Question

132. Public Policy Polling recently conducted a survey asking adults across the U.S. about music preferences. When asked, $80$ of the $571$ participants admitted that they have illegally downloaded music.

a. Create a $99\%$ confidence interval for the true proportion of American adults who have illegally downloaded music.

b. This survey was conducted through automated telephone interviews on May 6 and 7,2013. The error bound of the survey compensates for sampling error or natural variability among samples. List some factors that could affect the survey's outcome that is not covered by the margin of error.

c. Without performing any calculations, describe how the confidence interval would change if the confidence level changed from $99\%to90\%$.

Short answer

a. The $CL$of $99\%$for $p$is determined as $0.103\le p\le 0.177$

b. Many people will not reply to surveys including telephone interviews. You have no way of knowing who is responding to the surveys if they do respond.

c. The error bound decreases as the confidence level is reduced, narrowing the confidence interval.

Step-by-step solution

Introduction 

The given data is about a survey conducted to know the music preferences of the citizens

The objective is to find the confidence interval and error bound of the data

Step 1

A poll of music tastes was recently done by Public Policy Polling. When asked, $x=80$of the $n=571$participants acknowledged to downloading music illegally.

As a result, the sample's estimated proportion is

$p^{\text{'}}=\frac{x}{n}=\frac{80}{571}=0.14.$

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

$p^{\text{'}}-z_{\frac{a}{2}}\sqrt{\frac{p^{\text{'}}\left(1-p^{\text{'}}\right)}{n}}\le p\le p^{\text{'}}+z_{\frac{\alpha }{2}}\sqrt{\frac{p^{\text{'}}\left(1-p^{\text{'}}\right)}{n}}$

where $z_{\frac{\alpha }{2}}$is the upper $\frac{\alpha }{2}$percentage point of the standard normal distribution. For $99\%$ two-sided confidence interval,

role="math" localid="1650208287393" $\frac{\alpha }{2}=\frac{1-0.99}{2}=0.005$

Step 2

and

$z_{\frac{\alpha }{2}}=2.58$

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

$$\begin{gathered} 14-2.58\sqrt{\frac{0.14(1-0.14)}{571}}\le p\le 0.14+2.58\sqrt{\frac{0.14(1-0.14)}{571}}, \\ 0.14-0.0374\le p\le 0.14+0.0374. \end{gathered}$$

Therefore, a $99\%$ $CL$ for $p$ is

$0.103\le p\le 0.177.$

Step 3

b) Many people will not respond to telephone interviews surveys. If they do respond to the surveys, you can not be sure who is responding.

c) Decreasing the confidence level causes the error bound to decrease, making the confidence interval narrower.