You measure 33 textbooks' weights, and find they have a mean weight of 48 ounces. Assume the population standard deviation is 14.1 ounces. Based on this, construct a 90% confidence interval for the true population mean textbook weight. Round answers to at least 4 decimal places.
To construct a confidence interval for the true population mean textbook weight, we can use the formula:
\[ Confidence \, Interval = \bar{x} \pm z \cdot \frac{\sigma}{\sqrt{n}} \]
where:
- \(\bar{x}\) is the sample mean,
- \(z\) is the z-score corresponding to the desired level of confidence,
- \(\sigma\) is the population standard deviation, and
- \(n\) is the sample size.
In this case, the sample mean \(\bar{x}\) is 48 ounces, the population standard deviation \(\sigma\) is 14.1 ounces, and the sample size \(n\) is 33 textbooks.
First, we need to find the z-score corresponding to a 90% confidence level. Since this is a two-tailed test and the desired level of confidence is 90%, we need to find the z-score that leaves 5% in each tail. Using a z-table or calculator, we find the z-score to be approximately 1.645.
Plugging in the values into the formula, we have:
\[ Confidence \, Interval = 48 \pm 1.645 \cdot \frac{14.1}{\sqrt{33}} \]
Calculating the standard error, \(\frac{\sigma}{\sqrt{n}}\):
\[ SE = \frac{14.1}{\sqrt{33}} \approx 2.4500 \]
Calculating the lower and upper bounds of the confidence interval:
\[ Lower \, Bound = 48 - 1.645 \cdot 2.4500 \approx 43.6298 \]
\[ Upper \, Bound = 48 + 1.645 \cdot 2.4500 \approx 52.3702 \]
Therefore, the 90% confidence interval for the true population mean textbook weight is approximately (43.6298, 52.3702) ounces.