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Increasing joint extension is one goal of athletic trainers. In a study to investigate the effect of therapy that uses ultrasound and stretching (Trae Tashiro, Masters Thesis, University of Virginia, 2004 ), passive knee extension was measured after treatment. Passive knee extension (in degrees) is given for each of 10 study participants. $$ \begin{array}{llllllllll} 59 & 46 & 64 & 49 & 56 & 70 & 45 & 52 & 63 & 52 \end{array} $$ Which would you choose to describe center and variability the mean and standard deviation or the median and interquartile range? Justify your choice.

Short Answer

Expert verified
In this case, since the data points are not significantly skewed and there are no extreme outliers, the mean and standard deviation are appropriate measures to describe the center and variability of this dataset. The mean passive knee extension is 55.6 degrees, with a standard deviation of 7.86 degrees.

Step by step solution

01

Analyze the given data

First, let's list the data points in ascending order: $$ \begin{array}{llllllllll} 45 & 46 & 49 & 52 & 52 & 56 & 59 & 63 & 64 & 70 \end{array} $$
02

Check for normality and presence of outliers

By visually examining the data points, we can observe that there are no extreme outliers, and the data does not seem to be heavily skewed.
03

Calculate the mean and median

Calculate the mean by adding all of the values and dividing by the number of values: Mean = \(\frac{(45 + 46 + 49 + 52 + 52 + 56 + 59 + 63 + 64 + 70)}{10}\) = 55.6 Calculate the median by finding the middle value(s) of the dataset: Since there are 10 data points, the median is the average of the 5th and 6th values - in this case, 52 and 56. Median = \(\frac{52 + 56}{2}\) = 54
04

Calculate the standard deviation and interquartile range

Calculate the standard deviation: First, find the deviation from the mean for each data point, square it, and take the average of these squared deviations: \(\frac{(45-55.6)^2 + (46-55.6)^2 + \cdots + (70-55.6)^2}{10} = 61.84\) Now, take the square root of this average to obtain the standard deviation: Standard Deviation = \(\sqrt{61.84}\) = 7.86 Calculate the interquartile range: The interquartile range is the difference between the first quartile (Q1) and the third quartile (Q3). Here, Q1 is the median of the first five values, and Q3 is the median of the last five values: Q1 = Median of (45, 46, 49, 52, 52) = 49 Q3 = Median of (56, 59, 63, 64, 70) = 63 Interquartile Range = Q3 - Q1 = 63 - 49 = 14
05

Choose the appropriate center and variability measures

Based on our analysis, we can see that the data points are not significantly skewed and there are no extreme outliers. Therefore, the mean and standard deviation are appropriate measures to describe the center and variability of this dataset. The mean passive knee extension is 55.6 degrees, with a standard deviation of 7.86 degrees.

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Key Concepts

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

Mean and Median Calculation
Understanding the central tendency of a dataset is crucial in descriptive statistics. The mean, often referred to as the average, is calculated by summing all the numerical values in a dataset and dividing the sum by the count of the values.

For the given exercise, the mean passive knee extension is calculated as follows:
Mean = \( \frac{(45 + 46 + 49 + 52 + 52 + 56 + 59 + 63 + 64 + 70)}{10} \) = 55.6 degrees.

The mean is sensitive to outliers, which can skew the average. In the absence of significant outliers, as in the exercise, the mean gives a balanced center point of the data.

The median, on the other hand, is the middle value of a dataset when ordered from the least to the greatest, or the average of the two middle values in the case of an even number of observations. The median divides the dataset into two equal halves. For the dataset in the exercise, the median is the average of the 5th and 6th values after sorting the data in ascending order, resulting in a median of 54 degrees.

The median is less affected by outliers or a skewed distribution, making it a robust measure of central tendency when such conditions are present.
Standard Deviation
The standard deviation is a measure of variability or dispersion in a dataset. It indicates how much individual data points deviate from the mean. To calculate the standard deviation, we first compute the variance, which is the average of the squared differences from the mean. The standard deviation is then the square root of the variance.

For the exercise's dataset, each data point's deviation from the mean is squared, these squared deviations are averaged, and the square root of this average is taken, yielding a standard deviation of 7.86 degrees. This calculation is expressed as:
Standard Deviation = \( \sqrt{\frac{(45-55.6)^2 + (46-55.6)^2 + \cdots + (70-55.6)^2}{10}} \) = 7.86 degrees.

