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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**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**• Question 1 Answer: • M =ΣX n • M =1+8+7+5+9 5 • M =30 5 • M = 6**Mean**• Question 2 Answer: • 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**• 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**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**• 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.**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.**Mean**• Question 8 Answer: • μ = 7 • Question 9 Answer: 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**• 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**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**• 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) • 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**• 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.**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**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. • 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**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**• 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