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Chapter 8: Problem 97

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

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

Step by step solution

01

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.
02

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.
03

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.
04

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)) 鈮� (34 - 0.596, 34 + 0.596) 鈮� (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.

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Most popular questions from this chapter

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