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Many drugs used to treat cancer are expensive. \(B u\) siness Week reported on the cost per treatment of Herceptin, a drug used to treat breast cancer (BusinessWeek, January 30,2006 ). Typical treatment costs (in dollars) for Herceptin are provided by a simple random sample of 10 patients. \\[ \begin{array}{lllll} 4376 & 5578 & 2717 & 4920 & 4495 \\ 4798 & 6446 & 4119 & 4237 & 3814 \end{array} \\] a. Develop a point estimate of the mean cost per treatment with Herceptin. b. Develop a point estimate of the standard deviation of the cost per treatment with Herceptin.

Short Answer

Expert verified
Mean cost: $4440; Standard deviation: $711.2.

Step by step solution

01

Find the Sum of the Costs

Add all the given treatment costs: 4376, 5578, 2717, 4920, 4495, 4798, 6446, 4119, 4237, 3814. The sum of these costs is 44400 dollars.
02

Calculate the Mean Cost

Divide the sum of costs by the number of patients to find the mean. Here, \(\text{mean} = \frac{44400}{10} = 4440\) dollars.
03

Calculate Each Cost's Deviation from the Mean

Subtract the mean from each cost: 4376 - 4440, 5578 - 4440, ..., 3814 - 4440. These deviations are: -64, 1138, -1723, 480, 55, 358, 2006, -321, -203, -626.
04

Square the Deviations

Square each deviation: (-64)^2, (1138)^2, ..., (-626)^2. The squared deviations are: 4096, 1295044, 2964529, 230400, 3025, 128164, 4024036, 103041, 41209, 391876.
05

Sum the Squared Deviations

Find the sum of all squared deviations: \(4096 + 1295044 + 2964529 + ... + 391876 = 4552600\).
06

Calculate the Variance

Divide the sum of squared deviations by \(n-1\) (where \(n\) is the number of data points) for the sample variance: \(\text{variance} = \frac{4552600}{9} \approx 505844.44\).
07

Calculate the Standard Deviation

Take the square root of the variance to get the standard deviation: \(\text{standard deviation} = \sqrt{505844.44} \approx 711.2\) dollars.

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Key Concepts

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

Point Estimate
In descriptive statistics, a point estimate refers to the best single guess or estimate of a population parameter based on a sample of data. When working with large datasets, especially in practical situations like estimating treatment costs, point estimates are crucial for simplifying and summarizing the data.

For example, if you're trying to understand the average cost of Herceptin treatment per patient, the point estimate for the mean cost provides a concise summary of the data. In this specific case, using the sample data from 10 patients, we employ point estimation methods to gain insights. This approach is commonly used in various fields such as healthcare, economics, and research to draw conclusions from sample data about larger populations.

In summary, a point estimate serves as a quick and straightforward method to provide information about a population characteristic, and it's the primary step towards understanding complex data.
Mean Calculation
The mean, often called the average, is a measure that summarizes a set of data points. To calculate the mean, you sum all the data values and divide by the number of values in the dataset. This method provides a central value around which the different data points are distributed.

In the Herceptin treatment cost example, we found the mean cost by first summing up all patient costs, resulting in a total of 44400 dollars. Next, we divided this sum by the number of patients (10), giving us \[ \text{Mean} = \frac{44400}{10} = 4440 \text{ dollars} \].

This mean provides a useful point estimate of the average treatment cost, summarizing the dataset into a single representative value. The mean is especially helpful in indicating the center of a dataset, and it's often used to compare different datasets or interpret significant patterns in data.
Standard Deviation Calculation
Standard deviation is a statistic that measures the dispersion or spread of data points in a dataset relative to its mean. A smaller standard deviation indicates that data points tend to be close to the mean, while a larger one signals a wider spread of values.

To calculate the standard deviation of Herceptin treatment costs, each treatment cost's deviation from the mean (4440 dollars) was found. These deviations were squared to eliminate negative values, and then summed, resulting in 4552600. Since we are dealing with a sample, the sum of squared deviations was divided by one less than the number of data points (n-1), namely \[ \text{Variance} = \frac{4552600}{9} \approx 505844.44 \].

Finally, the square root of the variance was computed to determine the standard deviation: \[ \text{Standard deviation} \approx 711.2 \text{ dollars} \].

Standard deviation provides insights into the variability of data, helping us understand whether costs are consistently near the mean or if they fluctuate significantly. It is an essential indicator of reliability and risk in data analyses and decision-making processes.

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Most popular questions from this chapter

The mean tuition cost at state universities throughout the United States is \(\$ 4260\) per year (St. Petersburg Times, December 11,2002 ). Use this value as the population mean and assume that the population standard deviation is \(\sigma=\$ 900 .\) Suppose that a random sample of 50 state universities will be selected. a. Show the sampling distribution of \(\bar{x}\) where \(\bar{x}\) is the sample mean tuition cost for the 50 state universities. b. What is the probability that the simple random sample will provide a sample mean within \(\$ 250\) of the population mean? c. What is the probability that the simple random sample will provide a sample mean within \(\$ 100\) of the population mean?

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