91影视

Variable at random $\hat{p}$is the percentage of adults in the United States who are very concerned about the quality of education in our schools. As a result, the true population proportion's point estimate is

$\hat{p}=63\%=0.63$

The percentage of adult Americans who are very concerned about the quality of education in our schools is random variable $X$. Therefore,

$$\begin{gathered} x=1000脳0.63 \\ =630 \end{gathered}$$

Given $\hat{p}=0.63$and$n=1000$, the distribution we should use to estimate the fraction is

$\mathcal{N}\left(0.63,\sqrt{\frac{0.63(1-0.63)}{1000}}\right)$

If $\hat{p}$is the proportion of observations in a random sample size $n$that belong to a class of interest, a$100(1-伪)\%$ confidence interval on the proportion of the population that belongs to this class can be calculated.

$\hat{p}-z_{\frac{伪}{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}鈮�p鈮�\hat{p}+z_{\frac{伪}{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

where $z_{\frac{伪}{2}}$is the standard normal distribution's upper $\frac{伪}{2}$percentage point. assuming a two-sided confidence interval of $95\%$,

$$\begin{gathered} \frac{伪}{2}=\frac{1-0.95}{2} \\ =\frac{0.05}{2} \\ =0.025 \end{gathered}$$

and

$z_{\frac{伪}{2}}=1.96$

i) A $95\%$two-sided $CI$for the population proportion is calculated using the equations.

$0.63-1.96\sqrt{\frac{0.63(1-0.63)}{1000}}鈮�p鈮�0.63+1.96\sqrt{\frac{0.63(1-0.63)}{1000}}$,

$0.63-0.0299鈮�p鈮�0.63+0.0299$,

$0.60鈮�p鈮�0.66$

ii) The error bound is$EBM=0.0299$

iii) The graph for this problem is

The indicated $卤3\%$error bound is the absolute maximum error bound. This indicates that people conducting the research will make a maximum error of $3\%$. As a result, they estimate that between $60$and $66\%$of adult Americans are concerned about the quality of education in our schools.