91Ó°ÊÓ

The given data is about the quality of education in the schools that worries the citizens of America

The objective is to find the random variable, the distribution kind, and the confidence interval, and the sampling error occurs

a) Variable at random The proportion of adult Americans who are very concerned about the quality of education in our schools is $\hat{p}$. As a result, the true population proportion point estimate is

Variable at random $X$ is a group of adults in the United States who are concerned about the quality of education in our schools. Therefore,

b) Because we're estimating a percentage. Given $\hat{p}=0.63andn=1000$, the appropriate distribution is

c) If the proportion of observations in a random sample of size $n$that belong to a class of interest is $\hat{p}$, an approximate $100(1-\mathrm{Î\pm })\%$confidence interval on the proportion $p$of the population that belongs to this class is $\hat{p}$.

$\hat{p}-z_{\frac{\mathrm{Î\pm }}{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}≤p≤\hat{p}+z_{\frac{\mathrm{Î\pm }}{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

where $z\mathrm{Î\pm }$is the standard normal distribution's upper $\frac{\mathrm{Î\pm }}{2}$percentage point. For a two-sided confidence interval of $95\%$,

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

and

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

The bound of error is

$EBM=0.0299$

The $95\%$of CL for $p$is

$0.60≤p≤0.66$

The graph can be plotted as,

The maximum error bound is represented by $Â\pm 3\%$. The number of Americans who worries about the quality of education in our schools to be between $60\%$and$66\%$.