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Chapter 1: Problem 49

The first four deviations from the mean in a sample of \(n=5\) reaction times were \(.3, .9,1.0\), and 1.3. What is the fifth deviation from the mean? Give a sample for which these are the five deviations from the mean.

Short Answer

Expert verified
The fifth deviation is -3.5. A sample is [10.3, 10.9, 11.0, 11.3, 6.5].

Step by step solution

01

Understanding Deviations from the Mean

Deviations from the mean in a data set are the differences between each data point and the mean. For a sample with 5 numbers, if the deviations are known for 4 of them, we can find the fifth by ensuring the sum of all deviations is zero. This is because the sum of deviations from the mean must be zero.
02

Calculate the Sum of Given Deviations

The given deviations from the mean are 0.3, 0.9, 1.0, and 1.3. First, calculate their sum:\[ 0.3 + 0.9 + 1.0 + 1.3 = 3.5 \]
03

Find the Fifth Deviation

Since the sum of all deviations from the mean must be zero, we can find the fifth deviation by solving the equation:\[ \text{sum of deviations} = 0.3 + 0.9 + 1.0 + 1.3 + x = 0 \]Hence,\[ 3.5 + x = 0 \]\[ x = -3.5 \]
04

Verify the Solution

We found the fifth deviation to be -3.5. The sum of all five deviations is:\[ 0.3 + 0.9 + 1.0 + 1.3 - 3.5 = 0 \]Therefore, the deviation calculation validates our solution.
05

Construct a Sample

To create a sample, choose a mean, say 10. Add each deviation to the mean:- First number: \( 10 + 0.3 = 10.3 \)- Second number: \( 10 + 0.9 = 10.9 \)- Third number: \( 10 + 1.0 = 11.0 \)- Fourth number: \( 10 + 1.3 = 11.3 \)- Fifth number: \( 10 - 3.5 = 6.5 \)The sample is \([10.3, 10.9, 11.0, 11.3, 6.5]\).

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Key Concepts

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

Sample Size
In statistics, the sample size refers to the number of observations in a dataset. For our exercise, the sample size, denoted as \( n \), is 5. This simply means you have 5 different data points or values to work with, which here are specific deviations from a mean. Understanding the sample size is crucial because it helps us determine various statistical measures, like the mean and the sum of deviations. Knowing that our sample size is 5 also directly impacts how we find the fifth deviation, considering we already know the first four.
Mean Calculation
The mean is a measure of central tendency that represents the average of a data set. We calculate the mean by summing all values and dividing by the number of values in the dataset. In our case, since we are focusing on deviations from a mean, the actual mean value isn't provided, but the principles remain the same. In statistical problems dealing with deviations, the mean acts as the central reference point from which deviations are measured. For the mean to correctly represent the dataset, the sum of all deviations from it should be zero. This explains why the fifth deviation was necessary to achieve a balanced sum of zero in our problem.
Sum of Deviations
When calculating deviations from the mean, it is important to remember that the sum of these deviations should equal zero. This is a key concept in statistics. In our problem, we were given four deviations, adding up to 3.5.To maintain the statistical rule that the sum of deviations equals zero, the fifth deviation must counterbalance the sum of the known deviations. Thus, the fifth deviation was calculated by solving the equation: \[0.3 + 0.9 + 1.0 + 1.3 + x = 0\]Solving it gives us \( x = -3.5 \), ensuring the total sum of deviations is zero, validating our approach.
Statistical Analysis
Statistical analysis involves interpreting data to uncover patterns or insights. In our context, it means using mathematical techniques to assess the dataset's properties, such as finding deviations from the mean. This problem led us through several components of statistical analysis:
  • We applied algebra to calculate numerical relationships.
  • We verified the mathematical consistency of our results, for example, by confirming that deviations summed up to zero.
  • We constructed a hypothetical dataset using our calculations, which mimics how real sample data might be generated.
This exercise showcases the practical applications of statistical concepts such as sample size, mean calculation, and deviations, all pivotal in making informed decisions from data.

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Most popular questions from this chapter

A certain city divides naturally into ten district neighborhoods. A real estate appraiser would like to develop an equation to predict appraised value from characteristics such as age, size, number of bathrooms, distance to the nearest school, and so on. How might she select a sample of singlefamily homes that could be used as a basis for this analysis?

