PSY1101_Lecture3_Jan16and17

PSY1101_Lecture3_Jan16and17 - Experimentation...

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Unformatted text preview: Experimentation Experimentation Correlations will show us where a causal relationship possibly exists Breastfeeding and cognitive abilities Breastfeeding could cause higher cognitive abilities Later cognitive abilities could cause breastfeeding A third factor causes them both (e.g., socioeconomic status) 1) Breast Feeding Higher Cognitive Ability OR 2) Higher Cognitive Ability Breast Feeding OR Breast Feeding 3) Socioeconomic Status Higher Cognitive Skills Experimentation Experimentation Two ways an experiment differs from correlational research Placebo Placebo effect 1) Manipulate the factor in question 2) Random assignment Inert An improvement that is elicited by the expectation or anticipation of getting better Independent and Dependent Variables Statistics Statistics Data/Observations How many hours did you spend studying for the quiz? 40 students 6, 4, 3, 7, 6, 5, 5, 9, 8, 2, 7, 4, 6, 8, 7, 3, 4, 5, 6, 10, 5, 8, 5, 6, 4, 1, 3, 2, 6, 7, 5, 9, 8, 6, 4, 3, 5, 9, 7, 6 Frequency Distribution Frequency Distribution Frequency Distribution Table Shows the number of observations that fall into each category Hours Studied Frequency 1 1 2 2 3 4 4 5 5 7 6 8 7 5 8 4 9 3 10 1 Frequency Distribution Frequency Distribution • Frequency Distribution Graph Hours Studied Frequency 1 1 2 2 3 4 4 5 5 7 6 8 7 5 8 4 9 3 10 1 • Histogram Measures of Central Tendency Measures of Central Tendency Mean= 平平平 Arithmetic average Add all the scores together divide by the number of scores Most commonly reported measure of central tendency 6 + 4 + 3 + 7 + 6 + 5 + 5 + 9 + 8 + 2 + 7 + 4 + 6 + 8 + 7 + 3 + 4 + 5 + 6 + 10 + 5 + 8 + 5 + 6 + 4 + 1 + 3 + 2 + 6 + 7 + 5 + 9 + 8 + 6 + 4 + 3 + 5 + 9 + 7 + 6 = 224 224 ÷ 40 = 5.6 The mean number of hours studied is 5.6 hours. Measures of Central Tendency Measures of Central Tendency Median= 平平平 The midpoint of the data Put all data points in order from smallest to largest, and the data point in the middle is the median If there is an even number of data points, there will be no middle data point In that case, the median is the arithmetic average of the two points that are in the middle 6, 4, 3, 7, 6, 5, 5, 9, 8, 2, 7, 4, 6, 8, 7, 3, 4, 5, 6, 10, 5, 8, 5, 6, 4, 1, 3, 2, 6, 7, 5, 9, 8, 6, 4, 3, 5, 9, 7, 6 1, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 10 The 20th and the 21st data points will be the two in the middle = 6 and 6 Arithmetic mean of 6 and 6 = (6 + 6) ÷ 2 = 6 The median is 6 Measures of Central Tendency Measures of Central Tendency Mode 平平平平平平平 The most frequently occurring score The mode is 6 Measures of Central Tendency Mean = 5.6 Median = 6 Mode = 6 Hours Studied Frequency 1 1 2 2 3 4 4 5 5 7 6 8 7 5 8 4 9 3 10 1 Measures of Variation Measures of Variation again, calculate a single value that will represent the group a value that best represent the degree to which the scores vary the amount of difference amongst the scores Variability Variability 10m 10m 10m 10m 10m 10m 10m 10m #1 10m 12.5m 12.5m 7.5m 7.5m 10m 12.5m 7.5m Mean = 10m Mean = 10m #2 17.5m 15m 10m #3 5m 15m Mean = 7.5m 10m 7.5m 2.5m Measures of Variation Measures of Variation Range this is the gap/distance between the lowest and highest scores Subtract the lowest score from the highest score Standard Deviation Represents the degree to which the scores vary An estimation of the “average” difference amongst scores #1 10 – 10 = 0 #2 12.5 – 7.5 = 5 #3 17.5 – 2.5 = 15 Standard Deviation Standard Deviation 10m #2 12.5m 12.5m 7.5m Mean Individual Deviation Squared Scores From Mean Deviation 10 – 10 = 0 0 10 – 12.5 = ­2.5 6.25 10 – 7.5 = 2.5 6.25 10 – 12.5 = ­2.5 6.25 10 – 7.5 = 2.5 6.25 10 – 10 = 0 0 10 – 12.5 = ­2.5 6.25 10 – 7.5 = 2.5 6.25 7.5m 10m 12.5m 7.5m Mean = 10m 1) Calculate sample mean 2) Obtain individual score’s deviations from mean 3) Square the deviations 4) Add up all the squared deviations and divide by the number of scores 5) Take the square root of this number • 0 + 6.25 + 6.25 + 6.25 + 6.25 + 0 + 6.25 + 6.25 = 37.5 • 37.5 ÷ 8 = 4.7 • √4.7 = 2.2 • The standard deviation is 2.2 Standard Deviation Standard Deviation Mean Individual Deviation Squared Scores From Mean Deviation 10 – 10 = 0 0 10 – 17.5 = ­7.5 56.3 10 – 5 = 5 25.0 10 – 15 = ­5 25.0 10 – 7.5 = 2.5 6.25 10 – 2.5 = 7.5 56.3 10 – 15 = ­5 25.0 10 – 7.5 = 2.5 6.25 17.5m • 0 + 56.3 + 25.0 + 25.0 + 6.25 + 56.3 + 25.0 + 6.25 = 200.1 • 200.1 ÷ 8 = 25.0 • √25.0 = 5.0 • The standard deviation is 5.0 15m 10m #3 1) Calculate sample mean 2) Obtain individual score’s deviations from mean 3) Square the deviations 4) Add up all the squared deviations and divide by the number of scores 5) Take the square root of this number 5m 15m Mean = 7.5m 10m 7.5m 2.5m Standard Deviation Standard Deviation 10m 10m 10m 10m 10m 10m 10m 10m #1 Mean Individual Deviation Squared Scores From Mean Deviations 10 – 10 = 0 0 10 – 10 = 0 0 10 – 10 = 0 0 10 – 10 = 0 0 10 – 10 = 0 0 10 – 10 = 0 0 10 – 10 = 0 0 10 – 10 = 0 0 Mean = 10m 1) Calculate sample mean 2) Obtain individual score’s deviations from mean 3) Square the deviations 4) Add up all the squared deviations and divide by the number of scores 5) Take the square root of this number • 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 = 0 • 0 ÷ 8 = 0 • √0 = 0 • The standard deviation is 0 Variability Variability 10m 10m 10m 10m 10m 10m 10m 10m #1 Mean = 10m SD = 0 10m 12.5m 12.5m 7.5m 7.5m 10m 12.5m 7.5m #2 SD = 2.2 17.5m 15m 10m #3 Mean = 10m 5m 15m Mean = 7.5m 10m 7.5m 2.5m SD = 5.0 Normal Distribution Frequency 55 70 85 100 115 Wechsler Intelligence Score 130 145 The Normal Curve The Normal Curve Using Standard Deviation Using Standard Deviation Test #1 Your score: 75% Class average: 65% Standard Deviation: 10 You scored one SD above the mean You scored better than 84% of the class Test #2 Your score: 75% Class average: 65% Standard Deviation: 20 You scored one half of a SD above the mean You scored better than 67% of the class Statistical Significance Statistical Significance Usually looking at differences between groups A statistical test will determine if the difference is significant Taking into account the means (mean difference) and the standard deviation A difference is more likely to be found to be statistically significant if: 1) the difference is reliable i.e., likely to get the same result if repeat the procedure Sample is representative There is low variability Many participants 2) if the difference is large Statistical test will evaluate the probability of getting a difference that large If it is less than 5% probability, it is statistically significant Statistical significance does not mean practically significant ...
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This note was uploaded on 03/02/2012 for the course PSY 1101 taught by Professor Textbook during the Winter '08 term at University of Ottawa.

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