/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 The risk of developing iron defi... [FREE SOLUTION] | 91Ó°ÊÓ

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The risk of developing iron deficiency is especially high during pregnancy. Detecting such a deficiency is complicated by the fact that some methods for determining iron status can be affected by the state of pregnancy itself. Consider the following data on transferrin receptor concentration for a sample of women with laboratory evidence of overt iron-deficiency anemia ("Serum Transferrin Receptor for the Detection of Iron Deficiency in Pregnancy," American Journal of Clinical Nutrition [1991]: \(1077-1081):\) \(\begin{array}{lllllllll}15.2 & 9.3 & 7.6 & 11.9 & 10.4 & 9.7 & 20.4 & 9.4 & 11.5\end{array}\) \(\begin{array}{lll}16.2 & 9.4 & 8.3\end{array}\) Compute the values of the sample mean and median. Why are these values different here? Which one do you regard as more representative of the sample, and why?

Short Answer

Expert verified
The mean of the sample data is approximately 11.5 and the median of the sample data is approximately 10.1. The mean is higher due to the influence of the higher values in the data set. In this case, the median might be more representative of the sample because it is not influenced by the extreme values.

Step by step solution

01

Computing the Sample Mean

The sample mean is computed by adding all the data points and then dividing by the number of data points. So, first arrrange all data points sequentially: \(7.6, 8.3, 9.3, 9.4, 9.4, 9.7, 10.4, 11.5, 11.9, 15.2, 16.2, 20.4\). Now, add all the values and then divide by the total number of values (12).
02

Calculating the Median

To calculate the median, order the data points from smallest to largest, and pick the middle value. If there's an even number of data points, take the average of the two middle values. In our set of 12 data points, the average of 6th and 7th data point after arranging them in ascending order is \( (9.7 + 10.4) / 2\).
03

Analyzing and Comparing the Values

The mean and median are different because they measure two different types of 'centers' for the data. The mean is influenced by all values in the data set including the extremely low or high values also known as outliers. The Median, on the other hand, is the middle point of data set and is not influenced by outliers. Considering this, determine which one (the mean or the median) is more representative of the given data based on whether the data seems to be skewed or not.

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Key Concepts

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

Descriptive Statistics
Descriptive statistics, at its core, involve summarizing and describing the main features of a collection of data, often called a dataset. This field includes compiling measures of central tendency like the mean (average), median (middle value), mode (most frequent value), and other parameters such as variance and standard deviation that tell us about the spread of data.

By calculating these statistics, we gain a clearer understanding of where most data points lie (central tendency) and how much variation there is in the data (variability). In the context of the given exercise, descriptive statistics helped in describing the distribution of transferrin receptor concentration by calculating the sample mean and median.

  • The sample mean provides a sense of the overall level of transferrin receptor concentration among the sampled women with iron-deficiency anemia.
  • The median, on the other hand, gives a middle value that divides the sample data into two equally sized groups.
Descriptive statistics facilitate the interpretation of data in a way that's easier to understand and communicate, which is essential in many fields including health sciences.
Mean and Median Difference
Understanding the difference between the mean and the median is crucial for correctly interpreting data. The mean, calculated by dividing the sum of all values by their count, is sensitive to every data point in the dataset. As such, it can be heavily influenced by extreme scores or outliers. In contrast, the median, denoted as the middle value in an ordered list of numbers, is more resistant to outliers and skewed data.

In the example exercise, after ordering the data points, we identify the two central numbers to calculate the median, thus ensuring that this measure reflects the central tendency without the impact of outlier values. This is especially significant when we are dealing with data that are not symmetrically distributed.

Typically, when a dataset includes outliers or is not normally distributed, the median is considered a more reliable representation of its central location. This is the reason why healthcare professionals might favor the median value when discussing iron levels in patients, as it can represent a 'typical' patient without getting distorted by extreme cases.
Outliers in Data Analysis
Outliers are values in a dataset that differ significantly from other observations. They can arise due to measurement errors, variability in the measurement, or they may indicate a novel discovery or an unusual event worth further investigation. In data analysis, it is essential to identify and assess the impact of outliers because they can skew the results of statistical calculations, such as the mean.

In data like the transferrin receptor concentration, where outliers are present, simply calculating the mean may not be enough to get a true understanding of the underlying trend in the data. The presence of an unusually high or low observation, such as the value of 20.4 in our exercise, can inflate or deflate the mean, respectively.

Therefore, it's of utmost importance to consider both the mean and median in conjunction with each other to better grasp the nature of the data. In some cases, statisticians may also opt to adjust the dataset by removing outliers to obtain a mean value that's more in line with the rest of the data. However, this should always be done with caution and proper justification, as it could result in the loss of significant information.

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Most popular questions from this chapter

The article "Comparing the Costs of Major Hotel Franchises" (Real Estate Review [1992]: \(46-51\) ) gave the following data on franchise cost as a percentage of total room revenue for chains of three different types: \(\begin{array}{llllllll}\text { Budget } & 2.7 & 2.8 & 3.8 & 3.8 & 4.0 & 4.1 & 5.5 \\ & 5.9 & 6.7 & 7.0 & 7.2 & 7.2 & 7.5 & 7.5 \\ & 7.7 & 7.9 & 7.9 & 8.1 & 8.2 & 8.5 & \\ \text { Midrange } & 1.5 & 4.0 & 6.6 & 6.7 & 7.0 & 7.2 & 7.2 \\\ & 7.4 & 7.8 & 8.0 & 8.1 & 8.3 & 8.6 & 9.0 \\ \text { First-class } & 1.8 & 5.8 & 6.0 & 6.6 & 6.6 & 6.6 & 7.1 \\ & 7.2 & 7.5 & 7.6 & 7.6 & 7.8 & 7.8 & 8.2 \\ & 9.6 & & & & & & \end{array}\) Construct a boxplot for each type of hotel, and comment on interesting features, similarities, and differences.

Suppose that your statistics professor returned your first midterm exam with only a \(z\) score written on it. She also told you that a histogram of the scores was approximately normal. How would you interpret each of the following \(z\) scores? a. \(2.2\) d. \(1.0\) b. \(0.4\) e. 0 c. \(1.8\)

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.

The Los Angeles Times (July 17, 1995) reported that in a sample of 364 lawsuits in which punitive damages were awarded, the sample median damage award was \(\$ 50,000\), and the sample mean was \(\$ 775,000\). What does this suggest about the distribution of values in the sample?

An instructor has graded 19 exam papers submitted by students in a class of 20 students, and the average so far is 70 . (The maximum possible score is \(100 .\) ) How high would the score on the last paper have to be to raise the class average by 1 point? By 2 points?

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