Question

124. Refer to Exercise 8.123. 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.

$Xand{P}^{\text{'}}$

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 Yanklovich Partners, Inc. (which conducted the poll) is $\pm 3\%$. In one to three complete sentences, explain what the $\pm 3\%$ represents.

Short answer

a. Variable at random $X$ is a group of adults in the United States who are concerned about the quality of education in our schools. Variable at random The proportion of adult Americans who are very concerned about the quality of education in our schools is $\hat{p}$.

b. We use the normal distribution $\mathcal{N}\left(0.63,\sqrt{\frac{0.63(1-0.63)}{10n0}}\right)$

c. The population proportion has a $95\%$confidence interval of $0.60\le p\le 0.66andEBM=0.0299$. The maximum error bound is specified as $"\pm 3\%"$.

Step-by-step solution

Introduction

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

Step 1 

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

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

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,

$x=1000\times 0.63=630.$

Step 2

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

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

Step 3

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

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

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

$\frac{\alpha }{2}=\frac{1-0.95}{2}=0.025,$

and

$z_{\frac{\pi }{2}}=1.96.$

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

$$\begin{gathered} 0.63-1.96\sqrt{\frac{0.63(1-0.63)}{1000}}\le p\le 0.63+1.96\sqrt{\frac{0.63(1-0.63)}{1000}} \\ 0.63-0.0299\le p\le 0.63+0.0299. \end{gathered}$$

Step 4

The bound of error is

$EBM=0.0299$

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

$0.60\le p\le 0.66$

The graph can be plotted as,

Step 5

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