Question

The article "Doctors Cite Burnout in Mistakes" (San Luis Obispo Tribune, March 5,2002 ) reported that many doctors who are completing their residency have financial struggles that could interfere with training. In a sample of 115 residents, 38 reported that they worked moonlighting jobs and 22 reported a credit card debt of more than \(\$ 3000\). Suppose that it is reasonable to consider this sample of 115 as a random sample of all medical residents in the United States. a. Construct and interpret a \(95 \%\) confidence interval for the proportion of U.S. medical residents who work moonlighting jobs. b. Construct and interpret a \(90 \%\) confidence interval for the proportion of U.S. medical residents who have a credit card debt of more than \(\$ 3000\). c. Give two reasons why the confidence interval in Part (a) is wider than the confidence interval in Part (b).

Short answer

a. The \(95\%\) confidence interval for the proportion of medical residents working moonlighting jobs is calculated using the sample proportion and size, and the z-score for a \(95\%\) confidence level. b. Similarly, the \(90\%\) confidence interval for the proportion of residents with credit card debt exceeding $3000 is computed using the respective proportion, sample size and the \(90\%\) confidence level z-score. c. The \(95\%\) interval is wider due to a higher level of confidence and likely greater variability in the data.

Step-by-step solution

Find the Sample Proportions

For the two situations mentioned in the exercise, sample proportions are computed by dividing the number of 'successes' by the total sample size. Let's denote them as \(p_1\) for moonlighting jobs and \(p_2\) for credit card debt. So, \(p_1 = 38/115\) and \(p_2 = 22/115\).

Construct the 95% Confidence Interval for Moonlighting Jobs

A \(95\%\) confidence interval formula is \(p \pm Z(1- \alpha/2) \sqrt{ p(1-p) / n }\), where \(Z(1- \alpha/2)\) is a z-score for a \(95\%\) confidence level, which is approximately \(1.96\). And the sample proportion \(p\), and the sample size \(n\) are known. The interval can therefore be computed as \(p_1 \pm 1.96*\sqrt{p_1*(1-p_1)/115}\).

Construct the 90% Confidence Interval for Credit Card Debt

For a \(90\%\) confidence interval, the z-score is approximately \(1.645\). The interval can be obtained using the formula: \(p_2 \pm 1.645*\sqrt{p_2*(1-p_2)/115}\).

Comparing Confidence Intervals' Widths

The confidence interval's width depends on two things: the level of confidence and the variability in the data. As the confidence level increases, the interval becomes wider. In this case, the \(95\%\) interval is wider than \(90\%\) because we want to be more certain. Moreover, the interval width is inversely proportional to the square root of the sample size. While the sample sizes are the same in both cases, the variability differs, meaning the proportion with higher variability \(p_1\) will have a wider confidence interval compared to \(p_2\).

Key concepts

Proportion of U.S. Medical Residents

Understanding the proportion of U.S. medical residents who engage in certain activities or face particular challenges is essential for both policy-making and improving residency programs. The proportion is essentially a statistical measurement that represents the part of the residents that share a common characteristic, such as holding moonlighting jobs or incurring significant credit card debt. By analyzing these proportions, stakeholders can gain insights into the economic pressures and extracurricular workload medical residents cope with. The importance of reliable data cannot be overstated, as it impacts the development of support systems that help residents manage their stress and financial burdens effectively.

Proportion is calculated by taking the count of residents who display the characteristic of interest and dividing by the total number of residents surveyed. For instance, if a study found that 40 out of 100 sampled residents had over $3000 in credit card debt, the proportion would be 0.40, or 40%. Knowing this proportion is critical for illustrating the broader landscape of medical residency in the U.S.

Moonlighting Jobs in Medical Residency

Moonlighting jobs in medical residency can be a double-edged sword. While they provide an extra source of income, they can also contribute to burnout due to the additional workload on top of an already demanding residency schedule. It is essential to recognize the proportion of residents taking on these extra roles to address the implications of such decisions holistically.

For education leaders and hospital administrators, data on moonlighting can prompt revisions in work hour policies or lead to initiatives designed to provide better financial support to medical residents. These actions aim at alleviating the need for moonlighting and thus potentially improving resident well-being and patient care quality.

Credit Card Debt Among Medical Residents

Concerning the credit card debt among medical residents, this statistic is a poignant indicator of the financial stress that residents may experience. High levels of such debt can signify that residents are struggling to meet their day-to-day living costs or are unable to cope with the expenses incurred during their medical training. Highlighting this proportion serves more than an academic purpose; it can compel policy change, such as increasing resident salaries or offering additional financial counseling services to help residents manage their finances more effectively, thus reducing the potential impact on their performance and mental health.

Sample Proportions

Sample proportions are a fragment of a larger statistical puzzle, shedding light on how often an event occurs within a sample group. They are vital for conducting inferential statistics, like constructing confidence intervals, and we calculate them by dividing the number of successes or occurrences of a particular attribute by the total number of observations in the sample. For example, if a sample of 200 residents shows that 50 have taken moonlighting jobs, the sample proportion (\( p \)) would be 0.25, or 25%. This proportion is a stepping stone in estimating the characteristics of the entire population of medical residents and forms the basis for further statistical analysis, including hypothesis testing and the creation of prediction models.

Z-score

A z-score plays a crucial role in calculating confidence intervals in statistics. In essence, the z-score provides an idea of how far from the mean a data point is in terms of standard deviations. When constructing confidence intervals for proportions, z-scores allow us to set the desired level of confidence. For instance, a 95% confidence interval uses a z-score that corresponds to the point at which 95% of the data would fall under a normal distribution. In practical terms, the higher the z-score used (and therefore, the broader the interval), the more confident we can be that the true population proportion falls within that range.

The z-score is derived from the probability associated with the desired level of confidence. To find the z-score for a 90% confidence interval, you would look at a standard normal distribution and find the z-score that leaves 5% of the data on each tail, which is about 1.645. Whereas, for a 95% confidence interval, you would use 1.96, which leaves 2.5% in each tail. Understanding and applying the correct z-score is fundamental to accurately estimate confidence intervals and, by extension, to make informed decisions based on sample data.