1. The usual course of events for conducting scientific work “The Scientific Method” Reformulate or extend hypothesis Develop a Working Hypothesis Observation Conduct an experiment or a series of controlled systematic observations Appropriate statistical tests Confirm or reject hypothesis

2. The usual course of events for conducting scientific work “The Scientific Method” Reformulate or extend hypothesis Develop a Working Hypothesis Observation Conduct an experiment or a series of controlled systematic observations Appropriate statistical tests Confirm or reject hypothesis In the intertidal zone, algae seem to be confined to specific areas There will be a positive correlation of algal abundance and tide height Measure tide heights and count number of algae at each Product-moment correlation There is a positive correlation of tide height and algal abundance Algal will grow higher on the shore in areas of high wave action

3. Imagine that you are collecting samples (i.e. individuals) from a population of little ball creatures - Critterus sphericales Little ball creatures come in 3 sizes: Small = Medium = Large =

4. -sample 1 -sample 2 -sample 3 -sample 4 -sample 5 You take a total of five samples

5. The real population (all the little ball creatures that exist) Your samples

6. Each sample is a representation of the population BUT No single sample can be expected to accurately represent the whole population So……………

7. To be statistically valid, each sample must be: 1) Random: Thrown quadrat?? Guppies netted from an aquarium?

8. To be truly random: 20 15 Choose numbers randomly from 1 to 300

9. To be truly random: 20 15 Choose numbers randomly from 1 to 300

10. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Assign numbers from a random number table

11. To be statistically valid, each sample must be: 2) Replicated:

12. Bark Samples for levels of cadmium

13. Pseudoreplicated Sample size (n) =1 Not pseudoreplicated Sample size (n) =10 10 samples from 10 different trees 10 samples from the same tree

14. IF YOUR DATA ARE: 1. Continuous data 2. Ratio or interval 3. Approximately normal distribution 4. Equal variance (F-test) 5. Conclusions about population based on sample (inductive) 6. Sample size > 10 samplepopulation

15. CHARACTERIZING DATA

16. Variables -dependent – in any experiment, the dependent variable is the one being measured by the experimenter -also known as a reponse or test variable -independent – in any experiment, the independent variable is the one being changed by the experimenter -also known as a factor

17. Nominal data (nominal scales, nominal variables) Drosophila genetic traits - data are in categories Species Sex

18. Look at the distribution of lizards in the forests Tree branches Tree trunks Ground Species ASpecies BSpecies CSpecies D

19. - Both the dependent and independent variables are nominal/categorical Habitat GroundTree trunkTree branchSpecies totals Lizard Species Species A901524 Species B901221 Species C95014 Species D910322 Totals36153081

20. - data are in categories -grades Ordinal data (ordinal scales, ordinal variables) - categories are ranked -surveys -behavioural responses

21. Interval data (interval scales, interval variables) zero point depends on the scale used e.g. temperature - constant size interval - no true zero point - values can be treated arithmetically (only +, -) to give a meaningful result

22. Ratio data (or ratio scales or ratio variables) - constant size interval - a zero point with some reality height weight time - values can be treated arithmetically (+, -, x, ÷ ) to give a meaningful result

23. Ratio data (or ratio scales or ratio variables) - constant size interval - a zero point with some reality Can also be continuous - values can be treated arithmetically (+, -, x, ÷ ) to give a meaningful result Or discrete - counts, “number of …..”

24. Kinds of Variables Assignment as a discrete (= categorical) or continuous variable can depend on the method of measurement Dappled Full Open Continuous Discrete ( = categorical)

25. The kind of data you are dealing with is one determining factor in the kind of statistical test you will use.

26. IF YOUR DATA ARE: 1. Continuous data 2. Ratio or interval 3. Approximately normal distribution 4. Equal variance (F-test) 5. Conclusions about population based on sample (inductive) 6. Sample size > 10 samplepopulation

27. Two ways of arriving at a conclusion 2. Inductive inference sample population sample population 1. Deductive inference

28. IF YOUR DATA ARE: 1. Continuous data 2. Ratio or interval 3. Approximately normal distribution 4. Equal variance (F-test) 5. Conclusions about population based on sample (inductive) 6. Sample size > 10 samplepopulation

29. Imagine the following experiment: 2 groups of crickets Group 1 – fed a diet with extra supplements Group 2 – fed a diet with no supplements Weights 12.113.913.012.1 14.912.212.914.9 13.612.013.513.6 12.015.912.412.0 10.912.111.010.9 9.18.911.010.1 9.99.28.011.9 8.69.08.59.6 10.010.99.48.0 11.97.110.08.9 Mean = 12.8 Mean = 9.49

30. What you’re doing here is comparing two samples that, because you’ve not violated any of the assumptions we saw before, should represent populations that look like this: 9.4912.8 Are the means of these populations different?? Frequency Weight

31. Are the means of these populations different?? To answer this question – use a statistical test A statistical test is just a method of determining mathematically whether you definitively say ‘yes’ or ‘no’ to this question What test should I use??

