Question

you measure 37 textbooks weights, and find they have a mean weight of 54 ounces. assume the population standard deviation is 3 ounces. based on this, construct a 99% confidence interval for the true population mean textbook weight. round answers to at least 4 decimal places.

Explanation:

Step1: Identify the formula

The formula for a confidence interval when the population standard - deviation $\sigma$ is known is $\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z_{\alpha/2}$ is the z - score, $\sigma$ is the population standard deviation, and $n$ is the sample size.

Step2: Determine the values of $\bar{x}$, $\sigma$, and $n$

We are given that $\bar{x} = 54$ (sample mean), $\sigma=3$ (population standard deviation), and $n = 37$ (sample size).

Step3: Find the z - score for a 99% confidence interval

For a 99% confidence interval, $\alpha=1 - 0.99 = 0.01$, so $\alpha/2=0.005$. Looking up in the standard normal distribution table, $z_{\alpha/2}=z_{0.005}=2.576$.

Step4: Calculate the margin of error $E$

$E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.576\times\frac{3}{\sqrt{37}}\approx2.576\times\frac{3}{6.0828}\approx2.576\times0.4932\approx1.2605$.

Step5: Calculate the confidence interval

The lower limit is $\bar{x}-E=54 - 1.2605 = 52.7395$.
The upper limit is $\bar{x}+E=54 + 1.2605 = 55.2605$.

Answer:

$52.7395<\mu<55.2605$