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Cost-to-charge ratios (see Example \(4.9\) for a definition of this ratio) were reported for the 10 hospitals in California with the lowest ratios (San Luis Obispo Tribune, December 15,2002 ). These ratios represent the 10 hospitals with the highest markup, because for these hospitals, the actual cost was only a small percentage of the amount billed. The 10 cost-to-charge values (percentages) were $$ \begin{array}{rrrrrr} 8.81 & 10.26 & 10.20 & 12.66 & 12.86 & 12.96 \\ 13.04 & 13.14 & 14.70 & 14.84 & & \end{array} $$ a. Compute the variance and standard deviation for this data set. b. If cost-to-charge data were available for all hospitals in California, would the standard deviation of this data set be larger or smaller than the standard deviation computed in Part (a) for the 10 hospitals with the lowest cost-to-charge values? Explain. c. Explain why it would not be reasonable to use the data from the sample of 10 hospitals in Part (a) to draw conclusions about the population of all hospitals in California.

Short Answer

Expert verified
a. The variance and standard deviation are measures of dispersion and are computed using the given formulas. You would need to plug the values into the formulas using a calculator or spreadsheet software to get the precise numerical values. b. The standard deviation of cost-to-charge ratios for all hospitals in California is likely to be larger than for the selected 10 hospitals, as including more data, especially those not at extreme ends, will increase the variability in the dataset. c. The sample of 10 hospitals is not representative of all the hospitals in California, and using it to draw conclusions about the entire population could produce biased results.

Step by step solution

01

Calculate the Mean

Add all cost-to-charge values and divide by the number of values (10) to find the mean. The mean of the data set is \( \bar{x} = \frac{{8.81 + 10.26 + 10.20 + 12.66 + 12.86 + 12.96 + 13.04 + 13.14 + 14.70 + 14.84}}{10} = 12.357 \)
02

Find the Deviations

Compute the deviation of each data value from the mean. Deviation is the difference between each data value and the mean. For each data value, subtract the mean of 12.357.
03

Square the Deviations

Square each deviation computed in Step 2 to get the set of squared deviations.
04

Compute the Variance

Calculate the variance by adding all squared deviations computed in step 3 and divide them by the number of values. The variance, denoted by \( S^2 \), is \( S^2 = \frac{{\sum_{i=1}^{n} (x_i - \bar{x})^2}}{n-1} \) where \(n\) is the number of data value, \(x_i\) each data value and \(\bar{x}\) is the mean.
05

Compute the Standard Deviation

Compute the standard deviation by taking the square root of the variance computed in step 4. The standard deviation, denoted by \( S \), is \( S = \sqrt{S^2} \)
06

Analyze for Part B

In order to determine whether or not the standard deviation of cost-to-charge ratios for all hospitals in California would be larger or smaller than the standard deviation for the 10 hospitals with lowest cost-to-charge ratios, we need to understand the nature of the data. Because the data we have includes extremes (lowest cost-to-charge ratios), including data from all hospitals will most likely increase the range (highest minus lowest value), leading to a higher standard deviation.
07

Formulate Explanation for Part C

The sample of 10 hospitals, which were selected for having the lowest cost-to-charge ratios, may not be representative of all the hospitals in California. The data is skewed towards hospitals with low cost-to-charge ratios and does not take into consideration hospitals with higher cost-to-charge ratios. Using this sample to draw conclusion about all hospitals in California could likely result in biased or misleading findings.

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Key Concepts

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

Cost-to-Charge Ratios
Understanding the concept of cost-to-charge ratios is crucial in healthcare finance. These ratios represent the relationship between the cost incurred by a hospital to deliver services and the charges billed to the patients. Essentially, it measures how much more a hospital charges compared to the actual costs of service delivery. Typically, a lower cost-to-charge ratio indicates a higher markup. This is because the billed amount significantly surpasses the actual costs involved. Therefore, this is critical for patients, insurers, and policy-makers to understand how hospitals set their rates, and it is a vital tool for financial analysis in the healthcare industry.
Standard Deviation
Standard deviation is a key statistical tool used to measure the amount of variation or dispersion in a set of data values.
It tells us how spread out the data points are from the mean and provides insight into the consistency of data.
To calculate the standard deviation:
  • First, find the variance by calculating the average of squared deviations from the mean.
  • Then, take the square root of the variance.
For example, in the case of cost-to-charge ratios, a smaller standard deviation would indicate that the values are close to the mean, reflecting consistency among the hospitals in terms of their cost-to-charge practices. Conversely, a larger standard deviation would suggest more variability among the hospitals' pricing strategies, indicating less predictability.
Variance
Variance is another fundamental statistical concept, representing the degree of spread in a data set.
It gives the average of the squared differences from the mean, helping quantify the data's dispersion.
In essence, variance serves as a measure of how each number in a data set differs from the mean and thus from every other number in the set.
  • To find the variance, we calculate the mean of the data set.
  • Then, subtract the mean from each data point and square the result.
  • Finally, find the average of these squared differences.
Because variance uses the square of deviations, it is expressed in square units. When considering financial and statistical analysis, like cost-to-charge ratios, recognizing variance helps in understanding how dispersed the data points are, which is critical for making informed decisions.
Sample Representativeness
Sample representativeness is highly important when drawing conclusions from a statistical analysis.
It reflects how well a sample reflects the entire population from which it is drawn.
A sample is considered representative if it equally represents the various segments of the entire population, without bias.
  • If a sample is not representative, any conclusions drawn may be misleading.
  • In the case of cost-to-charge ratios from only ten hospitals, the sample specifically highlights those with low ratios.
  • This can lead to skewed results, as it doesn’t account for hospitals with higher ratios, thus not accurately representing the complete spectrum of hospital billing practices in California.
Thus, it's crucial to examine how a sample is chosen to ensure that it can reliably inform about the overall population.

