To calculate the maximal margin of error for a 90% confidence interval, we can use the formula:
\[ E = z \times \frac{\sigma}{\sqrt{n}} \]
where:
- \(E\) is the margin of error,
- \(z\) is the z-score corresponding to the desired confidence level,
- \(\sigma\) is the population standard deviation,
- \(n\) is the sample size.
Step 1: Determine the z-score for a 90% confidence level
For a 90% confidence interval, the z-score can be obtained from the standard normal distribution. The critical value \(z\) for a 90% confidence level is approximately:
\[ z = 1.645 \]
Step 2: Use the values provided
Given:
- \(\sigma = 8.6\) (population standard deviation),
- \(n = 37\) (sample size).
Step 3: Calculate the standard error (SE)
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{8.6}{\sqrt{37}} \]
Calculating \(\sqrt{37}\):
\[ \sqrt{37} \approx 6.0828 \]
Now calculate the SE:
\[ SE = \frac{8.6}{6.0828} \approx 1.416 \]
Step 4: Calculate the margin of error (E)
Now we can calculate \(E\):
\[ E = z \times SE \] \[ E = 1.645 \times 1.416 \approx 2.332 \]
Step 5: Round to two decimal places
Finally, rounding \(E\) to two decimal places, we have:
\[ E \approx 2.33 \]
Thus, the maximal margin of error associated with a 90% confidence interval for the true population mean backpack weight is 2.33 ounces.
1 answer