You measure 48 randomly selected textbooks' weights, and find they have a mean weight of 46 ounces. Assume the population standard deviation is 7.6 ounces. Based on this, construct a 90% confidence interval for the true population mean textbook weight.

Question

Give your answers as decimals, to two places

___ < μ < ___

Answer 1

To construct a 90% confidence interval for the true population mean textbook weight, we will use the formula for the confidence interval based on the normal distribution. The formula is:

$$\overline{x}\pm z\left(\frac{\sigma }{\sqrt{n}}\right)$$

Where:

• $\overline{x}$ is the sample mean
• $z$ is the z-score corresponding to the confidence level
• $\sigma$ is the population standard deviation
• $n$ is the sample size

Given Data:

• Sample mean, $\overline{x}=46$ ounces
• Population standard deviation, $\sigma =7.6$ ounces
• Sample size, $n=48$

Step 1: Find the z-score for a 90% confidence level

The z-score for a 90% confidence interval (which leaves 5% in each tail) is approximately $z=1.645$.

Step 2: Calculate the standard error (SE)

$$SE=\frac{\sigma }{\sqrt{n}}=\frac{7.6}{\sqrt{48}}\approx \frac{7.6}{6.928}\approx 1.098$$

Step 3: Calculate the margin of error (ME)

$$ME=z\cdot SE=1.645\cdot 1.098\approx 1.806$$

Step 4: Construct the confidence interval

Now we will compute the lower and upper bounds of the confidence interval: $$\text{Lower bound}=\overline{x}-ME=46-1.806\approx 44.19$$ $$\text{Upper bound}=\overline{x}+ME=46+1.806\approx 47.81$$

Final Result

Thus, the 90% confidence interval for the true population mean textbook weight is:

$$\boxed{{\displaystyle 44.19<\mu <47.81}}$$