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COURSE: JUST 3900 TIPS FOR APLIA Developed By: Ethan Cooper (Lead Tutor) John Lohman Michael Mattocks Aubrey Urwick Chapter 4: Variability.

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Presentation on theme: "COURSE: JUST 3900 TIPS FOR APLIA Developed By: Ethan Cooper (Lead Tutor) John Lohman Michael Mattocks Aubrey Urwick Chapter 4: Variability."— Presentation transcript:

1 COURSE: JUST 3900 TIPS FOR APLIA Developed By: Ethan Cooper (Lead Tutor) John Lohman Michael Mattocks Aubrey Urwick Chapter 4: Variability

2 Key Terms: Don’t Forget Notecards Variability (p. 104) Variability (p. 104) Range (p. 106) Range (p. 106) Deviation (p. 107) Deviation (p. 107) Population Variance (p. 108) Population Variance (p. 108) Standard Deviation (p. 108) Standard Deviation (p. 108) Sum of Squares (p. 111) Sum of Squares (p. 111) Sample Variance (p. 115) Sample Variance (p. 115) Degrees of Freedom (p. 117) Degrees of Freedom (p. 117) Unbiased (p. 119) Unbiased (p. 119) Biased (p. 119) Biased (p. 119)

3 Think Notecards: For Formulas Think Notecards: For Formulas Sum of Squares (definitional): Sum of Squares (computational): Population Variance: Population Standard Deviation:

4 Think Notecards: For Formulas Think Notecards: For Formulas Sum of Squares (definitional): Sum of Squares (computational): Sample Variance: Sample Standard Deviation:

5 Range Question 1: What is the range for the following set of scores? (Use all 3 definitions) Question 1: What is the range for the following set of scores? (Use all 3 definitions) 1, 9, 5, 8, 7 1, 9, 5, 8, 7

6 The Range Question 1 Answer: Question 1 Answer: 1. Range = URL for X max – LRL for X min Range = 9.5 – 0.5 Range = 9.5 – 0.5 Range = 9 Range = 9 2. Range = X max – X min + 1 Range = 9 – Range = 9 – Range = 9 Range = 9 3. Range = X max – X min Range = 9 – 1 Range = 9 – 1 Range = 8 Range = 8 Note: This formula is an alternative to using real limits, but it only works for whole numbers. Do NOT use for scores with decimals!

7 Standard Deviation and Variance for a Population Question 2: Briefly explain what is meant by the standard deviation and what is measured by the variance. Question 2: Briefly explain what is meant by the standard deviation and what is measured by the variance. Question 3: What is the standard deviation for the following set of N = 5 scores: 10, 10, 10, 10, and 10? (Note: This can be done without using any calculations) Question 3: What is the standard deviation for the following set of N = 5 scores: 10, 10, 10, 10, and 10? (Note: This can be done without using any calculations) Question 4: Calculate the variance and standard deviation for the following population of N = 5 scores: 4, 0, 7, 1, 3. Question 4: Calculate the variance and standard deviation for the following population of N = 5 scores: 4, 0, 7, 1, 3.

8 Question 2 Answer: Question 2 Answer: Standard deviation measures the standard distance from the mean and variance measures the average squared distance from the mean. Standard deviation measures the standard distance from the mean and variance measures the average squared distance from the mean. Question 3 Answer: Question 3 Answer: Because there is no variability (the scores are all the same), the standard deviation is zero. Because there is no variability (the scores are all the same), the standard deviation is zero. Standard Deviation and Variance for a Population

9 X(X - µ)(X - µ)

10 Standard Deviation and Variance for a Population 3. Find the sum of the squared deviations SS = SS = SS = 30 SS = Find the variance σ 2 = SS/N σ 2 = SS/N σ 2 = 30/5 σ 2 = 30/5 σ 2 = 6 σ 2 = 6 5. Find the standard deviation

11 Standard Deviation and Variance for a Sample Question 5: Calculate the variance and standard deviation for the following sample of N = 5 scores: 4, 0, 7, 1, 3. (Use the computational formula for SS) Question 5: Calculate the variance and standard deviation for the following sample of N = 5 scores: 4, 0, 7, 1, 3. (Use the computational formula for SS) Note: Although the mean is a whole number, please use the computational formula so that you can see that the formulas are mathematical equivalents. However, on the test and on Aplia, use the definitional formula when dealing with whole number means and the computational formula with means containing decimals. This will save lots of time. Note: Although the mean is a whole number, please use the computational formula so that you can see that the formulas are mathematical equivalents. However, on the test and on Aplia, use the definitional formula when dealing with whole number means and the computational formula with means containing decimals. This will save lots of time. Question 6: Explain why the formula for sample variance divides SS by n – 1 instead of dividing by n. Question 6: Explain why the formula for sample variance divides SS by n – 1 instead of dividing by n.