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Most popular questions from this chapter

The standard deviation alone does not measure relative variation. For example, a standard deviation of \(\$ 1\) would be considered large if it is describing the variability from store to store in the price of an ice cube tray. On the other hand, a standard deviation of \(\$ 1\) would be considered small if it is describing store-to-store variability in the price of a particular brand of freezer. A quantity designed to give a relative measure of variability is the \(\mathrm{co}\) efficient of variation. Denoted by \(\mathrm{CV}\), the coefficient of variation expresses the standard deviation as a percentage of the mean. It is defined by the formula \(C V=100\left(\frac{s}{\bar{x}}\right)\). Consider two samples. Sample 1 gives the actual weight (in ounces) of the contents of cans of pet food labeled as having a net weight of 8 oz. Sample 2 gives the actual weight (in pounds) of the contents of bags of dry pet food labeled as having a net weight of \(50 \mathrm{lb}\). The weights for the two samples are: \(\begin{array}{lrrrrr}\text { Sample 1 } & 8.3 & 7.1 & 7.6 & 8.1 & 7.6 \\ & 8.3 & 8.2 & 7.7 & 7.7 & 7.5 \\ \text { Sample 2 } & 52.3 & 50.6 & 52.1 & 48.4 & 48.8 \\ & 47.0 & 50.4 & 50.3 & 48.7 & 48.2\end{array}\) a. For each of the given samples, calculate the mean and the standard deviation. b. Compute the coefficient of variation for each sample. Do the results surprise you? Why or why not?

A sample of 26 offshore oil workers took part in a simulated escape exercise, resulting in the accompanying data on time (in seconds) to complete the escape ("Oxygen Consumption and Ventilation During Escape from an Offshore Platform," Ergonomics \([1997]: 281-292\) ): \(\begin{array}{lllllllll}389 & 356 & 359 & 363 & 375 & 424 & 325 & 394 & 402 \\\ 373 & 373 & 370 & 364 & 366 & 364 & 325 & 339 & 393\end{array}\) \(\begin{array}{llllllll}392 & 369 & 374 & 359 & 356 & 403 & 334 & 397\end{array}\) a. Construct a stem-and-leaf display of the data. Will the sample mean or the sample median be larger for this data set? b. Calculate the values of the sample mean and median. c. By how much could the largest time be increased without affecting the value of the sample median? By how much could this value be decreased without affecting the sample median?

The paper 'Relationship Between Blood Lead and Blood Pressure Among Whites and African Americans" (a technical report published by Tulane University School of Public Health and Tropical Medicine, 2000 ) gave summary quantities for blood lead level (in micrograms per deciliter) for a sample of whites and a sample of African Americans. Data consistent with the given summary quantities follow: \(\begin{array}{lrllrrrl}\text { Whites } & 8.3 & 0.9 & 2.9 & 5.6 & 5.8 & 5.4 & 1.2 \\ & 1.0 & 1.4 & 2.1 & 1.3 & 5.3 & 8.8 & 6.6 \\ & 5.2 & 3.0 & 2.9 & 2.7 & 6.7 & 3.2 & \\ \text { African } & 4.8 & 1.4 & 0.9 & 10.8 & 2.4 & 0.4 & 5.0 \\\ \text { Americans } & 5.4 & 6.1 & 2.9 & 5.0 & 2.1 & 7.5 & 3.4 \\ & 13.8 & 1.4 & 3.5 & 3.3 & 14.8 & 3.7 & \end{array}\) a. Compute the values of the mean and the median for blood lead level for the sample of African Americans. Which of the mean or the median is larger? What characteristic of the data set explains the relative values of the mean and the median? b. Construct a comparative boxplot for blood lead level for the two samples. Write a few sentences comparing the blood lead level distributions for the two samples.

The paper "Answer Changing on Multiple-Choice Tests" (Journal of Experimental Education \([1980]: 18-21)\) reported that for a group of 162 college students, the average number of responses changed from the correct answer to an incorrect answer on a test containing 80 multiplechoice items was \(1.4\). The corresponding standard deviation was reported to be \(1.5 .\) Based on this mean and standard deviation, what can you tell about the shape of the distribution of the variable number of answers changed from right to wrong? What can you say about the number of students who changed at least six answers from correct to incorrect?

The percentage of juice lost after thawing for 19 different strawberry varieties appeared in the article "Evaluation of Strawberry Cultivars with Different Degrees of Resistance to Red Scale" (Fruit Varieties Journal [1991]: \(12-17):\) \(\begin{array}{llllllllllll}46 & 51 & 44 & 50 & 33 & 46 & 60 & 41 & 55 & 46 & 53 & 53 \\ 42 & 44 & 50 & 54 & 46 & 41 & 48 & & & & & & \end{array}\) a. Are there any observations that are mild outliers? Extreme outliers? b. Construct a boxplot, and comment on the important features of the plot.

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