To construct a 90% confidence interval for the true population mean turtle weight, we can use the formula for the confidence interval based on the population standard deviation:
\[ CI = \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right) \]
Where:
- \(\bar{x}\) is the sample mean,
- \(z\) is the z-score corresponding to the desired confidence level,
- \(\sigma\) is the population standard deviation,
- \(n\) is the sample size.
Given:
- Sample mean (\(\bar{x}\)) = 72 ounces
- Population standard deviation (\(\sigma\)) = 13.5 ounces
- Sample size (\(n\)) = 22
First, let's find the z-score for a 90% confidence level. For a 90% confidence interval, the alpha level (\(\alpha\)) is 0.10, which means we want the z-score that corresponds to the upper 5% in a standard normal distribution table (since the total area in the tails is 10%, 5% is in each tail).
The z-score for 90% confidence is approximately 1.645.
Now, we can calculate the margin of error (ME):
\[ ME = z \left( \frac{\sigma}{\sqrt{n}} \right) = 1.645 \left( \frac{13.5}{\sqrt{22}} \right) \]
Calculating \(\sqrt{22}\):
\[ \sqrt{22} \approx 4.690 \]
Now, calculate the standard error (SE):
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{13.5}{4.690} \approx 2.88 \]
Now, calculate the margin of error:
\[ ME = 1.645 \times 2.88 \approx 4.73 \]
Finally, we can construct the confidence interval:
\[ CI = \bar{x} \pm ME = 72 \pm 4.73 \]
Calculating the lower and upper bounds:
Lower bound:
\[ 72 - 4.73 \approx 67.27 \]
Upper bound:
\[ 72 + 4.73 \approx 76.73 \]
Therefore, the 90% confidence interval for the true population mean turtle weight is:
\[ (67.27, 76.73) \text{ ounces} \]
These values should be correct if the calculations are accurate. Please ensure that you are formatting the inputs correctly if you're entering these answers into a system.