Chapter 8, Problem 55SE

Which would be hardest for you to give up: Your computer or your television? In a recent survey of 1677 U.S. Internet users, 74% of the young tech elite (average age of 22) say their computer would be very hard to give up (PC Magazine, February 3, 2004). Only 48% say their television would be very hard to give up.

1. a. Develop a 95% confidence interval for the proportion of the young tech elite that would find it very hard to give up their computer.
2. b. Develop a 99% confidence interval for the proportion of the young tech elite that would find it very hard to give up their television.
3. c. In which case, part (a) or part (b), is the margin of error larger? Explain why.

To determine

Find the 95% confidence interval for the population proportion of young tech elite that would find it very hard to give up their computer.

Answer to Problem 55SE

The 95% confidence interval for the population proportion of young tech elite that would find it very hard to give up their computer is (0.72, 0.76).

Explanation of Solution

Calculation:

In a sample size of 1,677 U.S. Internet users, 74% of the young tech elite say their computer would be very hard to give up and 48% of young tech elite say their television would be very hard to give up. That is, $\overline{p}=0.74$.

The 95% confidence interval for population proportion is given below:

$\begin{array}{c}\text{Confidence interval}=\overline{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}\\ =\overline{p}\pm \text{Margin of error}\end{array}$

The value of $z_{\frac{\alpha }{2}}$ is obtained below:

$\begin{array}{c}\alpha =1-0.95\\ =0.05\end{array}$

Therefore,

$\begin{array}{c}\frac{\alpha }{2}=\frac{0.05}{2}\\ =0.025\end{array}$

From “Table 8.1: Values of $z_{\frac{\alpha }{2}}$ for the most commonly used confidence levels”, the value of $z_{\frac{\alpha }{2}}$ for 95% is 1.96.

The 95% confidence interval for population proportion is calculated as follows:

$\begin{array}{c}\text{Confidence interval}=\overline{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}\\ =0.74\pm \left(1.96\times \sqrt{\frac{0.74\left(1-0.74\right)}{1,667}}\right)\\ =0.74\pm \left(1.96\times \sqrt{\frac{0.1924}{1,667}}\right)\\ =0.74\pm \left(1.96\times 0.0107\right)\end{array}$

                                $\begin{array}{c}=0.74\pm 0.020972\\ =\left(0.74-0.02,0.74+0.02\right)\\ =\left(0.72,0.76\right)\end{array}$

Thus, the 95% confidence interval for the proportion of the young tech elite that would find it very hard to give up their computer is (0.72, 0.76).

To determine

Find the 99% confidence interval for the population proportion of the young tech elite that would find it very hard to give up their television.

Answer to Problem 55SE

The 99% confidence interval for the population proportion of young tech elite that would find it very hard to give up their television is (0.45, 0.51).

Explanation of Solution

Calculation:

It is given that the value of $\overline{p}=0.48$.

The value of $z_{\frac{\alpha }{2}}$ is obtained below:

$\begin{array}{c}\alpha =1-0.99\\ =0.01\end{array}$

Therefore,

$\begin{array}{c}\frac{\alpha }{2}=\frac{0.01}{2}\\ =0.005\end{array}$

From “Table 8.1: Values of $z_{\frac{\alpha }{2}}$ for the most commonly used confidence levels”, the value of $z_{\frac{\alpha }{2}}$ for 99% is 2.576.

The 99% confidence interval for population proportion is calculated as follows:

$\begin{array}{c}\text{Confidence interval}=\overline{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}\\ =0.48\pm \left(2.576\times \sqrt{\frac{0.48\left(1-0.48\right)}{1,667}}\right)\\ =0.48\pm \left(2.576\times \sqrt{\frac{0.2496}{1,667}}\right)\\ =0.48\pm \left(2.576\times 0.01224\right)\end{array}$

                                 $\begin{array}{c}=0.48\pm 0.03\\ =\left(0.48-0.03,0.48+0.03\right)\\ =\left(0.45,0.51\right)\end{array}$

Thus, the 99% confidence interval for the population proportion of young tech elite that would find it very hard to give up their television is (0.45, 0.51).

To determine

Find the larger margin of error and explain the reason.

Explanation of Solution

In part b, the margin of error is 0.03, which is larger compared to the margin of error in part a. Thus, it can be concluded that the margin of error in part(b) is larger. Hence, the point estimate of population proportion and $z_{\frac{\alpha }{2}}$ is higher for television compared to computers. Therefore, part (b) has a larger margin of error.