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# Basic stat review

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### Basic stat review

1. 1. Level of Measurement: Nominal, Ordinal, Ratio, Interval Frequency tables to distributions Central Tendency: Mode, Median, MeanDispersion: Variance, Standard Deviation
2. 2.  Nominal – categorical data ; numbers represent labels or categories of data but have no quantitative value e.g. demographic data Ordinal – ranked/ordered data ; numbers rank or order people or objects based on the attribute being measured Interval – points on a scale are an equal distance apart ; scale does not contain an absolute zero point Ratio – points on a scale are an equal distance apart and there is an absolute zero point
3. 3.  Simple depiction of all the data Graphic — easy to understand Problems  Not always precisely measured  Not summarized in one number or datum
4. 4. Observation Frequency65 170 275 380 485 390 295 1
5. 5. 4 3Frequency 2 1 65 70 75 80 85 90 95 Test Score
6. 6. Two key characteristics of a frequency distribution are especially important when summarizing data or when making a prediction from one set of results to another: Central Tendency  What is in the “Middle”?  What is most common?  What would we use to predict? Dispersion  How Spread out is the distribution?  What Shape is it?
7. 7. Three measures of central tendency are commonly used in statistical analysis - the mode, the median, and the meanEach measure is designed to represent a typical scoreThe choice of which measure to use depends on: the shape of the distribution (whether normal or skewed), and the variable’s “level of measurement” (data are nominal, ordinal or interval).
8. 8.  Nominal variables Mode Ordinal variables Median Ratio/Interval level variables Mean
9. 9.  Most Common Outcome Male Female
10. 10.  Middle-most Value 50% of observations are above the Median, 50% are below it The difference in magnitude between the observations does not matter Therefore, it is not sensitive to outliers
11. 11. • first you rank order the values of X from low to high:  85, 94, 94, 96, 96, 96, 96, 97, 97, 98• then count number of observations = 10.• divide by 2 to get the middle score  the 5 score here 96 is the middle score
12. 12.  Most common measure of central tendency Best for making predictions Applicable under two conditions:1. scores are measured at the interval level, and2. distribution is more or less normal [symmetrical]. Symbolized as:  X for the mean of a sample  μ for the mean of a population
13. 13.  X = (Σ X) / N If X = {3, 5, 10, 4, 3} X = (3 + 5 + 10 + 4 + 3) / 5 = 25 / 5 = 5
14. 14. IF THE DISTRIBUTION IS NORMALMean is the best measure of central tendency  Most scores “bunched up” in middle  Extreme scores less frequent  don’t move mean around. But, central tendency doesn’t tell us everything !
15. 15. Dispersion/Deviation/Spread tells us a lot about how a variable is distributed.We are most interested in Standard Deviations (σ) and Variance (σ2)
16. 16.  Consider these means for weekly candy bar consumption. X = {7, 8, 6, 7, 7, 6, 8, 7} X = {12, 2, 0, 14, 10, 9, 5, 4} X = (7+8+6+7+7+6+8+7)/8 X = (12+2+0+14+10+9+5+4)/8 X=7 X=7 What is the difference?
17. 17. Once you determine that the variable of interest is normally distributed, ideally by producing ahistogram of the scores, the next question to beasked is its dispersion: how spread out are the scores around the mean.Dispersion is a key concept in statistical thinking.The basic question being asked is how much do the scores deviate around the Mean? The more “bunched up” around the mean the better the scores are.
18. 18. . If every X were very close to the Mean, the mean would be a very good predictor.If the distribution is very sharply peaked then the mean is a good measure of central tendency and if you were to use the mean to make predictions you would be right or close much of the time.
19. 19. The key concept for describing normal distributionsand making predictions from them is calleddeviation from the mean.We could just calculate the average distance between each observation and the mean. We must take the absolute value of the distance, otherwise they would just cancel out to zero!Formula: | X − Xi | ∑ n
20. 20. Data: X = {6, 10, 5, 4, 9, 8} X = 42 / 6 = 7X – Xi Abs. Dev. 1. Compute X (Average)7–6 1 2. Compute X – X and take7 – 10 3 the Absolute Value to get Absolute Deviations7–5 2 3. Sum the Absolute7–4 3 Deviations7–9 2 4. Divide the sum of the absolute deviations by N7–8 1 Total: 12 12 / 6 = 2
21. 21.  Instead of taking the absolute value, we square the deviations from the mean. This yields a positive value. This will result in measures we call the Variance and the Standard DeviationSample- Population-s: Standard Deviation σ: Standard Deviations2: Variance σ2: Variance
22. 22. Formulae: Variance: Standard Deviation:s =2 ∑(X − Xi ) 2 s= ∑(X − X ) i 2 N N
23. 23. Data: X = {6, 10, 5, 4, 9, 8}; N=6 Mean: X X−X (X − X ) 2 X= ∑X = 42 =7 6 -1 1 N 6 10 3 9 Variance: 5 -2 4 s = 2 ∑ ( X − X )2 = 28 = 4.67 4 -3 9 N 6 9 2 4 Standard Deviation: 8 1 1 s = s 2 = 4.67 = 2.16Total: 42 Total: 28