4.6: Putting All of the Measures Together
- Page ID
- 67040
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Phew! We’ve covered a lot in the last few chapters, and now we’re going to put it all together to calculate a standard deviation, then see if we can predict the shape of a distribution just based on the measures of central tendency and the standard deviation as the measure of variability.
Let’s start with the raw data in Table \(\PageIndex{1}\). The data are from 9 students in a face-to-face behavioral statistics class. The scores provided are each students’ score (out of 100 points) for all assignments related to research. These activities included smaller projects like worksheets identifying variables (which you could do now!), medium projects like calculating measures of central tendency and standard deviations as a measure of variability (which you could do now!), to larger projects like papers that statistically compared the means of different groups (which you’ll start learning in a few chapters).
|
Research Activities Score for Face-to-Face Class |
|---|
|
89 |
|
79 |
|
75 |
|
71 |
|
64 |
|
64 |
|
64 |
|
57 |
|
56 |
| \(\sum = 619.00\) |
Table \(\PageIndex{1}\) also includes the sum of the 9 scores in the last row because adding a bunch of scores in a calculator is hard!
Let’s start calculating the measures of central tendency.
- What is the mode for the Research Activities points in Table \(\PageIndex{1}\)?
- What is the median for the Research Activities points in Table \(\PageIndex{1}\)?
- What is the mean for the Research Activities points in Table \(\PageIndex{1}\)?
Solution
- The mode is 64 points as that is the only number that repeated, so it is the most frequent score.
- The median is also 64 points because that is the number that is in the exact middle of the distribution (four scores are higher and four scores are lower).
- The mean is rounded to 68.78 points.
\[\bar{X}=\frac{\sum X}{n} = \frac{619.00}{9} = 68.78 \nonumber\]
Once you have these three numbers that all represent the center of the frequency distribution, you can use them to make predictions about the shape of the distribution.
Based on the three measures of central tendency, what do you think the shape of the frequency distribution might be?
Solution
The mode will be the peak of the distribution, at 64 points. The median is the middle score in the whole distribution, and that is also 64 points. The mean is a little bigger at 68.78 points, which suggests that there are a few scores that are high (since the mean is higher than both the mode and the median). With a mean that’s higher than the mode and median, we’d expect a positively skewed distribution in which most scores are to the left, but there are one or a few higher scores.
We can now calculate the standard deviation.
What is the standard deviation for the Research Activities scores from Table \(\PageIndex{1}\)?
Solution
You know how to do this!
\[s=\sqrt{\dfrac{\sum(X-\overline {X})^{2}}{N-1}} \nonumber\]
Use Table \(\PageIndex{2}\) to calculate the numerator.
| \(\mathrm{X}\) | \(X-\overline{X}\) | Sums of Squares (\((X-\overline {X})^{2}\)) |
|---|---|---|
|
89 |
\(89-\overline{X} = 20.22 \) | \((X-\overline {X})^{2} = 20.22^{2} = 408.85 \) |
|
79 |
\(79-\overline{X} = 10.22\) | \((X-\overline {X})^{2} = 10.22^{2} = 104.45 \) |
|
75 |
\(75-\overline{X} = 6.22 \) | \((X-\overline {X})^{2} = 6.22^{2} = 38.69 \) |
|
71 |
\(71-\overline{X} = 2.22 \) | \((X-\overline {X})^{2} = 2.22^{2} = 4.93 \) |
|
64 |
\(64-\overline{X} = -4.78 \) | \((X-\overline {X})^{2} = -4.78^{2} = 22.85 \) |
|
64 |
\(64-\overline{X} = -4.78 \) | \((X-\overline {X})^{2} = -4.78^{2} = 22.85 \) |
|
64 |
\(64-\overline{X} = -4.78 \) | \((X-\overline {X})^{2} = -4.78^{2} = 22.85 \) |
|
57 |
\(57-\overline{X} = -11.78 \) | \((X-\overline {X})^{2} = -11.78^{2} = 138.77 \) |
|
56 |
\(56-\overline{X} = -12.78 \) | \((X-\overline {X})^{2} = -12.78^{2} = 163.33 \) |
| \(\Sigma = 619.00\) | \(\Sigma = 0\) (basically) | \(\Sigma = 927.56\) |
\[s=\sqrt{\dfrac{\sum(X-\overline {X})^{2}}{N-1}} = \sqrt{\dfrac{927.56}{9-1}} = \sqrt{\dfrac{927.56}{8}} = \sqrt{115.94} = 10.77 \nonumber\]
The standard deviation is 10.77 points.
Okay, here’s where it gets tricky (trickier?)! Now that we have the standard deviation (and the measures of central tendency), we can make predictions about the shape of the frequency distribution.
- Based on the mean, is this standard deviation small, medium or large?
- Based on the size of the standard deviation, what shape does this suggest that the frequency distribution will be?
Solution
Let’s start with deciding if the standard deviation is large. To determine that, we find out if the standard deviation of 10.77 points is larger than a quarter or a third of the mean of 68.78 points. Rounding to two decimal points, we find:
\[\frac{68.78}{4} = 17.20 \nonumber\]
\[\frac{68.78}{3} = 22.93 \nonumber\]
No, this is not a large standard deviation because 10.77 is not bigger than 17.20 or 22.93.
