1. Measures of Dispersion CUMULATIVE FREQUENCIES INTER-QUARTILE RANGE RANGE MEAN DEVIATION VARIANCE and STANDARD DEVIATION STATISTICS: DESCRIBING VARIABILITY Variability = Uncertainty Probabilities

2. STATISTICS: PROBABILITIES CALCULATING PROBABILITIES NORMAL DISTRIBUTION What are probabilities? z-DISTRIBUTION t-DISTRIBUTION Probabilities

3. Probability = 0.167 The chance of NOT throwing a = 5/6 = 0.833 = 1 – 0.167 The chance of throwing a = 1/6 The probability of an event A, symbolized by P(A), is a number between 0 and 1 The higher the P value the more probable the event The Total Number Of Possible Outcomes The Number Of Ways Event A Can Occur P(A) =

4. What is the probability of picking a student of 1.65 m high from the class? 0 2 4 6 8 10 12 14 16 1.21.31.41.51.61.71.81.9 2 2.12.2 Height (m) Frequency P(NOT 1.65) = 1 – P(1.65) = 1 – 0.094 = 0.906 Depends on how the data are distributed STATISTICS: PROBABILITY P(1.65) = 10/106 = 0.094 The probability is a number between 0 and 1 Class has 106 students (n=106) 10 are 1.65m tall

5. Frequency Distributions Height (m) 0 2 4 6 8 10 12 14 16 1.21.31.41.51.61.71.81.9 2 2.12.2 Height (m) Frequency (%) STATISTICS: PROBABILITY Area under graph = total number of observations Can display frequency distributions as % or proportion 0 2 4 6 8 10 12 14 16 1.21.31.41.51.61.71.81.9 2 2.12.2 No. of observations Area under graph = 1.0 (for proportion) = 100 (for percentage) 12 people 10 people 0.113 0.094

6. 0 2 4 6 8 10 12 14 16 1.21.31.41.51.61.71.81.9 2 2.12.2 Height (m) Frequency (%) STATISTICS: PROBABILITY Normal Distribution Data clustered around the mean Therefore good chance (high probability) of picking (at random) a student with a height close to the mean Small chance (low probability) of picking (at random) a student who is either very tall or very short Tails

7. POPULATION DYNAMICS Required background knowledge: Data and variability concepts  Data collection Measures of central tendency (mean, median, mode, variance, stdev) Normal distribution and Standard Error Student’s t-test and 95% confidence intervals Chi-Square tests MS Excel

8. Properties of a normal distribution: The mean, median and mode are the same The frequency distribution is completely symmetrical either side of the mean The area under the curve is proportional to number of observations Height (mm) Frequency (%) 0 2 4 6 8 10 12 0246 8 1012141618202224 STATISTICS: PROBABILITY s 2 = 4 s 2 = 8 s 2 = 12 s 2 = 16 x = 10 The shape of the curve depends on the variance or standard deviation: the spread of values about the mean Σ = 100%

9. Some dataset are normally distributed – but NOT all. Height (mm) Frequency (%) 0 2 4 6 8 10 12 02468 1012141618202224 Σ = 100% STATISTICS: PROBABILITY The normal curve has fixed mathematical properties, irrespective of: The scale on which it is drawn The magnitude or units of its mean The magnitude or units of its Standard Deviation …….and these render it susceptible to STATISTICAL ANALYSIS… Can use the normal distribution to calculate probabilities

10. STATISTICS: PROBABILITIES CALCULATING PROBABILITIES NORMAL DISTRIBUTION What are probabilities? z-DISTRIBUTION t-DISTRIBUTION Probabilities

11. To calculate the probability of a particular value x being drawn from a normally distributed population of data, you need to know the mean AND the standard deviation of the data X = value you are considering μ = population mean σ = population standard deviation Z = (x – μ) σ Equation 1 Z-values form the Z-DISTRIBUTION…. Z is based on data that are normally distributed, so the Z distribution is also normally distributed. STATISTICS: CALCULATING PROBABILITIES Z = how many standard deviations away from the mean is the value x If Z is small number then x ≈ mean If Z is large number then x ≈ mean Once we know the Z-value we use statistical tables to calculate the associated probability… 0 2 4 6 8 10 12 14 16 1.21.31.41.51.61.71.81.9 2 2.12.2 Height (m) Frequency

