To construct a 90% confidence interval for the true population mean of the textbook weights, we can use the formula for the confidence interval when the population standard deviation is known. The formula is:
\[ \text{Confidence Interval} = \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right) \]
Where:
- \(\bar{x}\) = sample mean
- \(z\) = z-score corresponding to the desired confidence level
- \(\sigma\) = population standard deviation
- \(n\) = sample size
Step 1: Identify the values given:
- Sample mean (\(\bar{x}\)) = 59 ounces
- Population standard deviation (\(\sigma\)) = 6.2 ounces
- Sample size (\(n\)) = 39
Step 2: Find the z-score for a 90% confidence level. For a 90% confidence interval, the z-score (z) can be found in a z-table or using statistical software. The z-score for a 90% confidence level is approximately 1.645.
Step 3: Calculate the standard error (SE). \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{6.2}{\sqrt{39}} \approx \frac{6.2}{6.244} \approx 0.993 \]
Step 4: Calculate the margin of error (ME). \[ ME = z \times SE = 1.645 \times 0.993 \approx 1.632 \]
Step 5: Construct the confidence interval. \[ \text{Confidence Interval} = \bar{x} \pm ME = 59 \pm 1.632 \]
Calculating the interval:
- Lower limit: \(59 - 1.632 \approx 57.37\)
- Upper limit: \(59 + 1.632 \approx 60.63\)
Final Result: The 90% confidence interval for the true population mean textbook weight is approximately:
\[ (57.37, 60.63) \]
Therefore, the confidence interval is (57.37, 60.63).