Presentation is loading. Please wait.

Presentation is loading. Please wait.

Preview Bem, 2011, claimed that through nine experiments he had demonstrated the existence of precognition Failure to replicate: “Across seven experiments.

Similar presentations


Presentation on theme: "Preview Bem, 2011, claimed that through nine experiments he had demonstrated the existence of precognition Failure to replicate: “Across seven experiments."— Presentation transcript:

1 Preview Bem, 2011, claimed that through nine experiments he had demonstrated the existence of precognition Failure to replicate: “Across seven experiments (N= 3,289), we replicate the procedure of Experiments 8 and 9 from Bem (2011), which had originally demonstrated retroactive facilitation of recall. We failed to replicate that finding.” Galek et. al. 2012 Sampling error, something unique about the participants in the Bem experiments. Related to inter-individual variability Research methodology

2 Inter- and Intra-individual Differences
“Among the most striking characteristics of human cognition is its variability, which is present both between people (inter-individual variability) and within a given person (intra-individual variability).” Siegler 2006 Intra-individual variability (within a person) an indicator of the functioning of the central nervous system inter-individual variability (between people) Source of sampling error Example: Reaction Time Test experiments Record a series of reaction times

3 Example data from reaction time test

4 Variability A measure of how spread out the scores are in a distribution Usually accompanies a measure of central tendency as basic descriptive statistics for a set of scores As a descriptive measure variability measures the degree to which the scores are spread out or clustered together in a distribution. As an important component of most inferential statistics variability provides a measure of how accurately any individual score or sample represents the entire population

5 Figure 4-1 Population distributions of adult heights and adult weights.

6 Selective breeding for taste preference
Selective breeding of dogs or cats for coat color, body size and behavior has occurred for centuries. Rats selectively bred for low versus high saccharin drinking. Low-saccharin-consuming (LoS) High-saccharin-consuming (HiS) Compare taste preference of male and female rats from HiS and LoS lines. Values in the table are phenotype scores which is taste preference adjusted for baseline water and body weight.

7 Measuring Variability
Variability can be measured with range is the total distance covered by the distribution, from the highest score to the lowest score variance average square distance from the mean standard deviation measures the standard distance between a score and the mean In each case, variability is determined by measuring distance.

8 The Range The range is the total distance covered by the distribution
The range tells us the number of measurement categories. from the highest score (MAX) to the lowest score (MIN) For continuous variables Upper Real Limit for Xmax – LRL for Xmin Scores from 1 to 5 ; 5.5 – 0.5 = 5 Alternative definition of range: When scores are whole numbers or discrete variables with numerical scores, Xmax – Xmin + 1 Scores from 0 to – = 5 8

9 The Range Commonly the range can be defined as the difference between the largest score (max) and the smallest score (min). Used by SPSS computer program Scores from 1 to = 4 OK for discrete but not continuous variables Which calculation is used usually does not matter Range usually is not an accurate measurement of variability. Completely determined by two scores (max – min) An extreme large or small score will inflate the range Can be used as an adjunctive measure with variance

10 Interquartile and Semi-interquartile Range
Interquartile range: of 16 numbers from 2 to 11 Range covered by the middle 50% of the distribution = (Q3–Q1) Q3 is the 75th percentile, Q1 is the 25th percentile Semi-interquartile range: half the interquartile range = (Q3–Q1) / 2 less likely to be influenced by extreme scores Note: not covered in the textbook

11 Deviation Scores Deviation is distance from the mean:
deviation score = X - µ If the mean on a quiz is 5 and your score on the quiz is 10 then your deviation from the mean is – 5 = +5 Every student would have a deviation score, their deviation from the mean. +5, +1, +1, -3, -4

12 Average of deviation scores is not useful
Each deviation score is X - µ The sum of the deviation scores is ∑(X-m) The average of all these deviation scores is ∑(X-m) /N However this calculation is not usable ∑(X-m) is always zero 12

13 Calculating Sum of Squares.
It is identified by the notation SS (for sum of squared deviations) Example 4.2 Population of N = 5 scores. Calculate the mean then Calculate deviation from the mean Calculate Squared Deviation then Sum of Squared Deviations ∑ X = 30 m = (∑ X)/N m = 30/5 = 6 ∑ (X – m) = 0 SS = ∑(X-m)2 = 40

14 Figure 4. 3 from example data 4
Figure from example data 4.2 A frequency distribution histogram for a population of N = 5 scores. The mean for this population is µ = 6. The smallest distance from the mean is 1 point, and the largest distance is 5 points. The standard distance (or standard deviation) should be between 1 and 5 points. See example 4.2 for calculation table Sum of Squares SS = ∑(X-m)2 = 40

15 Figure 4.2 The calculations of variance and standard deviation

16 Variance and Standard Deviation
Variance equals the mean of the squared deviations. Variance is the average squared distance from the mean. Standard deviation is the square root of the variance and provides a measure of the standard, or average distance from the mean.

