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Find the coefficient of variation for each of the two samples; then compare the variation. (The same data were used in Section 3-1.) Listed below are pulse rates (beats per minute) from samples of adult males and females (from Data Set 1 "Body Data" in Appendix B). Does there appear to be a difference? \(\begin{array}{llllllllllllllll}\text { Male: } & 86 & 72 & 64 & 72 & 72 & 54 & 66 & 56 & 80 & 72 & 64 & 64 & 96 & 58 & 66 \\ \text { Female: } & 64 & 84 & 82 & 70 & 74 & 86 & 90 & 88 & 90 & 90 & 94 & 68 & 90 & 82 & 80\end{array}\)

Short Answer

Expert verified
Males: 16.06%, Females: 11.46%. Males have more variation.

Step by step solution

01

Calculate the Mean

Determine the mean (average) pulse rate for each sample. The mean is calculated by adding all the values and then dividing by the number of values.
02

Mean for Males

\[ \text{Mean for Males} = \frac{86 + 72 + 64 + 72 + 72 + 54 + 66 + 56 + 80 + 72 + 64 + 64 + 96 + 58 + 66}{15} = 68.87 \]
03

Mean for Females

\[ \text{Mean for Females} = \frac{64 + 84 + 82 + 70 + 74 + 86 + 90 + 88 + 90 + 90 + 94 + 68 + 90 + 82 + 80}{15} = 81.87 \]
04

Calculate the Standard Deviation

Calculate the standard deviation for each sample. The standard deviation measures the amount of variation or dispersion from the mean.
05

Standard Deviation for Males

Use the formula for standard deviation: \[ s = \sqrt{\frac{1}{N-1} \sum_{i=1}^N (x_i - \bar{x})^2} \] Where \( N \) is the number of observations, \( x_i \) is each individual observation, and \( \bar{x} \) is the mean.\[ s_{\text{males}} = \sqrt{\frac{1}{14}((86 - 68.87)^2 + (72 - 68.87)^2 + \, \dots \, + (66 - 68.87)^2)} = 11.06 \]
06

Standard Deviation for Females

Using the same formula, calculate the standard deviation for the females.\[ s_{\text{females}} = \sqrt{\frac{1}{14}((64 - 81.87)^2 + (84 - 81.87)^2 + \, \dots \, + (80 - 81.87)^2)} = 9.39 \]
07

Calculate the Coefficient of Variation

The coefficient of variation (CV) is calculated by dividing the standard deviation by the mean and multiplying by 100.\[ CV = \left( \frac{s}{\bar{x}} \right) \times 100 \]
08

Coefficient of Variation for Males

Calculate the coefficient of variation for the males using their standard deviation and mean.\[ CV_{\text{males}} = \left( \frac{11.06}{68.87} \right) \times 100 = 16.06\% \]
09

Coefficient of Variation for Females

Calculate the coefficient of variation for the females using their standard deviation and mean.\[ CV_{\text{females}} = \left( \frac{9.39}{81.87} \right) \times 100 = 11.46\% \]
10

Compare the Variation

Compare the coefficients of variation for males and females. The group with the higher CV has more variability in their pulse rates.

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Key Concepts

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

Mean
Understanding the mean, or average, is key to many statistical calculations. The mean is obtained by summing all the values in a dataset and then dividing by the number of values.
For example, in the problem, we first calculate the mean pulse rate for both males and females:
For males: \( \text{Mean for Males} = \frac{86 + 72 + 64 + 72 + 72 + 54 + 66 + 56 + 80 + 72 + 64 + 64 + 96 + 58 + 66}{15} = 68.87 \)
For females: \( \text{Mean for Females} = \frac{64 + 84 + 82 + 70 + 74 + 86 + 90 + 88 + 90 + 90 + 94 + 68 + 90 + 82 + 80}{15} = 81.87 \)
The mean provides us with an average pulse rate, representing the center of our data. It's a useful metric to compare groups, but it doesn't tell us about the variability within the groups.
Standard Deviation
Standard deviation is a measure that indicates the amount of variation or dispersion within a dataset.
It is calculated using the formula:
\( s = \sqrt{\frac{1}{N-1} \sum_{i=1}^N (x_i - \bar{x})^2} \)
Where:
  • \( N \) is the number of observations
  • \( x_i \) is each individual observation
  • \( \bar{x} \) is the mean

In the provided exercise:
For males: \( s_{males} = \sqrt{\frac{1}{14}((86 - 68.87)^2 + (72 - 68.87)^2 + \dots + (66 - 68.87)^2)} = 11.06 \)
For females: \( s_{females} = \sqrt{\frac{1}{14}((64 - 81.87)^2 + (84 - 81.87)^2 + \dots + (80 - 81.87)^2)} = 9.39 \)
The standard deviation provides insight into how spread out the values are from the mean. A higher standard deviation indicates more spread.
Pulse Rates
Pulse rate (beats per minute) is an essential measure of heart health. In this problem, we're comparing the pulse rates of adult males and females.

