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The Oregon Department of Health web site provides information on cost-to- charge ratio (the percentage of billed charges that are actual costs to the hospital). The following table gives cost-to-charge ratios for both inpatient and outpatient care in 2002 for a random sample of six hospitals in Oregon. $$ \begin{array}{ccc} & \begin{array}{c} 2002 \\ \text { Inpatient } \end{array} & \begin{array}{c} 2002 \\ \text { Hospital } \end{array} & \begin{array}{c} \text { Ratio } \\ \text { Ratient } \end{array} \\ \hline 1 & 68 & \text { Ratio } \\ 2 & 100 & 54 \\ 3 & 71 & 75 \\ 4 & 74 & 53 \\ 5 & 100 & 56 \\ 6 & 83 & 74 \\ & 88 \end{array} $$ Is there evidence that the mean cost-to-charge ratio for Oregon hospitals is lower for outpatient care than for inpatient care? Use a significance level of \(0.05 .\)

Step by step solution

01

Determine Sample sizes, Means and Standard Deviations

First, calculate the sample sizes (n1 for inpatient and n2 for outpatient), sample means (\(\bar{x_1}\) for inpatient and \(\bar{x_2}\) for outpatient), and sample standard deviations (s1 for inpatient and s2 for outpatient). Use standard statistical formulas for these calculations.

02

State the Hypotheses

The null hypothesis (\(H_0\)) states that there is no difference in the means, i.e., the mean cost-to-charge ratio for outpatient care equals the mean cost-to-charge ratio for inpatient care. The alternative hypothesis (\(H_a\)) states that the mean cost-to-charge ratio for outpatient care is lower than the mean for inpatient care.

03

Calculate the Test Statistic

Next, calculate the test statistic (t) using the formula: \[t = \frac{\bar{x_1} - \bar{x_2}}{\sqrt{\frac{{s1^2}}{n1} + \frac{{s2^2}}{n2}}}\]

04

Determine the Critical Value and Make Decision

By using the t-distribution table, find the critical value corresponding to the given level of significance (0.05) and degrees of freedom, which is \(df = n1 + n2 - 2\). If the calculated t statistic is less than negative of the critical value, we reject the null hypothesis in favor of the alternative hypothesis, implying a significant difference with outpatient care lower than inpatient care. If it's otherwise, we fail to reject the null hypothesis, meaning there's no sufficient evidence to suggest outpatient care is lower than inpatient care.

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Key Concepts

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

Cost-to-Charge Ratio

The cost-to-charge ratio is an important metric used in healthcare to understand the true costs incurred by hospitals compared to what they charge patients. It is usually expressed as a percentage. For example, if a hospital's cost-to-charge ratio is 70%, it means the hospital's actual costs are 70% of what they charge patients. Understanding this ratio is crucial for analyzing hospital pricing practices and evaluating if charges reflect actual costs.
In the context of the exercise, compare the mean cost-to-charge ratios for inpatient versus outpatient care. Investigators want to see if outpatient care is less costly than inpatient care, urging an analysis through statistical testing. Look closely at values: higher ratios suggest a larger proportion of costs relative to charges, whereas lower ratios imply less cost-relevant charges.

Significance Level

A significance level, often denoted by \( \alpha \), is a threshold used in hypothesis testing to determine when to reject the null hypothesis. The significance level is usually set before testing starts and is commonly used values include 0.05, 0.01, and 0.10. In this exercise, the significance level is set at 0.05.
Choosing a significance level is about balancing Type I and Type II errors. A significance level of 0.05 means you are accepting a 5% chance of incorrectly rejecting the null hypothesis (a Type I error). This is generally considered a reasonable balance in the healthcare industry, where real-world implications and data analysis require thorough considerations before dismissing existing claims in favor of new findings.

Test Statistic

The test statistic is a numerical value that results from the test process. It is derived from sample data and used to decide whether to reject the null hypothesis. In this problem, it is calculated using the formula: \[ t = \frac{\bar{x_1} - \bar{x_2}}{\sqrt{\frac{{s1^2}}{n1} + \frac{{s2^2}}{n2}}} \] where \( \bar{x_1} \) and \( \bar{x_2} \) are the sample means for inpatient and outpatient care respectively, and \( s1 \) and \( s2 \) are their standard deviations.
The test statistic shows the number of standard errors the observed difference is away from the null hypothesis prediction (no difference). A large test statistic indicates stronger evidence against the null hypothesis. After calculating it, compare it with a critical t-value from the t-distribution table to infer the conclusion of the hypothesis test.

T-Distribution

The t-distribution is a probability distribution used when analysing small sample sizes, especially for estimating the mean. It is similar to a normal distribution but has fatter tails, making it more useful when data have higher variability, or the sample size is small.
In this analysis, the t-distribution is critical as it provides the basis for determining the critical value for our test statistic given the set degrees of freedom. Degrees of freedom, calculated by \( df = n1 + n2 - 2 \), adjust for sample size, and critical values are obtained from t-distribution tables.
Understanding the t-distribution helps in appropriately interpreting the test statistic's significance, especially when judging if inpatient and outpatient cost-to-charge ratios differ to a statistically significant extent.