In a famous experiment carried out in 1882, Michelson and Newcomb obtained 66 observations on the time it took for light to travel between two locations in Washington, D.C. A few of the measurements (coded in a certain manner) were 31,23 , \(32,36,22,26,27\), and 31 . a. Why are these measurements not identical? b. Does this study involve sampling an existing population or a conceptual population?

a. For what value of \(c\) is the quantity \(\sum\left(x_{i}-c\right)^{2}\) minimized? [Hint: Take the derivative with respect to \(c\), set equal to 0 , and solve.] b. Using the result of part (a), which of the two quantities \(\sum\left(x_{i}-\bar{x}\right)^{2}\) and \(\sum\left(x_{i}-\mu\right)^{2}\) will be smaller than the other (assuming that \(\bar{x} \neq \mu) ?\)

The accompanying data set consists of observations on shower-flow rate (L/min) for a sample of \(n=129\) houses in Perth, Australia ("An Application of Bayes Methodology to the Analysis of Diary Records in a Water Use Study," J. Amer. Statist. Assoc., 1987: 705-711): \(\begin{array}{rrrrrrrrrr}4.6 & 12.3 & 7.1 & 7.0 & 4.0 & 9.2 & 6.7 & 6.9 & 11.5 & 5.1 \\ 11.2 & 10.5 & 14.3 & 8.0 & 8.8 & 6.4 & 5.1 & 5.6 & 9.6 & 7.5 \\\ 7.5 & 6.2 & 5.8 & 2.3 & 3.4 & 10.4 & 9.8 & 6.6 & 3.7 & 6.4 \\ 8.3 & 6.5 & 7.6 & 9.3 & 9.2 & 7.3 & 5.0 & 6.3 & 13.8 & 6.2 \\ 5.4 & 4.8 & 7.5 & 6.0 & 6.9 & 10.8 & 7.5 & 6.6 & 5.0 & 3.3 \\ 7.6 & 3.9 & 11.9 & 2.2 & 15.0 & 7.2 & 6.1 & 15.3 & 18.9 & 7.2 \\ 5.4 & 5.5 & 4.3 & 9.0 & 12.7 & 11.3 & 7.4 & 5.0 & 3.5 & 8.2 \\ 8.4 & 7.3 & 10.3 & 11.9 & 6.0 & 5.6 & 9.5 & 9.3 & 10.4 & 9.7 \\ 5.1 & 6.7 & 10.2 & 6.2 & 8.4 & 7.0 & 4.8 & 5.6 & 10.5 & 14.6 \\ 10.8 & 15.5 & 7.5 & 6.4 & 3.4 & 5.5 & 6.6 & 5.9 & 15.0 & 9.6 \\ 7.8 & 7.0 & 6.9 & 4.1 & 3.6 & 11.9 & 3.7 & 5.7 & 6.8 & 11.3 \\ 9.3 & 9.6 & 10.4 & 9.3 & 6.9 & 9.8 & 9.1 & 10.6 & 4.5 & 6.2 \\ 8.3 & 3.2 & 4.9 & 5.0 & 6.0 & 8.2 & 6.3 & 3.8 & 6.0 & \end{array}\) a. Construct a stem-and-leaf display of the data. b. What is a typical, or representative, flow rate? c. Does the display appear to be highly concentrated or spread out? d. Does the distribution of values appear to be reasonably symmetric? If not, how would you describe the departure from symmetry? e. Would you describe any observation as being far from the rest of the data (an outlier)?

A sample of 26 offshore oil workers took part in a simulated escape exercise, resulting in the accompanying data on time \((\mathrm{sec})\) to complete the escape \(\left({ }^{*} \mathrm{Oxygen}\right.\) Consumption and Ventilation During Escape from an Offshore Platform," Ergonomics, 1997: 281-292): \(\begin{array}{lllllllll}389 & 356 & 359 & 363 & 375 & 424 & 325 & 394 & 402 \\\ 373 & 373 & 370 & 364 & 366 & 364 & 325 & 339 & 393 \\ 392 & 369 & 374 & 359 & 356 & 403 & 334 & 397 & \end{array}\) a. Construct a stem-and-leaf display of the data. How does it suggest that the sample mean and median will compare? b. Calculate the values of the sample mean and median. [Hint: \(\sum x_{i}=9638 .\) ] c. By how much could the largest time, currently 424 , be increased without affecting the value of the sample median? By how much could this value be decreased without affecting the value of the sample median? d. What are the values of \(\bar{x}\) and \(\bar{x}\) when the observations are reexpressed in minutes?
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