Question

You want to estimate the mean hourly yield for a process that manufactures an antibiotic. You observe the process for 100 hourly periods chosen at random, with the results \(\bar{x}=34\) ounces per hour and \(s=3\). Estimate the mean hourly yield for the process using a \(95 \%\) confidence interval.

Short answer

Answer: The estimated 95% confidence interval is between 33.404 and 34.596 ounces per hour.

Step-by-step solution

Determine the degrees of freedom

The degrees of freedom for the problem can be found using the formula df = n - 1, where df is the degrees of freedom and n is the sample size. In this case, n = 100, so we have df = 100 - 1 = 99.

Calculate the t-value

For a 95% confidence interval (meaning there is a 2.5% chance in each tail of the distribution), we look up the t-value in a t-distribution table, or using a calculator or statistical software. Using the degrees of freedom (99) and the desired confidence level (95%), we find the t-value to be approximately 1.984.

Compute the standard deviation of the sample mean

We can calculate the standard deviation of the sample mean (\(\frac{s}{\sqrt{n}}\)) using the sample standard deviation (s) and the sample size (n). Here, s = 3 and n = 100. Therefore, the standard deviation of the sample mean is \(\frac{3}{\sqrt{100}} = \frac{3}{10} = 0.3\) ounces per hour.

Calculate the confidence interval

We can now create the 95% confidence interval by multiplying the t-value (1.984) by the standard deviation of the sample mean (0.3) and adding/subtracting the result to the sample mean (34). The interval is given by: (34 - 1.984(0.3), 34 + 1.984(0.3)) ≈ (34 - 0.596, 34 + 0.596) ≈ (33.404, 34.596). Therefore, we estimate with 95% confidence that the true mean hourly yield for the antibiotic manufacturing process is between 33.404 and 34.596 ounces per hour.