32. IF YOU HAVEN’T VIOLATED ANY OF THE ASSUMPTIONS WE MENTIONED BEFORE…… Number of groups compared 2 other than 2 T -test Direction of difference specified? YesNo One-tailedTwo- tailed Does each data point in one data set (population) have a corresponding one in the other data set? YesNo Paired t-testUnpaired t-test Are the means of two populations the same? Are the means of more than two populations the same? Number of factors being tested 12>2 Does each data point in one data set (population) have a corresponding one in the other data sets? Two way ANOVA ANOVA YesNo One way ANOVA Repeated Measures ANOVA Other tests

33. A simple t-test 1. State hypotheses H o – there is no difference between the means of the two populations of crickets (i.e. the extra nutrients had no effect on weight) H 1 – there is a difference between the means of the two populations of crickets (i.e. the extra nutrients had an effect on weight)

34. A simple t-test 2. Calculate a t-value (any stats program does this for you) 3. Use a probability table for the test you used to determine the probability that corresponds to the t- value that was calculated. (for the truly masochistic)

35. A simple t-test 2. Calculate a t-value (any stats program does this for you) 3. Use a probability table for the test you used to determine the probability that corresponds to the t- value that was calculated. DataTest statisticProbability

36. Unpaired t test Do the means of Nutrient fed and No nutrient differ significantly? P value The two-tailed P value is < 0.0001, considered extremely significant. t = 7.941 with 38 degrees of freedom. 95% confidence interval Mean difference = -3.307 (Mean of No nutrient minus mean of Nutrient fed) The 95% confidence interval of the difference: -4.150 to -2.464 Assumption test: Are the standard deviations equal? The t test assumes that the columns come from populations with equal SDs. The following calculations test that assumption. F = 1.192 The P value is 0.7062. This test suggests that the difference between the two SDs is not significant. Assumption test: Are the data sampled from Gaussian distributions? The t test assumes that the data are sampled from populations that follow Gaussian distributions. This assumption is tested using the method Kolmogorov and Smirnov: Group KS P Value Passed normality test? =============== ====== ======== ======================= Nutrient fed 0.1676 >0.10 Yes No nutrient 0.1279 >0.10 Yes

37. Interpretation of p <.0001? This means that there is less than 1 chance in 10,000 that these two means are from the same population. In the world of statistics, that is too small a chance to have happened randomly and so the H o is rejected and the H 1 accepted

38. For all statistical tests that you’ll use, it is convention that the minimum probability that two samples can differ and still be from the same population is 5% or p =.05

39. What happens if you violate any of the assumptions? Step 1 - Panic

40. What happens if you violate any of the assumptions? Step 1 - Panic Step 2 - It depends on what assumptions have been violated. AssumptionOther testsAnother solution? 1. Continuous dataYes 2. Ratio/intervalYes 3. Normal distributionYesTransform the data 4. Equal varianceYes - Welch’s 5. Sample PopulationYes 6. N<10YesTake more samples

41. Nonparametric Tests These tests are used when the assumptions of t-tests and ANOVA have been violated They are called “nonparametric” because there is no estimation of parameters (means, standard deviations or variances) involved. Several kinds: 1)Goodness-of-Fit tests - when you calculate an expected value 2)Non-parametric equivalents of parametric tests

42. SUMMARY Problem - trying to determine the expected frequencies of any result in a particular experiment Type of data Discrete 2 categories & Bernoulli process > 2 categories Use a Binomial model to calculate expected frequencies Use a Poisson distribution to calculate expected frequencies

43. Consider the following problem: Sampling earthworms 25 plots 13 24 31 41 53 60 70 81 92 103 114 125 130 141 153 165 175 182 196 203 211 221 231 240 251 Quadrat# of worms

44. 13 24 31 41 53 60 70 81 92 103 114 125 130 141 153 165 175 182 196 203 211 221 231 240 251 Quadrat# of worms N = 25 X = 2.24 worms/quadrat

