# COURSE: JUST 3900 TIPS FOR APLIA Developed By: Ethan Cooper (Lead Tutor) John Lohman - PowerPoint PPT Presentation

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COURSE: JUST 3900 TIPS FOR APLIA Developed By: Ethan Cooper (Lead Tutor) John Lohman Michael Mattocks Aubrey Urwick. Chapter 3: Central Tendency. Key Terms: Don’t Forget Notecards. Central Tendency (p. 73) Mean (p. 74) Weighted Mean (p. 77) Median (p. 83) Mode (p. 87)

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COURSE: JUST 3900 TIPS FOR APLIA Developed By: Ethan Cooper (Lead Tutor) John Lohman

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#### Presentation Transcript

COURSE: JUST 3900

TIPS FOR APLIA

Developed By:

John Lohman

Michael Mattocks

Aubrey Urwick

Chapter 3:

Central Tendency

### Key Terms: Don’t Forget Notecards

• Central Tendency (p. 73)

• Mean (p. 74)

• Weighted Mean (p. 77)

• Median (p. 83)

• Mode (p. 87)

• Unimodal (p. 88)

• Bimodal (p. 88)

• Multimodal (p. 88)

• HINT: Review distribution shapes from Ch. 2!

### Mean

• Question 1: Find the mean for the sample of n=5 scores: 1, 8, 7, 5, 9

• Question 2: A sample of n=6 scores has a mean of M=8. What is the value of ΣX for this sample?

• Question 3: One sample has n=5 scores with a mean of M=4. A second sample has n=3 scores with a mean of M=10. If the two samples are combined, what is the mean for the combined sample?

### Mean

• M =ΣX

n

• M =1+8+7+5+9

5

• M =30

5

• M = 6

• M =ΣX

n

• 8 =ΣX

6

• ΣX = 48

### Mean

• Question 4: A sample of n=6 scores has a mean of M=40. One new score is added to the sample and the new mean is found to be M=35. What can you conclude about the value of the new score?

• It must be greater than 40.

• It must be less than 40.

• Question 5: Find the values for n, ΣX, and M for the following sample:

### Mean

• B) It must be less than 40. A score higher than 40 would have increased the mean.

• n = 1+2+3+5+1

• n = 12

• Σ X = 5+4+4+3+3+3+2+2+2+2+2+1

• Σ X = 33

• M =33

12

• M = 2.75

### Mean

• Question 6: Adding a new score to a distribution always changes the mean. True or False?

• Question 7: A population has a mean of μ = 40.

• If 5 points were added to every score, what would be the value for the new mean?

• If every score were multiplied by 3, what would be the value of the new mean?

### Mean

• False. If the score is equal to the mean, it does not change the mean.

• The new mean would be 45. When a constant is added to every score, the same constant is added to the mean.

• The new mean would be 120. When every score is multiplied (or divided) by a constant, the mean changes in the same way.

### Mean

• Question 8: What is the mean of the following population?

### Mean

• Question 9: Using the scores from question 8, fill in the following table.

• μ = 7

Below

4

1

Below

1

Below

2

Above

4

Above

### Median

• Question 10: Find the median for each distribution of scores:

• 3, 4, 6, 7, 9, 10, 11

• 8, 10, 11, 12, 14, 15

• Question 11:The following is a distribution of measurements for a continuous variable. Find the precise median that divides the distribution exactly in half.

### Median

• The median is X = 7

• The median is X = 11.5

2/3

6

1/3

5

Count 8 boxes

7

2

4

3

1

1

2

3

4

5

6

Median = 6.83

### Median

• Question 11 Explanation:

• To find the precise median, we first observe that the distribution contains n = 16 scores. The median is the point with exactly 8 boxes on each side. Starting at the left-hand side and moving up the scale of measurement, we accumulate a total of 7 boxes when we reach a value of 6.5. We need 1 more box to reach our goal of 8 boxes (50%), but the next interval contains 3 boxes. The solution is to take a fraction of each box so that the fractions combine to give you one box. The fraction is determined by the number of boxes needed to reach 50% (numerator) and the number that exists in the interval (denominator).

### Median

• Question 11 Explanation:

• For this example, we needed 1 out of the 3 boxes in the interval, so the fraction is 1/3. The median is the point located exactly one-third of the way through the interval. The interval for X = 7 extends from 6.5 to 7.5. The interval width is one point, so one-third of the interval corresponds to approximately 0.33 points. Starting at the bottom of the interval and moving up 0.33 points produces a value of 6.50 + 0.33 = 6.83. This is the median, with exactly 50% of the distribution (8 boxes) on each side.

### Mode

• Question 12: What is the mode(s) of the following distribution? Is the distribution unimodal or bimodal?

### Mode

The modes are 2 and 8

The distribution is bimodal.

Note: While this is a bimodal distribution,

both modes have the same frequency.

Thus, there is no “minor” or “major” mode.

### Selecting a Measure of Central Tendency

• Question 13: Which measure of central tendency is most affected if one extremely large score is added to a distribution? (mean, median, mode)

• Question 14: Why is it usually inappropriate to compute a mean for scores measured on an ordinal scale?

• Question 15: In a perfectly symmetrical distribution, the mean, the median, and the mode will all have the same value. (True or False)

• Question 16: A distribution with a mean of 70 and a median of 75 is probably positively skewed. (True or False)

### Selecting a Measure of Central Tendency

• Mean

• The definition of the mean is based on distances (the mean balances the distances) and ordinal scales do no measure distance.

• False, if the distribution is bimodal.

• False. The mean is displaced toward the tail on the left-hand side.

### Central Tendency and Distribution Shape

• Graphs make life so much easier!

Symmetrical Distributions

Negatively Skewed Distribution

Positively Skewed Distribution

Note: Median

usually falls

between mean

and mode.

Notice how the means follow the outliers

• Interpolation

• Real Limits

• Median for Continuous Variables

• Frequency Distribution

• Cumulative Distributions

• Weighted Mean

• How do I find the median for a continuous variable?

• Step 1: Count the total number of boxes.

• Step 2: How many boxes are necessary to reach 50%?

• Step 3: Count the necessary number of boxes starting from the left (in this case 8).

50% of 16

is 8.

14

16 boxes

10

13

16

7

9

12

11

15

4

5

6

8

3

1

2

Uh-oh!

What now?

1/3

2/3

7

1

2

3

4

5

6

6.5

7.5

• Step 4: We need one more box to reach 8, but there are

• three boxes over the interval spanning 6.5 – 7.5. Thus,

• we need 1/3 of each box to reach 50%.