To construct a 90% confidence interval for the true population mean turtle weight, we can use the formula for the confidence interval given by:
\[ \text{CI} = \bar{x} \pm z \left(\frac{\sigma}{\sqrt{n}}\right) \]
Where:
- \(\bar{x}\) is the sample mean (72 ounces)
- \(z\) is the z-score corresponding to the desired confidence level (90%)
- \(\sigma\) is the population standard deviation (13.5 ounces)
- \(n\) is the sample size (22 turtles)
Step 1: Find the z-score for a 90% confidence level.
For a confidence level of 90%, the z-score corresponds to the critical value that leaves 5% in each tail of the normal distribution. Using a standard normal distribution table or a calculator, the z-score for 90% confidence is approximately:
\[ z \approx 1.645 \]
Step 2: Calculate the standard error (SE).
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{13.5}{\sqrt{22}} \approx \frac{13.5}{4.690} \approx 2.88 \]
Step 3: Calculate the margin of error (ME).
\[ ME = z \times SE \approx 1.645 \times 2.88 \approx 4.73 \]
Step 4: Construct the confidence interval.
\[ \text{CI} = \bar{x} \pm ME = 72 \pm 4.73 \]
Calculating the two bounds of the interval:
- Lower bound: \(72 - 4.73 \approx 67.27\)
- Upper bound: \(72 + 4.73 \approx 76.73\)
Thus, the 90% confidence interval for the true population mean turtle weight is:
\[ (67.27, 76.73) \]
Finally, the answers to two decimal places are:
Lower bound: 67.27
Upper bound: 76.73