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Chapter 7: Problem 15

Many drugs used to treat cancer are expensive. Business Week reported on the cost per treatment of Herceptin, a drug used to treat breast cancer (Business Week, 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 estimate: $4560; Standard deviation estimate: $900.69.

Step by step solution

01

Understanding Point Estimate

A point estimate is a single value that approximates a parameter of the population, such as the mean or standard deviation. Here, we need to calculate the mean and standard deviation of the given sample costs.
02

Calculate Mean Cost

The mean of a dataset is calculated by summing all the values and then dividing by the number of values. For our sample:\[\text{Mean} = \frac{4376 + 5578 + 2717 + 4920 + 4495 + 4798 + 6446 + 4119 + 4237 + 3814}{10}\]First, add all the costs together. The sum is 45600, then divide by 10 (the number of samples):\[\text{Mean} = \frac{45600}{10} = 4560\]So, the point estimate of the mean cost is $4560.
03

Calculate Standard Deviation of Costs

The standard deviation provides a measure of how much variance there is from the mean. Use the formula for the sample standard deviation:\[ s = \sqrt{\frac{1}{n-1}\sum_{i=1}^n (x_i - \bar{x})^2}\] where \( \bar{x} \) is the mean, \( n \) is the number of samples, and \( x_i \) represents each individual sample. Calculate each \( (x_i - \bar{x})^2 \), sum them, divide by \( n-1 = 9 \), and find the square root.
04

Perform Calculations

First, calculate each deviation from the mean:\((4376-4560)^2 = 33856, (5578-4560)^2 = 1030416, \ldots\)The sum of these squared deviations is 7301142.8.Then divide by 9:\[\frac{7301142.8}{9} = 811238.089\]Finally, take the square root:\[s = \sqrt{811238.089} \approx 900.69\]The point estimate of the standard deviation is approximately $900.69.

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Key Concepts

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

Standard Deviation Calculation
Calculating the standard deviation is a fundamental aspect in statistics, as it provides insight into the spread or dispersion of a set of data points. In our context, it measures how much the individual costs for Herceptin treatment vary from the estimated mean cost.

To determine the standard deviation of our sample, we employ a well-known formula:
  • Firstly, we find the mean of the data set, which from previous calculations was $4560.
  • Next, we calculate the squared difference between each sample cost and the mean. This results in a series of squared differences like \((4376-4560)^2\) and so on.
  • The sum of these squared differences, in this scenario, is 7301142.8.
  • We then divide this sum by the number of samples minus one, which equals 9 here, to account for sample size.
  • The result, 811238.089, must then be square rooted to provide the standard deviation.
Finally, we arrive at a standard deviation of approximately 900.69, indicating a considerable variability in the costs per treatment.
Mean Calculation
The mean, or average, is central when analyzing data, as it summarizes a data set into a single value, representing a central point around which the other data points cluster. In the case of Herceptin treatment costs, the mean provides an estimation of typical expenses.

The calculation of the mean follows a straightforward process:
  • Add all the data points together. In this case, adding every individual treatment cost results in a total of 45600 dollars.
  • Divide this total sum by the number of data points. Here, it's 10 patients, creating the fraction \(\frac{45600}{10}\).
This equates to a mean cost of $4560. By establishing this mean, we obtained a useful point estimate that helps in understanding how much on average one might expect to pay for Herceptin treatment. While it doesn't reflect the variability, it provides a starting point for further analysis.
Sample Data Analysis
Sample data analysis is an invaluable step in making informed decisions based on incomplete information. Unlike analyzing the full population, which might be costly or time-consuming, sample analysis allows us to infer about the larger population's trends and parameters such as mean and standard deviation.

Within the context of Herceptin treatment costs:
  • A sample of 10 patients' treatment costs is analyzed instead of all patients receiving the drug.
  • This smaller, manageable data set is assumed to be representative of the broader population as it is selected randomly.
  • The calculated point estimates 鈥� mean and standard deviation 鈥� thus represent larger trend insights that health economists, policymakers, and clinical stakeholders can use.
Sample data analysis does, however, come with limitations. It's pivotal to ensure the sample accurately reflects the characteristics of the entire population to avoid biases. As such, sampling methods and size are crucial to the overall validity of the conclusions drawn.

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Most popular questions from this chapter

The Food Marketing Institute shows that \(17 \%\) of households spend more than \(\$ 100\) per week on groceries. Assume the population proportion is \(p=.17\) and a simple random sample of 800 households will be selected from the population. a. Show the sampling distribution of \(\bar{p}\), the sample proportion of households spending more than \(\$ 100\) per week on groceries. b. What is the probability that the sample proportion will be within 卤.02 of the population proportion? c. Answer part (b) for a sample of 1600 households.

BusinessWeek conducted a survey of graduates from 30 top MBA programs ( Business Week , September 22,2003 ). On the basis of the survey, assume that the mean annual salary for male and female graduates 10 years after graduation is \(\$ 168,000\) and \(\$ 117,000,\) respectively. Assume the standard deviation for the male graduates is \(\$ 40,000,\) and for the female graduates it is \(\$ 25,000\) a. What is the probability that a simple random sample of 40 male graduates will provide a sample mean within \(\$ 10,000\) of the population mean, \(\$ 168,000 ?\) b. What is the probability that a simple random sample of 40 female graduates will provide a sample mean within \(\$ 10,000\) of the population mean, \(\$ 117,000 ?\) c. In which of the preceding two cases, part (a) or part (b), do we have a higher probability of obtaining a sample estimate within \(\$ 10,000\) of the population mean? Why? d. What is the probability that a simple random sample of 100 male graduates will provide a sample mean more than \(\$ 4000\) below the population mean?

Lori Jeffrey is a successful sales representative for a major publisher of college textbooks. Historically, Lori obtains a book adoption on \(25 \%\) of her sales calls. Viewing her sales calls for one month as a sample of all possible sales calls, assume that a statistical analysis of the data yields a standard error of the proportion of .0625 a. How large was the sample used in this analysis? That is, how many sales calls did Lori make during the month? b. Let \(\bar{p}\) indicate the sample proportion of book adoptions obtained during the month. Show the sampling distribution of \(\bar{p}\) c. Using the sampling distribution of \(\bar{p},\) compute the probability that Lori will obtain book adoptions on \(30 \%\) or more of her sales calls during a one-month period.

A production process is checked periodically by a quality control inspector. The inspector selects simple random samples of 30 finished products and computes the sample mean product weights \(\bar{x}\). If test results over a long period of time show that \(5 \%\) of the \(\bar{x}\) values are over 2.1 pounds and \(5 \%\) are under 1.9 pounds, what are the mean and the standard deviation for the population of products produced with this process?

Roper ASW conducted a survey to learn about American adults' attitudes toward money and happiness (Money, October 2003). Fifty-six percent of the respondents said they balance their checkbook at least once a month. a. Suppose a sample of 400 American adults were taken. Show the sampling distribution of the proportion of adults who balance their checkbook at least once a month. b. What is the probability that the sample proportion will be within 卤.02 of the population proportion? c. What is the probability that the sample proportion will be within 卤.04 of the population proportion?
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