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

1 answer

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:

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