The bookstore at IRSC would like to estimate the proportion of students who prefer electronic textbooks (eBooks) over printed textbooks (hard copies). A random sample of 30 students was surveyed. Their preferences are listed below.
\begin{tabular}{|c|c|c|c|c|}
\hline eBook & eBook & eBook & eBook & eBook \\
\hline eBook & hard copy & hard copy & hard copy & hard copy \\
\hline hard copy & hard copy & hard copy & hard copy & hard copy \\
\hline hard copy & hard copy & hard copy & hard copy & hard copy \\
\hline hard copy & hard copy & hard copy & hard copy & hard copy \\
\hline hard copy & hard copy & hard copy & hard copy & hard copy \\
\hline
\end{tabular}
Determine the point estimate, $\widehat{p}$ and the sample standard deviation, $s_{\widehat{p}}$. Round the sample proportion to four decimal places and round the standard deviation to six decimal places, if necessary.
\[
\begin{array}{l}
\widehat{p}= \\
s_{\hat{p}}=
\end{array}
\]
Using a $96 \%$ confidence level, determine the margin of error, $E$, and a confidence interval for the proportion of all students at the college who work prefer ebooks over printed textbooks. Report the confidence interval using interval notation. Report the solutions in percent form, rounded to two decimal. places, if necessary.
The margln of error is given by $E=$
A $96 \%$ confidence interval is given by
Answer
\(\boxed{\text{So, a 96% confidence interval for the proportion of all students at the college who prefer eBooks over printed textbooks is (1.88%, 32.46%).}}\)
Step 1 ：Calculate the point estimate, \(\widehat{p}\), which is the sample proportion of students who prefer eBooks. It is calculated as the number of successes (students who prefer eBooks) divided by the total number of trials (total students surveyed). So, \(\widehat{p} = \frac{5}{30} = 0.1667\).
Step 2 ：Calculate the sample standard deviation, \(s_{\widehat{p}}\), using the formula \(\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}\), where \(n\) is the total number of students surveyed. So, \(s_{\widehat{p}} = \sqrt{\frac{0.1667(1-0.1667)}{30}} = 0.072168\).
Step 3 ：Calculate the margin of error, \(E\), using the formula \(E = Z_{\alpha/2} \cdot s_{\widehat{p}}\), where \(Z_{\alpha/2}\) is the Z-score corresponding to the desired confidence level. For a 96% confidence level, the Z-score is approximately 2.05. So, \(E = 2.05 \cdot 0.072168 = 0.147944\) or 14.79%.
Step 4 ：Calculate the confidence interval, which is given by \(\widehat{p} \pm E\), or \(0.1667 \pm 0.147944\). In percent form, this is 16.67% \(\pm\) 14.79%.
Step 5 ：\(\boxed{\text{So, a 96% confidence interval for the proportion of all students at the college who prefer eBooks over printed textbooks is (1.88%, 32.46%).}}\)
