Problem 39 - Basic Statistics for Business & Economics

Question

In 2003 , the Accreditation Council for Graduate Medical Education (ACGME) implemented new rules limiting work hours for all residents. A key component of these rules is that residents should work no more than 80 hours per week. The following is the number of weekly hours worked in 2017 by a sample of residents at the Tidelands Medical Center. $$ \left.\begin{array}{lllllllllll}84 & 86 & 84 & 86 & 79 & 82 & 87 & 81 & 84 & 78 & 74 & 86\end{array}\right)$$ a. What is the point estimate of the population mean for the number of weekly hours worked at the Tidelands Medical Center? b. Develop a \(90 \%\) confidence interval for the population mean. c. Is the Tidelands Medical Center within the ACGME quideline? Why?

Step-by-Step Solution

Answer

a. Point estimate is 82.58 hours. b. Confidence interval is (80.43, 84.73). c. No, the confidence interval shows hours could exceed 80.

Step 1: Calculate the Point Estimate of the Mean

To find the point estimate of the population mean, we calculate the sample mean \( \bar{x} \). The formula for the sample mean is \( \bar{x} = \frac{\sum x_i}{n} \), where \( x_i \) are the sample values and \( n \) is the sample size. Here, \( n = 12 \). So, sum the weekly hours worked: \( 84 + 86 + 84 + 86 + 79 + 82 + 87 + 81 + 84 + 78 + 74 + 86 = 991 \). Then, divide by \( n \): \( \bar{x} = \frac{991}{12} = 82.58 \).

Step 2: Calculate the Standard Deviation of the Sample

The sample standard deviation \( s \) is needed for the confidence interval. The formula is \[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}} \]. Calculate each \( (x_i - \bar{x})^2 \), find their sum, and divide by 11 (since \( n - 1 = 11 \)), and finally take the square root. The calculations result in \( s \approx 4.12 \).

Step 3: Determine the Confidence Interval Using t-Distribution

A \(90\%\) confidence interval uses the t-distribution since the sample size is small (\( n<30 \)). Find \( t_{0.05, 11} \) from the t-distribution table, which is approximately 1.796. The formula for the confidence interval is \[ \bar{x} \pm t \times \frac{s}{\sqrt{n}} \]. Substitute the values: \( 82.58 \pm 1.796 \times \frac{4.12}{\sqrt{12}} \). This results in an approximate interval of \( (80.43, 84.73) \).

Step 4: Check Compliance with ACGME Guidelines

Determine if the confidence interval obtained lies within the ACGME limit of 80 hours. The interval \((80.43, 84.73)\) indicates that the mean weekly hours could be above 80 hours, meaning that Tidelands Medical Center is not within ACGME guidelines.

Key Concepts

Sample Mean Calculation

To understand the sample mean calculation, think of it as a way to measure the "average" of the data you have. In our scenario, the sample includes the number of weekly hours residents worked.
To find the sample mean, sum all the recorded hours and divide by the number of observations.
This is mathematically represented as:

• \( \bar{x} = \frac{\sum x_i}{n} \)
• Where \( x_i \) is each individual data point, and \( n \) is the total number of data points.

In our example with 12 observations, we calculated the sum of weekly hours as 991. By dividing by 12, the sample size, we determined the sample mean to be approximately 82.58 hours. This value acts as our best guess or point estimate for the average hours worked by all residents if we cannot measure every single resident's working hours.

T-Distribution

The t-distribution is crucial when dealing with small sample sizes, typically less than 30. It helps build confidence intervals when normal distribution assumptions might not hold.
Unlike the normal distribution, the t-distribution has heavier tails, accounting for additional uncertainty with smaller samples. Thus, it provides a more reliable estimate of the interval where the true population mean might lie.
In this exercise, since our sample is only 12, we turn to the t-distribution for confidence intervals. Specifically, we looked up the critical value \( t_{0.05,11} \) and found it to be 1.796 for a 90% confidence level. This t-value is used to widen our confidence interval, accommodating the small sample's variability. Understanding t-distribution is essential for accurate statistical inference on limited data.

Sample Standard Deviation

The sample standard deviation provides insight into the data's spread or variability.
It shows how much individual data points deviate, on average, from the sample mean. It's a crucial component in calculating confidence intervals, which tell us how certain we are about our point estimate's reliability.

• The formula is: \( s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}} \).
• Here, \( x_i \) is each data point, \( \bar{x} \) is the sample mean, and \( n-1 \) is used instead of n for an unbiased estimation.

In our exercise, calculating the deviations and summing them, we found the sample standard deviation to be approximately 4.12. This variation reflects the residents' working hours' differences from our calculated mean of 82.58. Recognizing the standard deviation's role permits more accurate confidence interval estimations.

Point Estimate

A point estimate is a single value representing your best guess of the population parameter based on the sample data. In our case, this is represented by the sample mean.
It provides a quick snapshot of the average characteristic of the population, but since it's just one number, it's subject to sampling variability.
This is why we pair it with confidence intervals. These intervals account for that variability, giving a range of values where the true population mean is likely to lie. Here, our point estimate for the weekly working hours is 82.58, showing an average slightly higher than the ACGME 80-hour guideline. Thanks to the confidence interval, we see how confident we can be that the true mean isn't actually higher than regulatory standards.