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A researcher applied the carcinogenic (cancercausing) compound benzo(a)pyrene to the skin of five mice, and measured the concentration in the liver tissue after 48 hours. The results (nmol/gm) were as follows: \(^{21}\) $$\begin{array}{lllll}6.3 & 5.9 & 7.0 & 6.9 & 5.9\end{array}$$ Determine the mean and the median.

Short Answer

Expert verified
The mean concentration is 6.4 nmol/gm, and the median concentration is 6.3 nmol/gm.

Step by step solution

01

List the Data

Begin by listing the given concentrations of benzo(a)pyrene in the liver tissue of the mice, measured in nmol/g: 6.3, 5.9, 7.0, 6.9, 5.9.
02

Calculate the Mean

To calculate the mean (average) concentration, sum all the concentrations and divide by the number of data points. Mean = (6.3 + 5.9 + 7.0 + 6.9 + 5.9) / 5.
03

Organize the Data in Ascending Order

For calculating the median, the data must be ordered from smallest to largest. Here's the ordered set: 5.9, 5.9, 6.3, 6.9, 7.0.
04

Calculate the Median

The median is the middle value in the ordered set. Since there is an odd number of data points, the median is the third number: 6.3.

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Key Concepts

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

Mean Calculation
The mean, often referred to as the average, is a fundamental concept in statistics that provides a simple summary of a set of numerical data. To calculate it, you sum up all the values in a dataset and then divide that total by the number of values.

In the context of our exercise, we have five measurements of benzo(a)pyrene concentration in the liver tissue of mice: 6.3, 5.9, 7.0, 6.9, and 5.9 nmol/g. By adding these together, we get a total of 32.0 nmol/g. Dividing this sum by the number of measurements, which is 5, we get the mean concentration of 6.4 nmol/g.

The mean is especially useful for understanding the central tendency of data, but it's sensitive to outliers, which are atypical values that differ significantly from other observations in the dataset. Despite its susceptibility to extreme values, the mean is a reliable indicator of the middle of a dataset when the values are relatively evenly distributed.
Median Calculation
The median, another measure of central tendency, is the value at the center of an ordered dataset. Unlike the mean, it is not influenced by outliers, which makes it a more robust representation of a dataset's central point in certain cases. To compute the median, we first need to arrange the data in ascending or descending order.

If there is an odd number of observations, the median is the number that sits exactly in the middle. If instead there is an even number, the median is obtained by averaging the two central numbers. For our dataset, when we arrange the values (5.9, 5.9, 6.3, 6.9, 7.0) in order, we can see that the median is 6.3, as it is the third value in the list. This median value is particularly insightful when dealing with skewed data or when you want to find a typical value that isn't affected by outliers.
Data Analysis
Data analysis is an overarching term for the various methods and techniques used to inspect, clean, and model data with the aim of discovering useful information, drawing conclusions, and supporting decision making. Descriptive statistics, like the mean and median we've just computed, are vital tools in data analysis.

In analyzing our dataset, we started by summarizing the measurements of a carcinogenic compound in the liver tissues of mice. Through descriptive statistics, we are able to convert raw data into meaningful information—such as identifying the typical concentration level (using the mean) and the central tendency that is unaffected by extreme values (using the median). It's the initial stage of data analysis that can guide further inquiry and helps us understand what questions to ask next. Whether in biostatistics or other fields, data analysis lays the foundation for informed and evidence-based conclusions.
Biostatistics
Biostatistics is the application of statistics to a wide range of topics in biology, and it's particularly crucial in the fields of medicine, public health, and ecology. It includes the design of biological experiments and the collection, summarization, analysis, interpretation, and presentation of data related to biological phenomena.

In our exercise, we used biostatistical methods to determine the central tendency of concentrations of a chemical in animal tissue. Such analyses are essential in biostatistics as they can inform further research and policy decisions, like evaluating the safety and efficacy of new drugs or assessing the impact of environmental toxins on health. In our dataset, knowing the mean and median concentrations can help researchers understand distribution patterns and potential risks associated with the compound under study.

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Most popular questions from this chapter

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Dopamine is a chemical that plays a role in the transmission of signals in the brain. A pharmacologist measured the amount of dopamine in the brain of each of seven rats. The dopamine levels (nmoles/g) were as follows: \({ }^{41}\) $$ \begin{array}{lllllll} 6.8 & 5.3 & 6.0 & 5.9 & 6.8 & 7.4 & 6.2 \end{array} $$ (a) Calculate the mean and SD. (b) Determine the median and the interquartile range. (c) Replace the observation 7.4 by 10.4 and repeat parts (a) and (b). Which of the descriptive measures display robustness and which do not?

The two claws of the lobster (Homarus americanus) are identical in the juvenile stages. By adulthood, however, the two claws normally have differentiated into a stout claw called a "crusher" and a slender claw called a "cutter." In a study of the differentiation process, 26 juvenile animals were reared in smooth plastic trays and 18 were reared in trays containing oyster chips (which they could use to exercise their claws). Another 23 animals were reared in trays containing only one oyster chip. The claw configurations of all the animals as adults are summarized in the table. \({ }^{31}\) $$ \begin{array}{|lccc|} \hline&& {\text { Claw Configuration }} \\ \text { Treatment } & \begin{array}{c} \text { Right } \\ \text { crusher, } \\ \text { left cutter } \end{array} & \begin{array}{c} \text { Right } \\ \text { cutter, } \\ \text { left crusher } \end{array} & \begin{array}{c} \text { Right and } \\ \text { left cutter } \\ \text { (no crusher) } \end{array} \\ \hline \text { Oyster chips } & 8 & 9 & 1 \\ \text { Smooth plastic } & 2 & 4 & 20 \\ \text { One oyster chip } & 7 & 9 & 7 \\ \hline \end{array} $$ (a) Create a stacked frequency bar chart to display these data. (b) Create a stacked relative frequency bar chart to display these data. (c) Of the two charts you created in parts (a) and (b), which is more useful for comparing the claw configurations across the three treatments? Why?

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