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

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

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

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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

- 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!

- 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?

- Question 1 Answer:
- M =ΣX
n

- M =1+8+7+5+9
5

- M =30
5

- M = 6

- M =ΣX

- Question 2 Answer:
- M =ΣX
n

- 8 =ΣX
6

- ΣX = 48

- M =ΣX

- 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:

- Question 4 Answer:
- B) It must be less than 40. A score higher than 40 would have increased the mean.

- Question 5 Answer:
- 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

- 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?

- Question 6 Answer:
- False. If the score is equal to the mean, it does not change the mean.

- Question 7 Answer:
- 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.

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

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

- Question 8 Answer:
- μ = 7

- Question 9 Answer:

Below

4

1

Below

1

Below

2

Above

4

Above

- 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.

- Question 10 Answers:
- The median is X = 7
- The median is X = 11.5

- Question 11 Answer:

2/3

6

1/3

5

Count 8 boxes

7

2

4

3

1

1

2

3

4

5

6

Median = 6.83

- 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).

- 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.

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

- Question 12 Answers:

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.

- 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)

- Question 13 Answer:
- Mean

- Question 14 Answer:
- The definition of the mean is based on distances (the mean balances the distances) and ordinal scales do no measure distance.

- Question 15 Answer:
- False, if the distribution is bimodal.

- Question 16 Answer:
- False. The mean is displaced toward the tail on the left-hand side.

- 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%.

- Step 5: We stopped counting when we reached seven boxes at the interval X = 6, which has an upper real limit of 6.5. We want 1/3 of the boxes in the next interval, so we add 6.5 + (1/3) = 6.83.

Median = 6.83