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A sample of \(n=5\) college students yielded the following observations on number of traffic citations for a moving violation during the previous year: $$ x_{1}=1 \quad x_{2}=0 \quad x_{3}=0 \quad x_{4}=3 \quad x_{5}=2 $$ Calculate \(s^{2}\) and \(s\). 4.16 Give two sets of five numbers that have the same mean but different standard deviations, and give two sets of five numbers that have the same standard deviation but different means.

Short Answer

Expert verified
For part one, following the calculations, you get \(s^{2} = 1.7\) and \(s \approx 1.30\). Part two is open-ended, and possible responses are: (1) Sets with same mean, different standard deviations: [2,2,2,2,2] and [0,1,2,3,4], (2) Sets with same standard deviation, different means: [1,2,3,4,5] and [6,7,8,9,10]. Note that different sets of numbers could be generated as an answer to part two.

Step by step solution

01

Calculate the Mean

The mean (\(\bar{x}\)) is the sum of all the observations divided by the number of observations. For the given data, add 1, 0, 0, 3, and 2 and divide by 5.
02

Calculate Sample Variance

Sample variance (\(s^{2}\)) measures how dispersed the data is from the mean calculated in step 1. Subtract the mean from each observation, square the result, sum these squared deviations, then divide by n-1 (the number of observations minus one).
03

Calculate Standard Deviation

The standard deviation (\(s\)) is the square root of the variance. Calculate the square root of the answer from Step 2.
04

Create Sets With Same Mean, Different Standard Deviations

To maintain the same mean, adjust some values while keeping the total sum constant. Deviations from the mean should vary to result in different standard deviations.
05

Create Sets With Same Standard Deviation, Different Means

To maintain the same standard deviation, the spread or dispersion of the numbers should be the same. However, their actual values and total sum should change to result in different means.

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Key Concepts

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

Sample Variance
Sample Variance is a key concept in descriptive statistics, providing a measure of the dispersion within a dataset. It tells us how much the data points deviate from the mean, in a squared fashion. To calculate sample variance (\(s^2\)), follow these steps:
  • First, calculate the mean (\(\bar{x}\)) of your dataset, which will provide a central reference point.
  • Subtract the mean from each of the data points to find their deviation from the mean.
  • Square each of these deviations to eliminate negative values and emphasize larger disparities.
  • Sum up all these squared deviations.
  • Finally, divide this total by \(n-1\), where \(n\) is the number of observations. This division by \(n-1\) instead of \(n\) is known as Bessel’s correction, used because we are dealing with a sample rather than the full population.
This process yields a result in squared units of your data, providing a solid measure of how spread out the numbers are around their average.
Standard Deviation
The Standard Deviation is a vital statistical tool that tells us how spread out the numbers in a dataset are. Unlike variance, standard deviation has the same units as the original data, making it easier to interpret:
  • The standard deviation (\(s\)) is simply the square root of the sample variance (\(s^2\)).
  • Calculating the square root converts the units from squared back to the original units of the data.
This conversion is particularly useful because it allows us to understand the dispersion in the context of the original data set. A higher standard deviation means that the data points are more widely spread out from the mean. Conversely, a lower standard deviation indicates that they are closer to the mean. Understanding standard deviation is crucial because it helps us grasp the variability of data, which is often more intuitive to comprehend than variance. It can be used to compare the spread of two different datasets even if they have different means.
Mean Calculation
Understanding how to calculate the mean is foundational in statistics, as it provides the average value and a central point of a dataset. Here’s how you calculate the mean:
  • Add up all the numbers in the dataset to get a total sum.
  • Count the number of data points you have, denoted as \(n\).
  • Divide the total sum by \(n\) to find the mean (\(\bar{x}\)).
For example, using the provided dataset (1, 0, 0, 3, 2), the calculation would be: \[\bar{x} = \frac{1+0+0+3+2}{5} = \frac{6}{5} = 1.2\]The mean tells us where the center of the data lies and what can be considered an average observation. It's crucial for comparing other statistical measures such as variance or standard deviation since these often relate back to how data points deviate from the mean.

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Most popular questions from this chapter

The standard deviation alone does not measure relative variation. For example, a standard deviation of \(\$ 1\) would be considered large if it is describing the variability from store to store in the price of an ice cube tray. On the other hand, a standard deviation of \(\$ 1\) would be considered small if it is describing store-to-store variability in the price of a particular brand of freezer. A quantity designed to give a relative measure of variahility is the \(\mathrm{co-}\) efficient of variation. Denoted by CV, the coefficient of variation expresses the standard deviation as a percentage of the mean. It is defined by the formula \(C V=100\left(\frac{s}{\bar{x}}\right)\). Consider two samples. Sample 1 gives the actual weight (in ounces) of the contents of cans of pet food labeled as having a net weight of 8 oz. Sample 2 gives the actual weight (in pounds) of the contents of bags of dry pet food labeled as having a net weight of \(50 \mathrm{lb}\). The weights for the two samples are: \(\begin{array}{lrrrrr}\text { Sample 1 } & 8.3 & 7.1 & 7.6 & 8.1 & 7.6 \\ & 8.3 & 8.2 & 7.7 & 7.7 & 7.5 \\ \text { Sample 2 } & 52.3 & 50.6 & 52.1 & 48.4 & 48.8 \\ & 47.0 & 50.4 & 50.3 & 48.7 & 48.2\end{array}\) a. For each of the given samples, calculate the mean and the standard deviation. b. Compute the coefficient of variation for each sample. Do the results surprise you? Why or why not?

The ministry of Health and Long-Term Care in Ontario, Canada, publishes information on its web site (www.health.gov.on.ca) on the time that patients must wait for various medical procedures. For two cardiac procedures completed in fall of 2005 the following information was provided: a. The median wait time for angioplasty is greater than the median wait time for bypass surgery but the mean wait time is shorter for angioplasty than for bypass surgery. What does this suggest about the distribution of wait times for these two procedures? b. Is it possible that another medical procedure might have a median wait time that is greater than the time reported for " \(90 \%\) completed within"? Explain.

The article "Taxable Wealth and Alcoholic Beverage Consumption in the United States" (Psychological Reports [1994]: \(813-814\) ) reported that the mean annual adult consumption of wine was \(3.15\) gal and that the standard deviation was \(6.09\) gal. Would you use the Empirical Rule to approximate the proportion of adults who consume more than \(9.24\) gal (i.e., the proportion of adults whose consumption value exceeds the mean by more than 1 standard deviation)? Explain your reasoning.

A sample of 26 offshore oil workers took part in a simulated escape exercise, resulting in the accompanying data on time (in seconds) to complete the escape ("Oxygen Consumption and Ventilation During Escape from an Offshore Platform," Ergonomics[1997]: \(281-292\) ): \(\begin{array}{lllllllll}389 & 356 & 359 & 363 & 375 & 424 & 325 & 394 & 402\end{array}\) \(\begin{array}{lllllllll}373 & 373 & 370 & 364 & 366 & 364 & 325 & 339 & 393\end{array}\) $$ \begin{array}{llllllll} 392 & 369 & 374 & 359 & 356 & 403 & 334 & 397 \end{array} $$ a. Construct a stem-and-leaf display of the data. Will the sample mean or the sample median be larger for this data set? b. Calculate the values of the sample mean and median. c. By how much could the largest time be increased without affecting the value of the sample median? By how much could this value be decreased without affecting the sample median?

An instructor has graded 19 exam papers submitted by students in a class of 20 students, and the average so far is 70 . (The maximum possible score is 100 .) How high would the score on the last paper have to be to raise the class average by 1 point? By 2 points?

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