12. Z = (x – μ) σ Z = (1.95 – 1.55) 0.3 Z = (0.4) 0.3 Z = 1.33 ? What is the probability of a randomly drawing a student measuring more than 1.95 m from the population if μ = 1.55 m and σ = 0.3 m? -3.67-3.33-3.00-2.67-2.33-2.00-1.67-1.33-0.67-0.33 0.000.330.671.001.331.672.002.332.673.003.33 Z Frequency 0.0918 STATISTICS: CALCULATING PROBABILITIES – an example Step 1: calculate Z using known information P(0.0918) of randomly drawing a student measuring 1.95 m from the population [P > 1.95 = 0.0918] REMEMBER: The higher the P-value the more probable the event 1 st decimal place 2 nd decimal place PROBABILITY P(A) p > 1.95 Z = 1.33 Step 2: Look up Z in Z-tables

13. Now you try: A population of bone measurements is normally distributed with μ = 60 mm and σ = 10 mm. What is the probability of selecting a bone with a length greater than 66 mm? Z = (x – μ) σ Step 1: calculate Z using known information Step 2: Look up the P-value in the Z-tables Z = 0.60 Therefore p = 0.2743 ANSWER: In Excel: Enter x, μ and σ into different cells Formula: =(x – μ)/ σ

14. POPULATION DYNAMICS Required background knowledge: Data and variability concepts  Data collection Measures of central tendency (mean, median, mode, variance, stdev) Normal distribution and Standard Error Student’s t-test and 95% confidence intervals Chi-Square tests MS Excel

15. NOTE: Standard Deviation = tells you the variation around the mean Standard Error = tells you how well you’ve estimated the mean Standard Deviation vs. Standard Error

16. 0 2 4 6 8 10 12 14 16 1.21.3 1.4 1.51.61.7 1.8 1.9 2 2.1 2.2 Height (m) Frequency X2X2 X1X1 X3X3 X4X4 Population normal = distribution 0 2 4 6 8 10 12 14 16 1.21.3 1.4 1.51.61.7 1.8 1.9 2 2.1 2.2 Sample Means Frequency Sample mean = normal distribution Standard Error

17. X2X2 X1X1 X3X3 X4X4 X6X6 X5X5 X7X7 X8X8 X 10 X9X9 X 11 X 12 σ2σ2 n σ2σ2 x = Equation 2 As n increases ….So σ decreases Standard Error The variance of the mean

18. σ2σ2 n σ2σ2 x = Standard Error Square Root both sides σ2σ2 n σ x = σ x = Equation 2 σ n √ STANDARD ERROR

19. 0 2 4 6 8 10 12 14 16 1.21.3 1.4 1.51.61.7 1.8 1.9 2 2.1 2.2 Height (m) Frequency Population normal = distribution 0 2 4 6 8 10 12 14 16 1.21.3 1.4 1.51.61.7 1.8 1.9 2 2.1 2.2 Sample Means Frequency Sample mean = normal distribution Standard Error A normal deviate referring to the normal distribution of X i values Z = (x – μ) σ A normal deviate referring to the normal distribution of means Z = (x – μ) σ x

20. What is the probability of obtaining a random sample of nine measurements with a mean greater than 50.0 mm, from a population having a mean of 47 mm and a standard deviation of 12.0 mm? Standard Error N = 9, X = 50.0 mm, μ = 47.0 mm, σ = 12.0 mm σ x = 12.0 √ 9 = 4= 12 3 Z = (50.0 – 47.0) = 3 = 0.75 44 Z = (x – μ) σ x Step 1: calculate Z using known information σ x = Equation 2 σ n √ Step 2: Look up Z in Z-tables

21. STATISTICS: z-DISTRIBUTION Z = 0.75 Step 2: Look up Z in Z-tables P = 0.2266 So there is a 0.2266 is the probability of obtaining a random sample of nine measurements with a mean greater than 50.0 mm, from a population having a mean of 47 mm and a standard deviation of 12.0 mm. For probability values, always report 4 decimal places

22. σ x = σ n √ Now you try: What is the probability of obtaining a random sample of 5 measurements with a mean greater than 60.0 mm, from a population having a mean of 57 mm and a standard deviation of 7.0 mm? Step 1: calculate Z using known information Step 2: Look up the P-value in the Z-tables Z = 0.96 Therefore p = 0.1685 ANSWER: Z = (x – μ) σ x Formula: =(x - μ)/(σ /(SQRT(n))) In Excel: Enter x, μ, σ and n

23. In the last couple of equations for Z we have used the population parameters: μ, σ and BUT we don’t usually have access to population data and must make do with sample estimators x, s and σ x s x IF n is very, very large : we use Z distribution to calculate normal deviates Z = (x – μ) σ x STATISTICS: z-DISTRIBUTION σ x s x = t = (x – μ) s x Equation 3 If n is not large, we must use t distribution: σs PROXY