/*! 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 19 The data on the following page r... [FREE SOLUTION] | 91Ó°ÊÓ

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The data on the following page represent the pulse rates (beats per minute) of nine students enrolled in a section of Sullivan's course in Introductory Statistics. Treat the nine students as a population. $$ \begin{array}{lc} \text { Student } & \text { Pulse } \\ \hline \text { Perpectual Bempah } & 76 \\ \hline \text { Megan Brooks } & 60 \\ \hline \text { Jeff Honeycutt } & 60 \\ \hline \text { Clarice Jefferson } & 81 \\ \hline \text { Crystal Kurtenbach } & 72 \\ \hline \text { Janette Lantka } & 80 \\ \hline \text { Kevin MeCarthy } & 80 \\ \hline \text { Tammy Ohm } & 68 \\ \hline \text { Kathy Wojdyla } & 73 \end{array} $$ (a) Determine the population standard deviation. (b) Find three simple random samples of size 3 , and determine the sample standard deviation of each sample. (c) Which samples underestimate the population standard deviation? Which overestimate the population standard deviation?

Short Answer

Expert verified
Population standard deviation is 7.67. Samples 1 and 2 overestimate; Sample 3 underestimates.

Step by step solution

01

- List all Pulse Rates

Extract all the pulse rates from the given data. The nine students' pulse rates are: 76, 60, 60, 81, 72, 80, 80, 68, and 73.
02

- Calculate the Population Mean

Determine the mean pulse rate of the population using the formula: \[ \text{mean}(\mu) = \frac{\sum_{i=1}^{N}x_i}{N} \] \( N \) is the number of students. \( \mu = \frac{76+60+60+81+72+80+80+68+73}{9} = \frac{650}{9} \approx 72.22 \)
03

- Compute Each Deviation From the Mean

Calculate each deviation and then square it: i) \((76 - 72.22)^2 = 14.33\) ii) \((60 - 72.22)^2 = 149.38 \) iii) \((60 - 72.22)^2 = 149.38\) iv) \((81 - 72.22)^2 = 77.14\) v) \((72 - 72.22)^2 = 0.0484\) vi) \((80 - 72.22)^2 = 60.8\) vii) \((80 - 72.22)^2 = 60.8\) viii) \((68 - 72.22)^2 = 17.79\) ix) \((73 - 72.22)^2 = 0.60\)
04

- Calculate Population Variance

Determine the population variance by averaging the squared deviations: \[ \sigma^2 = \frac{\sum_{i=1}^{N}(x_i - \mu)^2}{N} \] Sum of squared deviations is: 14.33 + 149.38 + 149.38 + 77.14 + 0.0484 + 60.8 + 60.8 + 17.79 + 0.60 = 530.2584 Variance, \[ \sigma^2 = \frac{530.2584}{9} \approx 58.92 \]
05

- Calculate Population Standard Deviation

The population standard deviation is the square root of the variance: \[ \sigma = \sqrt{\sigma^2} \] \[ \sigma = \sqrt{58.92} \approx 7.67 \]
06

- Random Sampling and Sample Standard Deviation

Select three simple random samples of size 3 and compute their sample standard deviations using the formula: \[ s = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n-1}} \] **Sample 1:** (76, 60, 60) Mean is 65.33. Sample standard deviation: 9.28 **Sample 2:** (81, 72, 60) Mean is 71. Sample standard deviation: 10.68 **Sample 3:** (73, 80, 68) Mean is 73.67. Sample standard deviation: 6.11
07

- Compare Sample Standard Deviations

Compare sample standard deviations with the population standard deviation: Sample 1: 9.28 (Overestimates) Sample 2: 10.68 (Overestimates) Sample 3: 6.11 (Underestimates)

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Key Concepts

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

Mean Calculation
To find the mean, you simply add up all the values and then divide by the number of values you added. For the pulse rates provided for nine students, calculate the mean by summing their pulse rates and then dividing by 9. The equation for the mean (or average) is given by:

\[\text{mean}(\mu) = \frac{\sum_{i=1}^{N}x_i}{N}\text{,}\] where \(N\) represents the total number of values. Here, adding the pulse rates yields a sum of 650. Therefore, the mean pulse rate is:

\[\text{mean} = \frac{650}{9} \approx 72.22\text{.}\] This value is crucial because it represents the central tendency of the data.
Population Variance
Population variance measures how much the values in a population differ from the mean of the population. It gives us an idea of the overall spread of the data. To calculate it, you first compute the deviation (the difference between each value and the mean), square each deviation, and then find the average of these squared deviations.

The equation for population variance is:

\[\text{Population Variance} \(\sigma^2\) = \frac{\sum_{i=1}^{N}(x_i - \mu)^2}{N}\text{,}\] where \(x_i =\) individual values, and \(N =\) number of values.

In this exercise, you calculate the deviations for each pulse rate and then square them, e.g., \((76 - 72.22)^2 \approx 14.33\text{,}\...\text{.}\) Adding all the squared deviations gives about 530.26. Finally, the variance is:

\[\text{Population Variance} \(\sigma^2\) = \frac{530.2584}{9} \approx 58.92\text{.}\]
Sample Standard Deviation
A sample standard deviation measures the spread of data points within a sample, rather than an entire population. It estimates how much individual values deviate from the sample mean. This measure is often used when it is impractical or impossible to measure the entire population.

To calculate the sample standard deviation, first, find the mean of the sample. Then, calculate each deviation from this mean, square it, and sum these squared deviations. Lastly, divide by \(n - 1\) (where \(n\) is the number of values in the sample), and then take the square root of this quotient. The formula for sample standard deviation is:

\[\text{Sample Standard Deviation} \(s\) = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n-1}}\text{,}\] where \(\bar{x}\) represents the sample mean.

For example, Sample 1 with values (76, 60, 60) has a sample mean of 65.33 and a sample standard deviation of approximately 9.28. Comparing sample standard deviations against the population standard deviation (7.67), we see whether the samples overestimate or underestimate the spread.

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Most popular questions from this chapter

The following data represent exam scores in a statistics class taught using traditional lecture and a class taught using a "flipped" classroom. The "flipped" classroom is one where the content is delivered via video and watched at home, while class time is used for homework and activities. $$ \begin{array}{llllllll} \hline \text { Traditional } & 70.8 & 69.1 & 79.4 & 67.6 & 85.3 & 78.2 & 56.2 \\\ & 81.3 & 80.9 & 71.5 & 63.7 & 69.8 & 59.8 & \\ \hline \text { Fipped } & 76.4 & 71.6 & 63.4 & 72.4 & 77.9 & 91.8 & 78.9 \\ & 76.8 & 82.1 & 70.2 & 91.5 & 77.8 & 76.5 & \end{array} $$ (a) Which course has more dispersion in exam scores using the range as the measure of dispersion? (b) Which course has more dispersion in exam scores using the standard deviation as the measure of dispersion? (c) Suppose the score of 59.8 in the traditional course was incorrectly recorded as \(598 .\) How does this affect the range? the standard deviation? What property does this illustrate?

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