Question

you measure 23 textbooks weights, and find they have a mean weight of 45 ounces. assume the population standard deviation is 10.4 ounces. based on this, construct a 99% confidence interval for the true population mean textbook weight. give your answers as decimals, to two places \boxed{} < \mu < \boxed{}

Explanation:

Step1: Identify the formula for confidence interval

The formula for a confidence interval when the population standard deviation (\(\sigma\)) is known is:
\(\bar{x} - z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}} < \mu < \bar{x} + z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}\)
Where:

• \(\bar{x} = 45\) (sample mean),
• \(\sigma = 10.4\) (population standard deviation),
• \(n = 23\) (sample size),
• \(z_{\alpha/2}\) is the z-score for 99% confidence. For 99% confidence, \(\alpha = 0.01\), so \(\alpha/2 = 0.005\). The \(z\)-score for \(0.995\) (since \(1 - 0.005 = 0.995\)) is approximately \(2.576\).

Step2: Calculate the margin of error (E)

First, compute \(\frac{\sigma}{\sqrt{n}}\):
\(\frac{10.4}{\sqrt{23}} \approx \frac{10.4}{4.7958} \approx 2.168\)

Then, multiply by \(z_{\alpha/2}\):
\(E = 2.576 \times 2.168 \approx 5.585\)

Step3: Compute the confidence interval bounds

Lower bound: \(\bar{x} - E = 45 - 5.585 \approx 39.415\) (rounded to two decimals: \(39.42\))
Upper bound: \(\bar{x} + E = 45 + 5.585 \approx 50.585\) (rounded to two decimals: \(50.59\))

Answer:

\(39.42 < \mu < 50.59\)