17 Calculation of Standard Deviation for a Population
For example 4.2 Calculate mean m = (∑ X)/N m = 30/5 = 6 Calculate (SS) ∑(X-m)2 = 40 Compute variance which is mean square calculate mean of the squared deviation SS divided by N Variance s2 = SS/N = 40/5 = 8 Compute Standard Deviation Standard deviation is the square root of variance Standard deviation s = √8 = 2.83

18 Figure 4. 3 from example data 4
Figure 4.3 from example data 4.2 A frequency distribution histogram for a population of N = 5 scores. The mean for this population is µ = 6. The smallest distance from the mean is 1 point, and the largest distance is 5 points. The standard distance (or standard deviation) should be between 1 and 5 points. Sum of Squares SS = ∑(X-m)2 = 40 Variance s2 = SS/N = 40/5 = 8 Standard deviation s = √ s2 = √8 = 2.83

19 Population Variance and Standard Deviation Formulas
Definitional formulas Sum of Squares SS = ∑(X-m)2 Population Variance s2 = SS/N Population Standard deviation s = √ SS/N Computational formulas Sum of Squares Easier with a hand calculator (do not need to know for exam) Using example 4.2 x 1 9 5 8 7 x2 1 81 25 64 49 SS = /5 SS = SS = 40

20 Relationship with Other Statistical Measures
Variance and standard deviation are mathematically related to the mean. They are computed from the squared deviation scores (squared distance of each score from the mean). Median and semi-interquartile range are both based on percentiles and therefore are used together. When the median is used to report central tendency, semi-interquartile range is often used to report variability. Range has no direct relationship to any other statistical measure.

21 Population Variability
Effects on sampling When the population variability is small “homogeneous” all of the scores are clustered close together individual score or sample will necessarily provide a good representation of the entire set. When population variability is large “heterogeneous” scores are widely spread one or two extreme scores can give a distorted picture of the general population

22 Figure 4-4 (p. 114) The population of adult heights forms a normal distribution. If you select a sample from this population, you are most likely to obtain individuals who are near average in height. As a result, the scores in the sample will be less variable (spread out) than the scores in the population. F R E Q U N C Y 48 inches 84 inches

23 Calculation of Standard Deviation for a Sample
1. Compute the deviation (distance from the mean) for each score. 2. Square each deviation. Compute the mean of the squared deviation called the variance or mean square for samples sum of the squared deviations (SS) dividing by n - 1, rather than N n - 1, is know as degrees of freedom (df) used so that the sample variance will provide an unbiased estimate of the population variance take the square root of the variance to obtain the standard deviation

24 Frequency Distribution Histogram and Standard Deviation
Example 4.6 for a sample of n = 8 scores. The sample mean is M = 6.5. The smallest distance from the mean is 0.5 points, and the largest distance from the mean is 4.5 points. The standard distance (standard deviation) should be between 0.5 and 4.5 points, or about 2.5. FIGURE 4.5 The frequency distribution histogram for a sample of n = 8 scores. The sample mean is M = 6.5. The smallest distance from the mean is 0.5 points, and the largest distance from the mean is 4.5 points. The standard distance (standard deviation) should be between 0.5 and 4.5 points, or about 2.5. Sum of Squares SS = ∑(X-M)2 = see next slide for calculation of SS Sample Variance s2 = SS/n-1 = 48/7 = 6.86 Sample Standard Deviation s = √ s2 = √ = 2.62 APA uses SD for standard deviation so it would be SD = 2.62 Copyright © 2017 Cengage Learning. All Rights Reserved.

25 Example 4.6 showing deviation scores and squared deviation in a table for the sample of n = 8 scores
M =(∑ X)/n M = 52/8 = 6.5 ∑ (X – M) = 0 ∑(X-M)2 = 48

26 Degrees of Freedom (df)
Sample variability is biased, tends to under report population variability Using n – 1 corrects for that bias Sample variance is an unbiased estimate of population variance For a single sample of size n; degrees of freedom is n-1 Used as the denominator when calculating sample variance s2 = SS/n-1 BTW: Degrees of freedom refers to number of scores that are free to vary For a sample of n = 3 and a M = 5, where the first two scores are X = 4 and X = 5, the third score must be? X = 6, it has to be, i.e. has no freedom to vary So when n = 3 there are only 2 degrees of freedom


Download ppt "Preview Bem, 2011, claimed that through nine experiments he had demonstrated the existence of precognition Failure to replicate: “Across seven experiments."

Similar presentations


Ads by Google