- The margin of error is calculated using the formula \(E = z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}\).
- With a \(90\%\) confidence level, the significance level is \(\alpha = 0.10\), leading to a critical z-value of \(z_{0.05} \approx 1.645\).
- Using \(\sigma = 12.1\), \(n = 35\), and the critical value, the margin of error is \(E = 1.645 \cdot \frac{12.1}{\sqrt{35}}\).
- The final result is \(E \approx \boxed{3.36}\) ounces.
Explanation
- Problem analysis and formula identification
We are analyzing the weights of \(n = 35\) backpacks. The population standard deviation is given as \(\sigma = 12.1\) ounces, and we assume the distribution of weights is normal. We aim to find the maximal margin of error \(E\) for a \(90\%\) confidence interval for the true population mean weight \(\mu\). The margin of error is defined by the formula:
\[E = z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}\]
where \(z_{\alpha/2}\) is the critical value from the standard normal distribution corresponding to the desired confidence level.
- Determine the critical z-value
The confidence level is \(90\%\), which means that \(\alpha = 1 - 0.90 = 0.10\). To find the critical value \(z_{\alpha/2}\), we look for the z-score such that the area in each tail of the normal distribution is \(\alpha/2 = 0.10 / 2 = 0.05\). This is equivalent to finding the z-score such that the area to the left is \(1 - 0.05 = 0.95\). Using standard normal distribution tables, the critical value \(z_{0.05} \approx 1.645\).
- Perform the calculation
Now we plug the values into our formula:
- \(\sigma = 12.1\)
- \(n = 35\)
- \(z_{0.05} \approx 1.64485\)
Substituting these into the margin of error equation:
\[E = 1.64485 \cdot \frac{12.1}{\sqrt{35}}\]
First, calculating the standard error of the mean: \(\frac{12.1}{\sqrt{35}} \approx \frac{12.1}{5.91608} \approx 2.04527\).
Now, multiply by the critical z-value:
\[E \approx 1.64485 \cdot 2.04527 \approx 3.36418\]
Rounding the result to two decimal places, we get \(3.36\).
- Final conclusion
The maximal margin of error associated with a \(90\%\) confidence interval for the population mean weight of the backpacks is approximately \(3.36\) ounces. Thus, \(E = \pm 3.36\) ounces.
Examples
When manufacturing products like backpacks, companies need to estimate the average weight to determine shipping costs accurately. By taking a sample and calculating the margin of error, they can establish a range within which the true average weight is likely to fall. For instance, if the sample mean is \(43\) ounces with a margin of error of \(3.36\) ounces, they can be \(90\%\) confident the true average weight lies between \(39.64\) and \(46.36\) ounces, which helps in planning logistics and budget.