45. What is the expected number of worms/quadrat? OR What is the probability of x worms being in a particular quadrat?

46. Use a Poisson distribution ->2 mutually exclusive categories -N is relatively large and p is relatively small The distribution of worms in space is expected to be random

47. Formula for a Poisson distribution P x = e -µ µ x X! Probability of observing X individuals in a category Base of natural logarithms (= 2.71828….) True mean of the population (approximated by sample mean) An integer (number of indviduals)

48. Formula for a Poisson distribution P x = e -µ µ x X! Probability of observing X worms in a quadrat Base of natural logarithms (= 2.71828….) µ = X = 2.24 Number of worms)

49. # of worms Probability of finding X worms in a quadrat Calculation 0Po = e -µ (µ x /0!)=e -2.24 =.1065 1Po = e -µ (µ 1 /1!)=e -2.24 (2.24/1) =.2385 2Po = e -µ (µ 2 /2!)=e -2.24 (2.24 2 /2) =.2671 3Po = e -µ (µ 3 /3!)=e -2.24 (2.24 3 /6) =.1994 4Po = e -µ (µ 4 /4!)=e -2.24 (2.24 4 /24) =.1117 5Po = e -µ (µ 5 /5!)=.05 6Po = e -µ (µ 6 /6!)=.0187 7Po = e -µ (µ 7 /7!)=.006 Could go on forever or to ∞ - whichever comes first!

50. Practically…. P 0 + P 1 + P 2 + P 3 + P 4 + P 5 + P 6 + P 7 =.998 And P 8 + P 9 ……=.002 For convenience - P 8 =.002

51. Other kinds of Poisson problems 1. Cell counts in a hemocytometer 2. Number of parasitic mites per fly in a population 3. Number of fish per seine 4. Number of animals in a particular subdivision of the habitat Poisson Distributions are very common in biological work!

52. Goodness-of-Fit Tests Use with nominal scale data e.g. results of genetic crosses Also, you’re using the population to deduce what the sample should look like

53. Classic example - genetic crosses Do they conform to an “expected’ Mendelian ratio? Back to our little ball creatures - Critterus sphericales Phenotypes: A_B_ A_bb aaB_ aabb Mendelian inheritance -Predict a 9:3:3:1 ratio

54. -sampled 320 animals A_B_A_bbaaB_aabb Observed (o)19453676

55. -sampled 320 animals A_B_A_bbaaB_aabb Observed (o)19453676 Expected (e)18060 20

56. -sampled 320 animals A_B_A_bbaaB_aabb Observed (o)19453676 Expected (e)18060 20 o - e14-77-14

57. -sampled 320 animals A_B_A_bbaaB_aabb Observed (o)19453676 Expected (e)18060 20 o - e14-77-14 (o - e) 2 19649 196

58. -sampled 320 animals A_B_A_bbaaB_aabb Observed (o)19453676 Expected (e)18060 20 o - e14-77-14 (o - e) 2 19649 196 (o - e) 2 e 1.08.82 9.8

59. -sampled 320 animals A_B_A_bbaaB_aabb Observed (o)19453676 Expected (e)18060 20 o - e14-77-14 (o - e) 2 19649 196 (o - e) 2 e 1.08.82 9.8 (o -e) 2 e   2 = = 1.08 +.82 +.82 + 9.8 = 12.52 df = number of classes -1 = 3

60. X 2 = 12.52Critical value for 3 degrees of freedomat.05 level is7.82 X 2 Table Conclusion: Probability of these data fitting the expected distribution is <.05, therefore they are not from a Mendelian population The actual probability of X 2 =12.52 and df = 3 is.01 > p >.001

61. A little X 2 wrinkle - the Yates correction Formula is (o -e) 2 e   2 = Except of df = 1 (i.e. you’re using two categories of data) Then the formula becomes (|o -e| - 0.5) 2 e   2 =

62. Type of dataNumber of samples Are data related? Test to use Nominal2YesMcNemar Nominal2NoFisher’s Exact Nominal>2YesCochran’s Q Summary!

63. Type of dataNumber of samplesAre data related?Test to use Nominal2YesMcNemar Nominal2NoFisher’s Exact Nominal>2YesCochran’s Q Ordinal1NoKomolgorov- Smirnov Ordinal+2YesWilcoxon (paired t-test analogue) Ordinal+2NoMann Whitney U (unpaired t-test analogue) Ordinal+>2NoKruskal Wallis (analogue of one- way ANOVA Ordinal>2YesFriedman two-way ANOVA All of the parametric tests (remember the big flow chart!) have non-parametric equivalents (or analogues)