PSYC 2317 Mark W. Tengler, M.S. Assignment #3 3.1 A population has a mean of µ = 52 and a standard deviation of σ = 5. a. For this population, find the z-score corresponding to each of the following scores. Plot each point on a graph. b. 3.2 X = 55 X = 40 X = 35 X = 48 X = 70 X = 65 For the same population, find the score (X value) corresponding to each of the following z-scores. Plot each Z score on a graph. Z = -2.00 Z = 1.50 Z = -0.50 Z = 0.60 Z = 1.00 Z = 0.00 A distribution with a mean of µ = 69 and a standard deviation of σ = 7 is being transformed into a standardized distribution of µ = 112 and σ = 10. Find the new, standardized score for each of the following values from the original population (plot each point on a graph): a. X = 80 b. X = 70 c. X = 65 d. X = 87 Transforming the Scores from One Scoring System to Another Using Z Scores Original Scoring System SAT Scores σ = 100 X = 730 New Scoring System: ACT Scores σ=3 µ = 500 µ = 18 Z= X-µ σ X = μ + Zσ When converting from one scoring system to another, the position of the two scores is The same (i.e. the same Z score). X=? Descriptive Statistics Formula Sheet Sample Population Characteristic statistic Parameter raw scores x, y, . . . . . X, Y, . . . . . mean (central tendency) M= μ= range (interval/ratio data) ∑x n highest minus lowest value deviation (distance from mean) Deviation = (x − M ) Deviation = (X − μ ) average deviation (average ∑(x − M ) =0 n ∑(X − μ ) N distance from mean) sum of the squares (SS) (computational formula) variance ( average deviation2 or standard deviation2) (computational formula) (∑ x)2 n (∑ x)2 ∑ x2 − n = SS s2 = n−1 df SS = ∑ x 2 − standard deviation (average deviation or distance from mean) (computational formula) Z scores (standard scores) mean = 0 standard deviation = ± 1.0 Area Under the Normal Curve s= Z= ∑X N highest minus lowest value (∑ X)2 N (∑ X)2 ∑ X2 − N σ2 = N SS = ∑ X 2 − (∑ x)2 n n−1 σ= x−M deviation = s stand. dev. Z= 2 √∑ x − X = M + Zs (∑ X)2 N N 2 √∑ X − X−μ σ X = μ + Zσ -1s to +1s = 68.3% -2s to +2s = 95.4% -3s to +3s = 99.7% ~ lA' N t(J \ Z Scores ) r--~~~ ~~~~~~----~~==---------------------1= I ,.,o,z.- ll5'ai1~ ;} 1 = 1 ID 111012. - /1')341·" { r .;_____/ { ..-J t:!~ I ~ \._____./ .._.; ; JAN KJ \ I ~ '1 ~--'-.__/ ~-=-~ 2 ~ JLbtlJ: ~~---f---..L: W:. . ·. L.. . . .t. . .--"-DY1V/r+~llny ~ -~~~-- scuri ng ryrttm tv tnt J±ttnli6\ui. r-...J~~- _8ill !lefu fD ttu z -J'coru_pr~LL)WLhlllYJLJbL-t+) ,--' = Brittany’s notes - 2/2/17 Lecture 3: Z Scores  Review: The Scientific Method  Everything has to be: a. OBSERVABLE: 5 senses b. MEASURABLE: attach numbers  Nominal: categorical (eg. Yes/no)  Ordinal: ranking (eg. On a scale of 1 to 10, how do you feel?); doesn’t give equal intervals (b/c it’s often used for subjective data)  Discreet data: only WHOLE numbers  Interval/Ratio: requires some sort of technology that will give you equal intervals  Continuous data: in between numbers (eg.: decimals) c. REPEATABLE: to ensure it didn’t happen by chance  Why do we use the scientific method? To learn about POPULATIONS a. Knowledge that can be generalized and isn’t just applicable to ONE person  How do we do that? Collect SAMPLES that represent the population a. Why do we use statistics?  Descriptive Statistics: to tell us what the sample data LOOKS like  Inferential Statistics: to tell us how it applies back to the population  Tools of DESCRIPTIVE statistics: a. Frequencies: when you want to look at the WHOLE data  How often does something happen?  Looking for patterns in the data b. Summarize: organize the data into GROUPS  Independent variable: one that’s being MANIPULATED  Dependent variable: one that’s being MEASURED  Nominal, Ordinal, Interval/Ratio  2 ways to summarize:  Central Tendency: calculating ONE score that represents the whole group; the MIDDLE of the group (tends toward the center); measuring how SIMILAR everyone is  Nominal data: calculate mode  Ordinal data: calculate median  Interval/Ratio data: calculate mean  Variability: how DIVERSE everyone is  Calculate standard deviation: how different the scores are from a reference point (central tendency: mean, median, or mode depending on the type of data) c. Location: where a score falls in the data  Calculate: Z score  What is the Bell/Normal Curve?  What data generally looks like on the POPULATION level Brittany’s notes - 2/2/17  Z Score  What: Where a score falls (location) on the normal/bell curve distribution; tell you what % are above & below the score a. Counting how many standard deviations the score falls about or below the MEAN  Why? To standardize different scoring systems (eg. SAT/ACT) so that they can be compared Class Demo Example 1: A standardized test of aptitude scores has a mean (μ) of 60 and a standard (σ) deviation of 7. What is the Z score for the following test scores? a. X = 65 b. X = 72 c. X = 81 d. X = 44 Step 1: “standardized” tells you what equation and symbols to use  population: Z= Step 2: Draw normal curve graph and plot Xs X = 65 X = 44 σ=7 X = 72 X = 81 39 46 53 Step 3: plug Xs into equation and solve a. Z = b. Z = c. Z = d. Z = 65−60 7 72−60 7 81−60 7 44−60 7 5 = = .71 = = = 7 12 7 21 = 1.71 =3 7 −16 7 = −2.29 60 67 74 81 X− μ σ Brittany’s notes - 2/2/17 Step 4: Graph each Z score on normal curve a. b. etc…. Class Demo Example 2: A standardized IQ test has a mean (μ) of 100 and a standard deviation (σ) of 15. What IQ would someone have for the following Z scores? a. Z = 1.0 b. Z = 0 c. Z = -.5 d. Z = 1.5 Step 1: “standardized” tells you what equation and symbols to us  population: X = μ + Zσ Step 2: Plug Zs into equation and solve a. X = 100 + (1 ∗ 15) = 100 + 15 = 115 b. X = 100 + (0 ∗ 15) = 100 + 0 = 100 c. X = 100 + (−.5 ∗ 15) = 100 − 7.5 = 92.5 d. X = 100 + (1.5 ∗ 15) = 100 + 22.5 = 122.5
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