Question

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.

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:

\[
\bar{x} \pm z \left(\frac{\sigma}{\sqrt{n}}\right)
\]

Where:
- \(\bar{x}\) 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, \(\bar{x} = 46\) 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:
\[
\text{Lower bound} = \bar{x} - ME = 46 - 1.806 \approx 44.19
\]
\[
\text{Upper bound} = \bar{x} + ME = 46 + 1.806 \approx 47.81
\]

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

\[
\boxed{44.19 < \mu < 47.81}
\]