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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.
 Winter '08
 TEXTBOOK

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