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COURSE: JUST 3900 INTRODUCTORY STATISTICS FOR CRIMINAL JUSTICE Chapter 3: Central Tendency Peer Tutor Slides Instructor: Mr. Ethan W. Cooper, Lead Tutor © 2013 - - PLEASE DO NOT CITE, QUOTE, OR REPRODUCE WITHOUT THE WRITTEN PERMISSION OF THE AUTHOR. FOR PERMISSION OR QUESTIONS, PLEASE EMAIL MR. COOPER AT THE FOLLWING: coopere07@students.ecu.edu 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! Key Formulas Population Mean: = Sample Mean: = Overall (Weighted) Mean: = 1 + 2 1 +2 Mean Question 1: Find the mean for the sample of n=5 scores: 1, 8, 7, 5, 9 Mean Question 1 Answer: M=ΣX n M = 1+8+7+5+9 5 M = 30 5 M=6 Mean Question 2: A sample of n=6 scores has a mean of M=8. What is the value of Σ X for this sample? Mean Question 2 Answer: M=ΣX n 8=ΣX 6 Σ X = 48 Mean 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 Question 3 Answer: 1 + 2 1 +2 20+30 5+3 50 8 = = = = 6.25 Remember: 1 is calculated by 1 ∗ 1 2 is calculated by 2 ∗ 2 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? a) b) It must be greater than 40. It must be less than 40. Mean Question 4 Answer: B) It must be less than 40. A score higher than 40 would have increased the mean. Mean Question 5: Find the values for n, Σ X, and M for the following sample: X f 5 1 4 2 3 3 2 5 1 1 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 Mean Question 6: Adding a new score to a distribution always changes the mean. True or False? Mean Question 6 Answer: False. If the score is equal to the mean, it does not change the mean. Mean Question 7: A population has a mean of μ = 40. a) b) 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 Question 7 Answer: a) The new mean would be 45. When a constant is added to every score, the same constant is added to the mean. b) 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 8 Answer: μ=7 Mean Question 9: Using the scores from question 8, fill in the following table. Mean Question 9 Answer: 4 Below 1 Below 1 Below 2 4 Above Above Median Question 10: Find the median for each distribution of scores: a) b) 3, 4, 6, 7, 9, 10, 11 8, 10, 11, 12, 14, 15 Median Question 10 Answers: a) b) The median is X = 7 The median is X = 11.5 Median 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 Question 11 Answer: 2/3 6 1/3 5 Count 8 boxes 1 2 3 4 5 7 4 2 6 3 1 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 onethird 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 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. 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) Selecting a Measure of Central Tendency Question 13 Answer: Mean Selecting a Measure of Central Tendency Question 14: Why is it usually inappropriate to compute a mean for scores measured on an ordinal scale? Selecting a Measure of Central Tendency Question 14 Answer: The definition of the mean is based on distances (the mean balances the distances) and ordinal scales do no measure distance. Selecting a Measure of Central Tendency Question 15: In a perfectly symmetrical distribution, the mean, the median, and the mode will all have the same value. (True or False) Selecting a Measure of Central Tendency Question 15 Answer: False, if the distribution is bimodal. Selecting a Measure of Central Tendency 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 Question 16 Answer: 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 Notice how the means follow the outliers Note: Median usually falls between mean and mode. Frequently Asked Questions Interpolation Real Limits Median for Continuous Variables Frequency Distribution Cumulative Distributions Weighted Mean Frequently Asked Question FAQs How do I find the median for a continuous variable? Frequently Asked Questions FAQs Step 1: Count the total number of boxes. 50% of 16 is 8. 1 2 3 4 14 16 boxes 10 13 5 7 9 12 16 6 8 11 15 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). Frequently Asked Questions FAQs 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%. Frequently Asked Questions FAQs Median = 6.83 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. Frequently Asked Questions FAQs Frequently Asked Questions FAQs How do I calculate the weighted mean? Compute the weighted mean for two sets of scores. The first set consists of n = 12 scores and has a mean of M = 6. The second set consists of n = 8 scores and has a mean of M = 7. Step 1: Find for both samples. 1 = 12 ∗ 6 = 72 2 = 8 ∗ 7 = 56 Plug numbers into the formula = 72+56 12+8 128 20 = = = 6.4 1 + 2 1 +2