A larger standard deviation indicates greater variability among the data points. When the data points closely cluster around the mean, the standard deviation is smaller. In the context of the exercise, the standard deviation suggests moderate variability in passive knee extension among study participants.
Interquartile Range
The interquartile range (IQR) is another statistical measure that describes the spread of the middle 50% of a dataset. It is the difference between the third quartile (Q3) and the first quartile (Q1) of the dataset. Quartiles divide the sorted dataset into four equal parts. The IQR is preferred over range because it is not affected by extreme values, known as outliers.

In our exercise, the data is first arranged in ascending order to find the quartiles. The first quartile (Q1) is the median of the lower half, while the third quartile (Q3) is the median of the upper half of the dataset. The IQR is calculated as:
Interquartile Range = Q3 - Q1 = 63 degrees (Q3) - 49 degrees (Q1) = 14 degrees.

The IQR provides a clearer picture of the dataset's central tendency and variability, especially when the mean and standard deviation are less reliable due to skewed data or outliers. In the absence of such issues, the IQR still complements these measures by offering insights into the dataset's range where most data points lie.

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Most popular questions from this chapter

The accompanying data are a subset of data read from a graph in the paper "Ladies First? A Field Study of Discrimination in Coffee Shops" (Applied Economics [April, 2008] . The data are the waiting times (in seconds) between ordering and receiving coffee for 19 female customers at a Boston coffee shop. $$ \begin{aligned} &\begin{array}{llllllll} 60 & 80 & 80 & 100 & 100 & 100 & 120 & 120 \end{array}\\\ &\begin{array}{llllllll} 120 & 140 & 140 & 150 & 160 & 180 & 200 & 200 \end{array}\\\ &\begin{array}{lll} 220 & 240 & 380 \end{array} \end{aligned} $$ a. Calculate the mean and standard deviation for this data set. b. Delete the observation of 380 and recalculate the mean and standard deviation. How do these values compare to the values calculated in Part (a)? What does this suggest about using the mean and standard deviation as measures of center and variability for a data set with outliers?

The mean playing time for a large collection of compact discs is 35 minutes, and the standard deviation is 5 minutes. a. What value is 1 standard deviation above the mean? One standard deviation below the mean? What values are 2 standard deviations away from the mean? b. Assuming that the distribution of times is mound shaped and approximately symmetric, approximately what percentage of times are between 25 and 45 minutes? Less than 20 minutes or greater than 50 minutes? Less than 20 minutes? (Hint: See Example \(3.19 .\) )

Although bats are not known for their eyesight, they are able to locate prey (mainly insects) by emitting high-pitched sounds and listening for echoes. A paper appearing in Animal Behaviour ("The Echolocation of Flying Insects by Bats" [1960]: 141-154) gave the following distances (in centimeters) at which a bat first detected a nearby insect: $$ \begin{array}{lllllllllll} 62 & 23 & 27 & 56 & 52 & 34 & 42 & 40 & 68 & 45 & 83 \end{array} $$ a. Calculate and interpret the mean distance at which the bat first detects an insect. b. Calculate the sample variance and standard deviation for this data set. Interpret these values.

In a study investigating the effect of car speed on accident severity, the vehicle speed at impact was recorded for 5000 fatal accidents. For these accidents, the mean speed was 42 mph and the standard deviation was 15 mph. A histogram revealed that the vehicle speed distribution was mound shaped and approximately symmetric. a. Approximately what percentage of the vehicle speeds were between 27 and \(57 \mathrm{mph} ?\) b. Approximately what percentage of the vehicle speeds exceeded \(57 \mathrm{mph} ?\)

The report titled "State of the News Media \(2013^{\text {" }}\) (Pew Research Center, May 7,2013 ) included the weekday circulation numbers for the top 20 newspapers in the country. Here are the data for the 6 months ending September 2012: $$ \begin{array}{rrrrr} 2,293798 & 1,713,833 & 1,613,866 & 641,369 & 535,875 \\ 529,999 & 522,868 & 462,228 & 432,455 & 412,669 \\ 411,960 & 410,130 & 392,989 & 325,814 & 313,003 \\ 311,504 & 300,277 & 296,427 & 293,139 & 285,088 \end{array} $$ a. Calculate and interpret the value of the median of this data set. b. Explain why the median is preferable to the mean for describing center for this data set. c. Explain why it would be unreasonable to generalize from this sample of 20 newspapers to the population of all daily newspapers in the United States.

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