Question

You measure 42 randomly selected textbooks' weights, and find they have a mean weight of 79 ounces. Assume the population standard deviation is 11.6 ounces. Based on this, construct a 90% confidence interval for the true population mean textbook weight.

Give your answers as decimals, to two places

Answer 1

To construct a confidence interval for the true population mean, we can use the formula for the confidence interval:

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

where:
- \(\bar{x}\) is the sample mean,
- \(z\) is the z-score corresponding to the desired confidence level,
- \(\sigma\) is the population standard deviation, and
- \(n\) is the sample size.

Given the information provided:
- \(\bar{x} = 79\) ounces
- \(\sigma = 11.6\) ounces
- \(n = 42\)

First, we need to determine the z-score for a 90% confidence level. For a 90% confidence interval, the alpha level (\(\alpha\)) is 0.10, and we divide this by 2 to find the critical value for each tail:

\[
z = z_{\alpha/2} = z_{0.05} \approx 1.645
\]

Next, we calculate the standard error (SE):

\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{11.6}{\sqrt{42}} \approx \frac{11.6}{6.4807} \approx 1.79
\]

Now we can calculate the margin of error (ME):

\[
ME = z \cdot SE = 1.645 \cdot 1.79 \approx 2.94
\]

Finally, we construct the confidence interval:

\[
\text{Lower bound} = \bar{x} - ME = 79 - 2.94 \approx 76.06
\]
\[
\text{Upper bound} = \bar{x} + ME = 79 + 2.94 \approx 81.94
\]

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

\[
\boxed{(76.06, 81.94)}
\]