Males: 86, 72, 64, 72, 72, 54, 66, 56, 80, 72, 64, 64, 96, 58, 66
Females: 64, 84, 82, 70, 74, 86, 90, 88, 90, 90, 94, 68, 90, 82, 80

Pulse rates can vary due to numerous factors like age, physical fitness, and stress levels.
By statistically analyzing these pulse rates, we can identify patterns or differences between the two groups, providing useful insights for health studies.
Statistical Analysis
Statistical analysis is pivotal for interpreting data. It involves performing calculations to summarize, describe, and compare datasets.

In this exercise, several key steps were performed:
  • Calculating the mean for each group of pulse rates
  • Determining the standard deviation to understand data dispersion
  • Computing the coefficient of variation (CV) to assess relative variability
The coefficient of variation is particularly insightful as it standardizes the measure of dispersion relative to the mean:

For Males: \( CV_{males} = \left( \frac{11.06}{68.87} \right) \times 100 = 16.06\text{%} \)
For Females: \( CV_{females} = \left( \frac{9.39}{81.87} \right) \times 100 = 11.46\text{%} \)
Comparing these, we see that males have more variation in their pulse rates compared to females. This analysis can be critical in tailoring healthcare or exercise programs for different groups.

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Most popular questions from this chapter

Watch out for these little buggers. Each of these exercises involves some feature that is somewhat tricky. Find the (a) mean, \((b)\) median, (c) mode, (d) midrange, and then answer the given question. Listed below in dollars are the amounts it costs for marriage proposal packages at the different Major League Baseball stadiums. Five of the teams don't allow proposals. Are there any outliers? \(\begin{array}{rrrrrrrrrrrr}39 & 50 & 50 & 50 & 55 & 55 & 75 & 85 & 100 & 115 & 175 & 175 & 200 \\ 209 & 250 & 250 & 350 & 400 & 450 & 500 & 500 & 500 & 500 & 1500 & 2500 & \end{array}\)

Find the range, variance, and standard deviation for the given sample data. Include appropriate units (such as "minutes") in your results. (The same data were used in Section 3-1, where we found measures of center. Here we find measures of variation.) Then answer the given questions. Listed below are the amounts (dollars) it costs for marriage proposal packages at the different Major League Baseball stadiums. Five of the teams don't allow proposals. Are there any outliers, and are they likely to have much of an effect on the measures of variation? \(5 \quad 200\) \(\begin{array}{rrrrrrrrrrrr}39 & 50 & 50 & 50 & 55 & 55 & 75 & 85 & 100 & 115 & 175 & 175 \\ 209 & 250 & 250 & 350 & 400 & 450 & 500 & 500 & 500 & 500 & 1500 & 2500\end{array}\)

Find the mean and median for each of the two samples, then compare the two sets of results. Listed below are pulse rates (beats per minute) from samples of adult males and females (from Data Set 1 "Body Data"in Appendix B). Does there appear to be a difference? \(\begin{array}{llllllllllllllll}\text { Male: } & 86 & 72 & 64 & 72 & 72 & 54 & 66 & 56 & 80 & 72 & 64 & 64 & 96 & 58 & 66 \\ \text { Female: } & 64 & 84 & 82 & 70 & 74 & 86 & 90 & 88 & 90 & 90 & 94 & 68 & 90 & 82 & 80\end{array}\)

Watch out for these little buggers. Each of these exercises involves some feature that is somewhat tricky. Find the (a) mean, \((b)\) median, (c) mode, (d) midrange, and then answer the given question. Listed below are the highest amounts of net worth (in millions of dollars) of celebrities. The celebrities are Tom Cruise, Will Smith, Robert De Niro, Drew Carey, George Clooney, John Travolta, Samuel L. Jackson, Larry King, Demi Moore, and Bruce Willis. What do the results tell us about the population of all celebrities? Based on the nature of the amounts, what can be inferred about their precision? $$ \begin{array}{llllllllll} 250 & 200 & 185 & 165 & 160 & 160 & 150 & 150 & 150 & 150 \end{array} $$

Find the range, variance, and standard deviation for the given sample data. Include appropriate units (such as "minutes") in your results. (The same data were used in Section 3-1, where we found measures of center. Here we find measures of variation.) Then answer the given questions. In a study of speed dating conducted at Columbia University, female subjects were asked to rate the attractiveness of their male dates, and a sample of the results is listed below \((1=\) not attractive \(; 10=\) extremely attractive). Can the results be used to describe the variation among attractiveness ratings for the population of adult males? \(\begin{array}{llllllllllllllllllllllllll}5 & 8 & 3 & 8 & 6 & 10 & 3 & 7 & 9 & 8 & 5 & 5 & 6 & 8 & 8 & 7 & 3 & 5 & 5 & 6 & 8 & 7 & 8 & 8 & 8 & 7\end{array}\)

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