Question

A telephone poll of $1000$adult Americans was reported in an issue of Time Magazine. Another question in the poll was “[How much are] you worried about the quality of education in our schools?” Sixty-three percent responded “a lot”. We are interested in the population proportion of adult Americans who are worried a lot about the quality of education in our schools.

a. Define the random variables $X$and $P\prime$in words.

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

c. Construct a $95\%$confidence interval for the population proportion of adult Americans who are worried a lot about the quality of education in our schools.

i. State the confidence interval.

ii. Sketch the graph.

iii. Calculate the error bound.

d. The sampling error given by Yankelevich Partners, Inc. (which conducted the poll) is $\pm 3\%$. In one to three complete sentences, explain what the $\pm 3\%$ represents.

Short answer

(a) Random variable $X$is of adult Americans who are worried a lot about quality of education in our schools, $x=630$. Random variable $\hat{p}$is the proportion of American adults who are worried lot about the quality of education in our schools, $\hat{p}=0.63$

(b) We use the distribution $\mathcal{N}\left(0.63,\sqrt{\frac{0.63(1-0.63)}{1000}}\right)$in this problem.

(c) $95\%$confidence interval for the population proportion is $0.60\le p\le 0.66$

(d) $\pm 3\%$represents the maximum error bound

Step-by-step solution

Explain the random variable X and p^ (part a)

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,

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

Estimate the distribution (part b)

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

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

Find the population proportion for 95% of confidence level (part c)

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

$\hat{p}-z_{\frac{\alpha }{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\le p\le \hat{p}+z_{\frac{\alpha }{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

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

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

and

$z_{\frac{\alpha }{2}}=1.96$

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

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

$0.63-0.0299\le p\le 0.63+0.0299$,

$0.60\le p\le 0.66$

ii) The error bound is$EBM=0.0299$

iii) The graph for this problem is

Give the representation for ±3% (part d)

The indicated $\pm 3\%$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.