Question

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 \(0.05 .\)

Short answer

The answer depends on the calculated sample sizes, means, standard deviations, and test statistic. Without the actual computations and comparison of the test statistic with the critical value, a conclusive statement cannot be made.

Step-by-step solution

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.

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.

Calculate the Test Statistic

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

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.