Post A - Word Response To The Following Discussion Question
Post A -word response to the following discussion question by
A random sample of 12 health maintenance organizations (HMOs) was selected, and the co-payment amounts for a doctor’s office visit were recorded as follows: 6, 5, 12, 6, 8, 11, 6, 10, 12, 12, 10, 5. Assuming these co-payments are normally distributed, we are asked to find a 95% confidence interval for the true mean co-payment amount. Additionally, we need to determine the degrees of freedom, the sample mean, the standard deviation, and the confidence interval limits. Finally, the discussion should relate the application of such statistics in a professional environment and their advantages.
Paper For Above instruction
The first step in constructing a 95% confidence interval for the mean co-payment is to organize and analyze the sample data. The recorded co-payment amounts are 6, 5, 12, 6, 8, 11, 6, 10, 12, 12, 10, 5. To proceed, we calculate the sample mean (\(\bar{x}\)), the sample standard deviation (s), and the degrees of freedom, which are essential for determining the confidence interval.
The degrees of freedom (\(df\)) for a sample mean in a t-distribution is calculated as \(n - 1\), where \(n\) is the sample size. Here, \(n = 12\), so \(df = 11\).
Next, we calculate the sample mean:
\[
\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}
\]
Summing the values: \(6 + 5 + 12 + 6 + 8 + 11 + 6 + 10 + 12 + 12 + 10 + 5 = 93\).
Thus,
\[
\bar{x} = \frac{93}{12} = 7.75
\]
Calculating the sample standard deviation involves determining the deviations of each data point from the mean, squaring these deviations, summing them, dividing by \(n - 1\), and taking the square root:
\[
s = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n - 1}}
\]
Applying this:
Deviations: (6 - 7.75) = -1.75; (5 - 7.75) = -2.75; (12 - 7.75) = 4.25; (6 - 7.75) = -1.75; (8 - 7.75) = 0.25; (11 - 7.75) = 3.25; (6 - 7.75) = -1.75; (10 - 7.75) = 2.25; (12 - 7.75) = 4.25; (12 - 7.75) = 4.25; (10 - 7.75) = 2.25; (5 - 7.75) = -2.75.
Squaring these deviations and summing:
Sum of squared deviations ≈ 30.75.
Therefore,
\[
s = \sqrt{\frac{30.75}{11}} \approx \sqrt{2.795} \approx 1.672
\]
The standard error (SE) of the mean is:
\[
SE = \frac{s}{\sqrt{n}} = \frac{1.672}{\sqrt{12}} \approx \frac{1.672}{3.464} \approx 0.482
\]
Using the t-distribution table for 11 degrees of freedom at a 95% confidence level, the critical value (\(t^*\)) ≈ 2.201.
The margin of error (ME):
\[
ME = t^* \times SE \approx 2.201 \times 0.482 \approx 1.060
\]
Lower limit:
\[
7.75 - 1.060 = 6.69
\]
Upper limit:
\[
7.75 + 1.060 = 8.81
\]
Thus, the 95% confidence interval for the mean co-payment amount is approximately ($6.69, $8.81).
Applying such statistical analyses in a professional healthcare environment allows administrators and policymakers to understand cost variability and set appropriate co-payment strategies. For instance, confidence intervals can inform negotiations with providers or aid in designing equitable payment systems. Moreover, understanding the mean and variability of costs assists in budgeting and financial planning, ultimately leading to improved resource allocation. Similar statistical tools can also assess the effectiveness of interventions aimed at reducing costs or improving access to care. Emphasizing data-driven decision-making enhances efficiency and ensures policies are grounded in objective evidence rather than assumptions or anecdotal evidence.
In conclusion, calculating confidence intervals for estimated costs provides valuable insights into the financial landscape of healthcare services. Rigorous statistical analysis supports evidence-based strategies that can optimize resource management, improve patient care, and facilitate transparent decision-making processes.
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