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Chapter 5 Questions
Complete the following questions. The questions may include pictures or graphics to illustrate or aid in solving the problem. You can check your answer by clicking View Answer. If the question is unclear, confusing, or if you need further clarification, send me an email.
1. Variability activity (Click on arrows in the media activity to navigate the activity)
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2. Given the data below, compute the following measures of variability (questions 3-8).
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See next questions
3. Range.
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Range is highest score minus lowest score: 80-20 = 60.
4. Semi-interquartile range (i.e. subtract the score at the 75th percentile from the score at the 25th, find the midpoint of this middle 50% of the data, and compare this value to the mean).
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5. Sum the deviations about the mean.
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6. Sum the squared deviations about the mean.
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7. Find the variance.
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8. Find the standard deviation.
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9. What weaknesses do each of the above measures of variability have that are compensated for by the standard deviation?
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a. The range is simply the difference between the highest and lowest scores in a data set. It does not give any information about the spread of the scores between those two bounds.
b. The semi-interquartile range is too unwieldy. A better measure of the variability of scores in a data set would be a single numeric index.
c. Summing the deviations about the mean always results in zero. Arbitrarily changing the negative deviation values to positives and then summing them has no mathematical basis.
d. The sum of the squared deviations about the mean is a good index of the variability around the mean, but suffers two shortcomings. First, the measurement is in squared units rather than the original units of measurement. Second, this sum is sensitive to sample size. In other words, it is likely that the more scores there are in the sample, the larger this sum will be, regardless of the amount of variability in the sample.
e. The variance avoids the problem of sensitivity to sample size by dividing the sum of the squared deviations by N-1, but still suffers from the problem of measuring variability in squared units.
f. The standard deviation avoids the problem of measurement in squared units by square-rooting the average squared deviation about the mean (the variance).
b. The semi-interquartile range is too unwieldy. A better measure of the variability of scores in a data set would be a single numeric index.
c. Summing the deviations about the mean always results in zero. Arbitrarily changing the negative deviation values to positives and then summing them has no mathematical basis.
d. The sum of the squared deviations about the mean is a good index of the variability around the mean, but suffers two shortcomings. First, the measurement is in squared units rather than the original units of measurement. Second, this sum is sensitive to sample size. In other words, it is likely that the more scores there are in the sample, the larger this sum will be, regardless of the amount of variability in the sample.
e. The variance avoids the problem of sensitivity to sample size by dividing the sum of the squared deviations by N-1, but still suffers from the problem of measuring variability in squared units.
f. The standard deviation avoids the problem of measurement in squared units by square-rooting the average squared deviation about the mean (the variance).