.822≤\mathrm{p}≤0.898$

Graph is,

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