COURSE: JUST 3900 - PowerPoint Presentation

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COURSE: JUST 3900TIPS FOR APLIADeveloped By: Ethan Cooper (Lead Tutor) John LohmanMichael MattocksAubrey Urwick

Chapter 4:

VariabilitySlide2

Key Terms: Don’t Forget NotecardsVariability (p. 104)Range (p. 106)Deviation (p. 107)Population Variance (p. 108)Standard Deviation (p. 108)Sum of Squares (p. 111)Sample Variance (p. 115)Degrees of Freedom (p. 117)Unbiased (p. 119)Biased (p. 119)Slide3

Think Notecards: For Formulas

Sum of Squares (definitional):

Sum of Squares (computational):

Population Variance:

Population Standard Deviation:

Slide4

Think Notecards: For Formulas Sum of Squares (definitional): Sum of Squares (computational): Sample Variance: Sample Standard Deviation: Slide5

RangeQuestion 1: What is the range for the following set of scores? (Use all 3 definitions)1, 9, 5, 8, 7Slide6

The RangeQuestion 1 Answer:Range = URL for Xmax – LRL for XminRange = 9.5 – 0.5Range = 9Range = Xmax – Xmin + 1Range = 9 – 1 + 1Range = 9Range = Xmax – XminRange = 9 – 1Range = 8Note: This formula is

a

n alternative to using

r

eal limits, but it only

works

for whole numbers

. Do

NOT

use

for

scores with

decimals

!Slide7

Standard Deviation and Variance for a PopulationQuestion 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 4: Calculate the variance and standard deviation for the following population of N = 5 scores: 4, 0, 7, 1, 3.Slide8

Question 2 Answer:Standard deviation measures the standard distance from the mean and variance measures the average squared distance from the mean.Question 3 Answer:Because there is no variability (the scores are all the same), the standard deviation is zero.Standard Deviation and Variance for a PopulationSlide9

Standard Deviation and Variance for a Population X(X - µ)(X - µ)24110-3974161-2

4

3

0

0Slide10

Standard Deviation and Variance for a PopulationFind the sum of the squared deviationsSS = 1 + 9 + 16 + 4 +0 SS = 30Find the varianceσ2 = SS/Nσ2 = 30/5σ2 = 6Find the standard deviationSlide11

Standard Deviation and Variance for a SampleQuestion 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.Question 6: Explain why the formula for sample variance divides SS by n – 1 instead of dividing by n.Slide12

Standard Deviation and Variance for a SampleQuestion 5 Answer:Find the meanM = 3Find dfdf = n - 1df = 5 – 1df = 4Find SS.XX241600

7

49

1

1

3

9

∑X = 15

∑X

2

= 75Slide13

Standard Deviation and Variance for a SampleFind SS.SS = 75 – [(152 )/5]SS = 75 – (225/5)SS = 75 – 45SS = 30Find s2s2 = SS/df = SS/(n-1)s2 = 30/4s2 = 7.5Find sSlide14

Standard Deviation and Variance for a SampleQuestion 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.Slide15

Things to Consider Correct!Incorrect, but will be ananswer choice!Slide16

Things to ConsiderOn 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 canget complicated!

Save yourself some trouble by using the computational formula for SS.Slide17

Transformations of ScaleQuestion 7: A population has a mean of µ = 70 and a standard deviation of σ = 5.If 10 points were added to every score in the population, what would be the new values for the population mean and standard deviation?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?Slide18

Transformations of ScaleQuestion 7 Answer: The new mean would be µ = 80 but the standard deviation would still be σ = 5.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 remainsconstant. Thus, the standard deviationremains 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.Slide19

Transformations of ScaleQuestion 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.X = 87

X = 87

3

X = 88

X = 83

X = 86Slide20

Frequently Asked Questions: FAQs Slide21

Frequently Asked Questions: FAQsHow do I use the definitional formula when calculating sum of squares?The best way to use this formula is to create a chart:Find SS for the following population: 4, 0, 7, 1, 3µ =ΣX/N = 15/5 = 3X(X - µ)(X - µ)244 - 3 = 112 = 100 – 3 = (-3)(-3)2 = 97

7 – 3 = 4

4

2

= 16

1

1 – 3 = (-2)

(-2)

2

=

4

3

3 – 3 = 0

02 = 0

 Slide22

Frequently Asked Questions: FAQsHow do I use the computational formula when calculating sum of squares?The best way to use this formula is to create a chart:Find SS for the following sample: 4, 0, 7, 1, 3XX2416007491139

∑X = 15

∑X

2

= 75Slide23

Frequently Asked Questions: FAQs