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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

Expert verified
The 95% confidence interval for the proportion of U.S. medical residents who work moonlighting jobs is calculated as \(p \pm z\sqrt{\frac{p(1-p)}{n}}\) giving a range. For the 90% confidence interval for the proportion of U.S. medical residents who have a credit card debt of more than $3000, we also can use the formula \(p \pm z\sqrt{\frac{p(1-p)}{n}}\) resulting in another range. The interval in Part (a) is wider than that in Part (b) due to a higher level of confidence and a larger calculated proportion.

Step by step solution

01

Calculate the proportion for moonlighting jobs

First, we need to calculate the proportion of medical residents who work moonlighting jobs. This is done by dividing the number of residents who work moonlighting jobs (38) by the total number of residents (115). This gives a proportion of \(p = \frac{38}{115}.\)
02

Construct a 95% confidence interval for moonlighting jobs

Next, we can construct the confidence interval. For a 95% confidence level, the value of \(z\) is approximately 1.96. The formula for the confidence interval is \(p \pm z\sqrt{\frac{p(1-p)}{n}}\) where \(p = \frac{38}{115}\), \(n = 115\), and \(z = 1.96.\)
03

Calculate the proportion for credit card debt

We can now calculate the proportion of residents who have a credit card debt of over $3000. This is done in the same way as step 1, giving \(p = \frac{22}{115}.\)
04

Construct a 90% confidence interval for credit card debt

As for the 90% confidence interval, the value of \(z\) is approximately 1.645. We can plug the values into the same formula as step 2, giving \(p \pm z\sqrt{\frac{p(1-p)}{n}}\) where \(p = \frac{22}{115}, n = 115,\) and \(z = 1.645.\)
05

Compare the two confidence intervals

The confidence interval in Part (a) is wider than in Part (b) for two reasons: First, the level of confidence is higher in Part (a) (95% compared to 90%), which implies a wider interval. Also, the proportion calculated in Part (a) is larger than that in Part (b), which also results in a wider confidence interval.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Proportion Calculation
Understanding proportion calculation is critical when trying to derive insights from statistical data. In the context of confidence interval estimations for a population parameter, the sample proportion serves as an initial point estimate for what we are trying to determine.

Take, for example, the exercise where 38 out of 115 medical residents were found to work moonlighting jobs. Here, we calculate the sample proportion of residents with moonlighting jobs by dividing the number of favorable outcomes (residents with moonlighting jobs) by the total number of trials (residents surveyed). This proportion, denoted as \(p = \frac{38}{115}\), translates to a percentage of the sample that experienced the event of interest.

The accuracy of the proportion is inherently linked to the sample size and the variability of the data. A larger sample size, assuming it is randomly selected, would usually offer a proportion that more accurately reflects the population's true proportion.
Confidence Level
The confidence level is a measure of certainty or reliability associated with a statistical estimate. In the case of constructing a confidence interval, such as in our problem, the confidence level indicates the degree of assurance that the calculated interval contains the true population parameter.

Common confidence levels include 90%, 95%, and 99%. If we say we're 95% confident, as in the exercise to estimate the proportion of U.S. medical residents working moonlighting jobs, we mean that if we were to take 100 different samples and compute a confidence interval for each sample, we would expect about 95 of those intervals to contain the true proportion.

To establish a specific confidence level, we use the Z-distribution to find our critical value (denoted as \(z\)). A higher confidence level demands a higher critical value, resulting in a wider interval. This concept explains why a 95% confidence interval is broader than a 90% confidence interval for the same sample data.
Sample Size
Sample size plays a crucial role in constructing confidence intervals and in overall statistical analysis. It refers to the number of subjects included in a sample that represents a population. The larger the sample, the more reliable are the statistical estimates derived from it.

In our exercise, the total sample size is 115 medical residents. The formula used to calculate the confidence interval includes the sample size in the denominator of the margin of error term, \(\sqrt{\frac{p(1-p)}{n}}\). Intuitively, this relationship demonstrates that as the number of observations in the sample (\(n\)) increases, the margin of error decreases, leading to a narrower confidence interval.

This characteristic of sample size is vital since it determines how precisely we can estimate the population parameter. A sufficiently large sample size is necessary to achieve a specific level of precision and reliable results. However, it is essential to balance such needs with the potential costs and practical considerations of data collection.

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