Our next try is to see if the standard deviation of 10.77 points is smaller than about 10% of the mean of 68.78 points. Rounding to two decimal points, we find:
\[68.78\times 0.10= 6.88 \nonumber\]
No, this is not a small standard deviation because 10.77 is not smaller than 6.88.
That means that the standard deviation is medium!
A medium standard deviation suggests a mesokurtic frequency distribution, or a bell-shaped curve.
Okay, now we have one set of numbers (measures of central tendency) suggesting a skewed distribution and the standard deviation suggesting a bell-shaped curve. Which is it? Let’s see!
It looks like the measures of central tendency were slightly better at predicting the shape of the graph than the standard deviation. As you’ll learn in the next section on distributions, these predictions will be more accurate the more scores that we have, but we used data with a sample size of only 9 so our predictions were not robust.
Table \(\PageIndex{2}\) provides another set of points earned on Research Activities, but this time for an online class. The sum of these 12 scores is also provided in the last row. Are you ready? Yes, you are! Go through the following Exercises to see how well you can put it all together.
|
Research Activities Score for Online Class |
|---|
|
99 |
|
93 |
|
92 |
|
88 |
|
88 |
|
83 |
|
79 |
|
77 |
|
76 |
|
69 |
|
67 |
|
56 |
|
\(\sum = 967\) |
What do we start with? Yes, let’s calculate the three measures of central tendency.
- What is the mode for the Research Activities points in Table \(\PageIndex{2}\)?
- What is the median for the Research Activities points in Table \(\PageIndex{2}\)?
- What is the mean for the Research Activities points in Table \(\PageIndex{2}\)?
- Answer
-
- The mode is 88 points as that is the only number that repeated, so it is the most frequent score.
- The median is also 81 points because that is the number that would be in the exact middle if there was one number in the exact middle. This is the score that would be between 83 points and 79 points.
- The mean is rounded to 80.58 points.
\[\bar{X}=\frac{\sum X}{n} = \frac{967.00}{12} = 80.58 \nonumber\]
Once you have these three numbers that all represent the center of the frequency distribution, you can use them to make predictions about the shape of the distribution.
Based on the three measures of central tendency, what do you think the shape of the frequency distribution might be?
- Answer
-
The mode will be the peak of the distribution, at 88 points. The median is a little lower than the mode, showing that the middle score in the whole distribution is lower than the most frequent score. The mean is a little lower still, at 80.58 points. This suggests a negatively skewed frequency distribution, especially as the mode is a much higher score.
We can now calculate the standard deviation.
What is the standard deviation for the Research Activities scores from Table \(\PageIndex{2}\?.
Solution
You know how to do this!
\[s=\sqrt{\dfrac{\sum(X-\overline {X})^{2}}{N-1}} \nonumber\]
Use Table \(\PageIndex{2}\) to calculate the numerator.
| \(\mathrm{X}\) | \(X-\overline{X}\) | Sums of Squares (\((X-\overline {X})^{2}\)) |
|---|---|---|
|
99 |
||
|
93 |
||
|
92 |
||
|
88 |
||
|
88 |
||
|
83 |
||
|
79 |
||
|
77 |
||
|
76 |
||
|
69 |
||
|
67 |
||
|
56 |
||
| \(\Sigma = 967.00\) | \(\Sigma = 0\) (basically) | \(\Sigma = 1698.92\) |
\[s=\sqrt{\dfrac{\sum(X-\overline {X})^{2}}{N-1}} = \sqrt{\dfrac{1698.72}{12-1}} = \sqrt{\dfrac{1698.92}{11}} = \sqrt{154.45} = 12.43 \nonumber\]
The standard deviation is 12.43 points.
Using the standard deviation (and the measures of central tendency), make additional predictions about the shape of the frequency distribution.
- Based on the mean, is this standard deviation small, medium or large?
- Based on the size of the standard deviation, what shape does this suggest that the frequency distribution will be?
- Answer
-
Let’s start with deciding if the standard deviation is large. To determine that, we find out if the standard deviation of 12.43 points is larger than a quarter or a third of the mean of 80.58 points.
\[\frac{80.58}{4} = 20.15 \nonumber\]
\[\frac{80.58}{3} = 26.86 \nonumber\]No, this is not a large standard deviation because 12.43 is not bigger than 20.15 or 26.86.
Our next try is to see if the standard deviation of 12.43 points is smaller than about 10% of the mean of 80.58 points.
\[80.58\times 0.10= 8.06 \nonumber\]
No, this is not a small standard deviation because 12.43 is not smaller than 8.06. That means that the standard deviation is medium!
A medium standard deviation suggests a mesokurtic frequency distribution, or a bell-shaped curve.
Again, we have one set of numbers (measures of central tendency) suggesting a skewed distribution and the standard deviation suggesting a bell-shaped curve.
It looks like the measures of central tendency were slightly better at predicting the shape of the graph again! Figure \(\PageIndex{2}\) is negatively skewed.
Next, let’s learn more about different kinds of distributions and theories that may explain why our predictions based on the measures of central tendencies didn’t match our predictions based on the standard deviation.