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Chapter 20: Problem 2

Researchers at Texas A\&M studied the effect of using standing height desks in an elementary school on the energy expended by nine students. In their paper about the study, they reported descriptive statistics for the nine students. These descriptive statistics were expressed as a mean plus or minus a standard deviation. \({ }^{3}\) One such descriptive statistic was the weight of students before using the standing desks, which was reported as \(27.0 \pm 7.9\) kilograms. What are \(\mathrm{x}^{-} \bar{x}\) and the standard error of the mean for these students? (This exercise is also a warning to read carefully: that \(27.0 \pm\) \(7.9\) is not a confidence interval, yet summaries in this form are common in scientific reports.)

Short Answer

Expert verified
Mean is 27.0 kg, standard error is 2.63 kg.

Step by step solution

01

Identify the Mean

The mean weight of the students before using standing desks is given directly in the problem as \( \bar{x} = 27.0 \) kilograms.
02

Identify the Standard Deviation

The problem also provides the standard deviation of the students' weights, which is \( s = 7.9 \) kilograms.
03

Calculate the Standard Error of the Mean

The standard error (SE) of the mean is calculated using the formula:\[ SE = \frac{s}{\sqrt{n}} \]where \( s \) is the standard deviation and \( n \) is the number of students, which is 9. Substituting the values:\[ SE = \frac{7.9}{\sqrt{9}} = \frac{7.9}{3} \approx 2.63 \]

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Key Concepts

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

Mean
The mean, often symbolized as \( \bar{x} \), represents the average of a set of numbers. To find the mean, you add up all the numbers in a set and then divide by the total number of values in that set.

In the context of the exercise, the mean weight of students before using standing desks equals 27.0 kilograms. This is a simplistic and easily understood statistic, as it provides a central value around which the data points are distributed. Using mean allows researchers to summarize the data with a single value, which is incredibly useful when dealing with large data sets.

The key takeaway from using the mean is to get a sense of the "typical" value in your dataset, knowing that it can be affected by extremely large or small numbers.
Standard Deviation
Standard deviation, denoted as \( s \), is a measure indicating how spread out the numbers in a dataset are. A smaller standard deviation indicates that the data points are close to the mean, while a larger one shows more spread.

In the exercise, the standard deviation of student weights before using standing desks was reported as 7.9 kilograms. This implies that the students' weights vary on average by 7.9 kilograms from the mean of 27.0 kilograms.

Standard deviation is crucial in statistics because it allows researchers to understand the variability within a set. This is important to determine how consistent or inconsistent the data is. In scientific contexts, knowing the standard deviation helps to measure the reliability of the mean.
Standard Error
The standard error (SE) quantifies the accuracy with which a sample mean represents the population mean. It is calculated by dividing the standard deviation \( s \) by the square root of the sample size \( n \). In formula terms:
  • \( SE = \frac{s}{\sqrt{n}} \)


The exercise provided a calculation of this standard error. With a standard deviation of 7.9 and 9 students (\( n = 9 \)), the standard error was found to be approximately 2.63 kilograms.
  • \( SE = \frac{7.9}{\sqrt{9}} = \frac{7.9}{3} \approx 2.63 \)


The smaller the standard error, the more closely the sample mean approaches the population mean. It's a vital tool to judge how much statistical noise might be throwing your sample off from the true population average. This helps in determining the precision of the sample mean, making it significant for researchers in drawing conclusions from their data.

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Most popular questions from this chapter

A study of commuting times reports the travel times to work of a random sample of 1000 employed adults in Seattle. \({ }^{2}\) The mean is \(\mathrm{x}^{-} \bar{x}=\) 37.9 minutes and the standard deviation is \(s=27.2\) minutes. What is the standard error of the mean?

You are testing \(H_{0}: \mu=100\) against \(H_{a}: \mu<100\) based on an SRS of 16 observations from a Normal population. The data give \(x^{-} \bar{x}=98\) and \(s=4\). The value of the \(t\) statistic is (a) \(-8\). (b) \(-2\). (c) \(-0.5\).

Velvetleaf is a particularly annoying weed in corn fields. It produces lots of seeds, and the seeds wait in the soil for years until conditions are right. How many seeds do velvetleaf plants produce? Here are counts from 28 plants that came up in a corn field when no herbicide was used: 28 \(\begin{array}{ll}245025042114111021378015 & 1623\end{array}\) \(\begin{array}{lllllllllll}721 & 863 & 1136 & 2819 & 1911 & 2101 & 1051 & 218 & 1711 & 164\end{array}\) \(22283635973 \quad 105019611809 \quad 130 \quad 880\) We would like to give a confidence interval for the mean number of seeds produced by velvetleaf plants. Alas, the \(t\) interval can't be safely used for these data. Why not?

Do wearable devices that monitor diet and physical activity help people lose weight? Researchers had 237 subjects, already involved in a program of diet and exercise, use wearable technology for 24 months. They measured their weight (in kilograms) before using the technology and 24 months after using the technology. 18 (a) Explain why the proper procedure to compare the mean weight before using the wearable technology and 24 months after using the wearable technology is a matched pairs \(t\) test. (b) The 237 differences in weight (weight after 24 months minus weight before using the wearable technology) had \(\mathrm{x}^{-} \bar{x}=-3.5\) and \(s=7.8\). Is there significant evidence of a reduction in weight after using the wearable technology?

Fortunately, we aren't really interested in the number of seeds velvetleaf plants produce (see Exercise \(20.41\) ). The velvetleaf seed beetle feeds on the seeds and might be a natural weed control. Here are the total seeds, seeds infected by the beetle, and percent of seeds infected for 28 velvetleaf plants: $$ \begin{array}{l|rrrrrrrrrr} \hline \text { Seeds } & 2450 & 2504 & 2114 & 1110 & 2137 & 8015 & 1623 & 1531 & 2008 & 1716 \\ \text { Infected } & 135 & 101 & 76 & 24 & 121 & 189 & 31 & 44 & 73 & 12 \\ \text { Percent } & 5.5 & 4.0 & 3.6 & 2.2 & 5.7 & 2.4 & 1.9 & 2.9 & 3.6 & 0.7 \\\ \hline \text { Seeds } & 721 & 863 & 1136 & 2819 & 1911 & 2101 & 1051 & 218 & 1711 & 164 \\ \text { Infected } & 27 & 40 & 41 & 79 & 82 & 85 & 42 & 0 & 64 & 7 \\ \text { Percent } & 3.7 & 4.6 & 3.6 & 2.8 & 4.3 & 4.0 & 4.0 & 0.0 & 3.7 & 4.3 \\\ \hline \text { Seeds } & 2228 & 363 & 5973 & 1050 & 1961 & 1809 & 130 & 880 & & \\ \text { Infected } & 156 & 31 & 240 & 91 & 137 & 92 & 5 & 23 & & \\ \text { Percent } & 7.0 & 8.5 & 4.0 & 8.7 & 7.0 & 5.1 & 3.8 & 2.6 & & \\ \hline \end{array} $$ Do a complete analysis of the percent of seeds infected by the beetle. Include a \(90 \%\) confidence interval for the mean percent infected in the population of all velvetleaf plants. Do you think that the beetle is very helpful in controlling the weed? Why is analyzing percent of seeds infected more useful than analyzing number of seeds infected?
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