12 Standard Deviation and Variance for a Sample Question 5 Answer: Question 5 Answer: 1. Find the mean M = 3 M = 3 2. Find df df = n - 1 df = n - 1 df = 5 – 1 df = 5 – 1 df = 4 df = 4 3. Find SS. XX2X ∑X = 15 ∑X 2 = 75

13 Standard Deviation and Variance for a Sample 3. Find SS. SS = 75 – [(15 2 )/5] SS = 75 – [(15 2 )/5] SS = 75 – (225/5) SS = 75 – (225/5) SS = 75 – 45 SS = 75 – 45 SS = 30 SS = Find s 2 s 2 = SS/df = SS/(n-1) s 2 = SS/df = SS/(n-1) s 2 = 30/4 s 2 = 30/4 s 2 = 7.5 s 2 = Find s

14 Standard Deviation and Variance for a Sample Question 6 Answer: Question 6 Answer: Without some correction, sample variability consistently underestimates the population variability. Dividing by a small number (n – 1 instead of n) increases the value of the sample variance and makes it an unbiased estimate of the population variance. Without some correction, sample variability consistently underestimates the population variability. Dividing by a small number (n – 1 instead of n) increases the value of the sample variance and makes it an unbiased estimate of the population variance.

15 Things to Consider Correct! Incorrect, but will be an answer choice!

16 Things to Consider On this week’s Aplia assignment, Aplia asks you to calculate SS using the definitional formula for a set of scores in which the mean has 3 decimal places. This makes the ensuing calculations more complicated. Use the computational formula to avoid this difficulty. On this week’s Aplia assignment, Aplia asks you to calculate SS using the definitional formula for a set of scores in which the mean has 3 decimal places. This makes the ensuing calculations more complicated. Use the computational formula to avoid this difficulty. These calculations can get complicated! Save yourself some trouble by using the computational formula for SS.

17 Transformations of Scale Question 7: A population has a mean of µ = 70 and a standard deviation of σ = 5. Question 7: A population has a mean of µ = 70 and a standard deviation of σ = 5. a) If 10 points were added to every score in the population, what would be the new values for the population mean and standard deviation? b) If every score were in the population were multiplied by 2, what would be the new values for the population mean and standard deviation? Question 8: In a population with a mean of µ = 80 and a standard deviation of σ = 8, would a score of X = 87 be considered an extreme value (far out in the tail of the distribution)? What if the standard deviation were σ = 3? Question 8: In a population with a mean of µ = 80 and a standard deviation of σ = 8, would a score of X = 87 be considered an extreme value (far out in the tail of the distribution)? What if the standard deviation were σ = 3?

18 Transformations of Scale Question 7 Answer: Question 7 Answer: a) The new mean would be µ = 80 but the standard deviation would still be σ = 5. b) The new mean would be µ = 140 and the new standard deviation would be σ = 10. In part a),the distribution moves, but the distance between scores remains constant. Thus, the standard deviation remains the same. In part b), the distribution moves, and the distance between scores is doubled. Thus, the standard deviation is also doubled. Ex: Suppose that our original distribution contained scores of X = 71 and X = 72. After adding 10 points to every score, these scores would become X = 81 and X = 82, respectively. While the scores themselves have changed, the distance between them remained the same, one point. Thus, the standard deviation remains σ = 5. Ex: Again, suppose our distribution contained scores of X = 71 and X = 72. If we multiplied every score by 2, these scores would become X = 142 and X = 144, respectively. In this instance, the distance between scores has increased from 1 point to 2 points. This increase in variability increases the standard deviation from σ = 5 to σ = 10.

19 Transformations of Scale Question 8 Answers: Question 8 Answers: With σ = 8, a score of X = 87 would be located in the central section of the distribution (within one standard deviation). With σ = 3, a score of X = 87 would be an extreme value, located more than two standard deviations above the mean. With σ = 8, a score of X = 87 would be located in the central section of the distribution (within one standard deviation). With σ = 3, a score of X = 87 would be an extreme value, located more than two standard deviations above the mean. X = 87 3 X = 88 X = 83 X = 86

20 Frequently Asked Questions: FAQs

21 Frequently Asked Questions: FAQs How do I use the definitional formula when calculating sum of squares? How do I use the definitional formula when calculating sum of squares? The best way to use this formula is to create a chart: The best way to use this formula is to create a chart: Find SS for the following population: 4, 0, 7, 1, 3 Find SS for the following population: 4, 0, 7, 1, 3 µ =ΣX/N = 15/5 = 3 µ =ΣX/N = 15/5 = 3 X(X - µ)(X - µ) = 11 2 = 1 00 – 3 = (-3)(-3) 2 = 9 77 – 3 = 44 2 = – 3 = (-2)(-2) 2 = 4 33 – 3 = 00 2 = 0

22 Frequently Asked Questions: FAQs How do I use the computational formula when calculating sum of squares? How do I use the computational formula when calculating sum of squares? The best way to use this formula is to create a chart: The best way to use this formula is to create a chart: Find SS for the following sample: 4, 0, 7, 1, 3 Find SS for the following sample: 4, 0, 7, 1, 3 XX2X ∑X = 15 ∑X 2 = 75

23 Frequently Asked Questions: FAQs


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