Question

A random sample of 100 patients treated in a program for alcoholism and drug dependency over the past 10 years was selected. It was determined that 53 of the patients had been readmitted to the program at least once. At the \(95 \%\) level, construct an estimate of the population proportion.

Short answer

The \(95\%\) confidence interval for the population proportion is \(0.434\) to \(0.626\).

Step-by-step solution

Identify the Given Data

From the problem, the sample size is given as \(n = 100\) and the number of patients readmitted is \(X = 53\). The sample proportion is thus \(\hat{p} = \frac{X}{n} = \frac{53}{100} = 0.53\).

Determine the Confidence Level

The confidence level given is \(95\%\), which corresponds to a significance level of \(\alpha = 0.05\). For a \(95\%\) confidence interval, the critical value \(z\) is approximately \(1.96\).

Calculate the Error Margin

The formula for the error margin (E) for a population proportion is \[E = z \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]Substituting the values, we have \[E = 1.96 \cdot \sqrt{\frac{0.53 \cdot 0.47}{100}} \approx 0.096\]

Construct the Confidence Interval

The confidence interval for the population proportion is given by\[\hat{p} \pm E\]So, substituting the values, we get \[0.53 \pm 0.096\]Therefore, the confidence interval is \[ (0.434, 0.626) \]

Key concepts

Population Proportion

Understanding the population proportion is crucial in statistical analysis. In the context of the given exercise, the population proportion represents the fraction of patients who were readmitted to the program. To find the population proportion, we often rely on a sample since it's usually impractical to measure the entire population. Here, a random sample of 100 patients was taken, and 53 of them had been readmitted at least once. This gives us a sample proportion, denoted as \(\frac{X}{n}\), where \(X = 53\) and \(n = 100\). So, the sample proportion (\(\hat{p}\)) is \(0.53\). This sample proportion is used to estimate the true population proportion.

Sample Size

Sample size is the number of observations included in the sample. It is denoted by \(n\). A larger sample size generally leads to more accurate estimates because it tends to better represent the population. In our example, the sample size of 100 patients is used to infer the population's behavior regarding readmission rates. A good sample size helps in reducing the magnitude of variation in the estimates and increases confidence that the sample proportion approximates the true population proportion.

Error Margin

The error margin, denoted by \(E\), reflects the range within which the true population proportion is expected to lie based on the sample data. It accounts for the variability within the sample. The formula to calculate the error margin for a population proportion is: \[ E = z \cdot \sqrt{ \frac{ \hat{p} (1 - \hat{p}) }{ n } } \] where \(z\) is the z-value corresponding to the confidence level, \(\hat{p}\) is the sample proportion, and \(n\) is the sample size. In our exercise, with a 95% confidence level, \(z = 1.96\), \(\hat{p} = 0.53\), and \(n = 100\), we calculate \(E\) as: \[ E = 1.96 \cdot \sqrt{ \frac{ 0.53 \cdot 0.47 }{ 100 } } \approx 0.096 \] This shows that our sample proportion is within \(0.096\) of the true population proportion with 95% confidence.

95% Confidence Level

A confidence level indicates the degree of certainty in a statistical estimate. A 95% confidence level means that if we were to take many samples and build a confidence interval from each one, about 95% of those intervals would contain the true population proportion. The z-value associated with a 95% confidence level is approximately 1.96. This z-value helps in calculating the error margin. Using this confidence level assures us that the calculated interval \[ (0.434, 0.626) \] includes the true population proportion 95 times out of 100. This way, we can be reasonably confident in the reliability and accuracy of our sample-based estimates.