Chapter 8.2, Problem 53E

a)

To determine

To estimate confidence level.

Answer to Problem 53E

The confidence level = 95.44%

Explanation of Solution

Given:

  $\begin{array}{l}n=1028\\ \widehat{p}=64\%=0.64\\ E=3\%=0.03\end{array}$

Formula:

Margin of error:

  $E=Z_{\alpha /2}\times \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}$

Using the formula,

  $\begin{array}{l}0.03=Z_{\alpha /2}\times \sqrt{\frac{0.64(1-0.64)}{1028}}\\ Z_{\alpha /2}=\frac{0.03}{0.01497}=2\end{array}$

Therefore,

We need to determine the probability that,

  $\begin{array}{l}P(-2<Z<2)=P(Z<2)-P(Z<-2)\\ P(-2<Z<2)=0.9772-0.0228\\ P(-2<Z<2)=0.9544=95.44\%\end{array}$

Hence, confidence level is 95.44%

b)

To determine

To give example of bias that could “Practical difficulty”

Answer to Problem 53E

The non-response bias.

Explanation of Solution

Given:

  $\begin{array}{l}n=1028\\ \widehat{p}=64\%=0.64\\ E=3\%=0.03\end{array}$

The non-sampling biases are: selection bias, response bias and non-response bias. From this biases, non-response bias is the results of not having a data for everybody in the sample. Therefore, this is most likely the common among teens as most of the teens will be unwilling to participate in surveys.