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 \(\bar{x}=1020\) grams per hour and \(s=90 .\) Estimate the mean hourly yield for the process and calculate the margin of error.

Short answer

Answer: The 95% confidence interval for the mean hourly yield is (1002.144, 1037.856) grams per hour.

Step-by-step solution

Compute the point estimate (sample mean) of the hourly yield

The sample mean \(\bar{x}\) is given to be 1020 grams per hour. This will be used as our point estimate for the true mean hourly yield.

Determine the sample size and standard deviation

The sample size, denoted \(n\), is given as 100 hourly periods. The sample standard deviation, denoted \(s\), is given as 90 grams per hour.

Choose a confidence level

We are not given a specific confidence level for this problem, so we will assume a 95% confidence level. This indicates that 95% of the time, the true mean hourly yield will fall within our confidence interval. Note that other confidence levels could be used depending on the desired level of certainty in the estimate.

Find the degrees of freedom and critical t-value

The degrees of freedom are equal to the sample size minus one: \(df = n - 1 = 100 - 1 = 99\). For a 95% confidence interval and 99 degrees of freedom, we can look up the critical t-value, denoted \(t_{\alpha/2}\), in a t-table or use a calculator to find it: \(t_{\alpha/2} = 1.984\). This means that 95% of the area under the t-distribution with 99 degrees of freedom falls within 1.984 standard deviations of the mean.

Calculate the standard error of the mean

The standard error of the mean, denoted \(SE\), is computed using the sample standard deviation \(s\) and the sample size \(n\), using the following formula: \(SE = \frac{s}{\sqrt{n}}\). Substituting our values: \(SE = \frac{90}{\sqrt{100}} = \frac{90}{10} = 9\) grams per hour.

Calculate the margin of error

The margin of error, denoted \(E\), is calculated using the critical t-value \(t_{\alpha/2}\) and the standard error of the mean \(SE\): \(E = t_{\alpha/2} \times SE\). Substituting our values: \(E = 1.984 \times 9 = 17.856\) grams per hour.

Determine the confidence interval for the true mean hourly yield

The confidence interval is determined by adding and subtracting the margin of error from the sample mean: \(\bar{x} \pm E = 1020 \pm 17.856\). This gives us a final confidence interval: (1002.144, 1037.856) grams per hour. The estimated mean hourly yield for the process is between 1002.144 and 1037.856 grams per hour, with a margin of error of 17.856 grams per hour.

Key concepts

Sample Mean

The sample mean is a crucial statistic that represents the average value of a set of numbers. It serves as an estimate of the true population mean, which is often unknown in real-world situations. In our exercise, we calculated the sample mean \(\bar{x}\) as 1020 grams per hour by observing the process over 100 hourly periods. This suggests that, on average, 1020 grams of antibiotic are produced each hour, based on the sample data. For this estimation to be reliable, it's vital that the sample be representative of the population; otherwise, the sample mean might not accurately reflect the true mean.

In educational or professional settings, understanding how to calculate the sample mean can help students and professionals make informed decisions using real data. It's a fundamental concept in statistics that paves the way for more advanced topics such as confidence intervals and hypothesis testing.

Standard Deviation

Standard deviation measures the amount of variation or spread in a set of values. It tells us how much the individual measurements in a dataset deviate from the mean, on average. In our example, we calculated the sample standard deviation, denoted as \(s\), and found it to be 90 grams per hour. A smaller standard deviation indicates that the data points tend to be close to the mean, while a larger standard deviation signifies that the data points are spread out over a wider range.

Understanding standard deviation is important because it provides context for the mean. For instance, two processes could have the same mean yield, but one might produce more consistent results (smaller standard deviation), while the other varies greatly (larger standard deviation). In critical applications such as the production of pharmaceuticals, consistency often matters as much as the average yield.

Sample Size

Sample size, denoted as \(n\), is the number of observations or measurements in a sample. The size of the sample has a direct impact on the accuracy and precision of statistical estimates. In the exercise, our sample size is 100 hourly periods, which provides a substantial amount of data to estimate the mean hourly yield.

A larger sample tends to give a more precise estimate because it reduces the effect of anomalies or outliers. This, in turn, affects the standard error and the width of the confidence interval. Understanding the role of sample size is vital for statistical reliability and is a key factor in determining how confident we can be about our estimates. It's essential for students to grasp the implications of sample size on their statistical calculations.

T-Distribution

The t-distribution is a probability distribution that accounts for the extra variability introduced when working with small samples or when the population standard deviation is unknown. Unlike the normal distribution, which is used when the population standard deviation is known and sample sizes are large, the t-distribution is more spread out and has heavier tails, accounting for the increased uncertainty.

For the 95% confidence level in our antibiotic yield estimation, the critical t-value reflects how many standard deviations from the sample mean will incorporate the central 95% of the data, based on the sample's degrees of freedom (in our case, 99). It's vital for students and professionals alike to understand the differences between the normal and t-distributions to apply the correct statistical methods for their data.