1. PCB 3043L - General Ecology Data Analysis

2. OUTLINE Organizing an ecological study Basic sampling terminology Statistical analysis of data –Why use statistics? –Describing data Measures of central tendency Measures of spread Normal distributions Using Excel –Producing tables –Producing graphs –Analyzing data –Statistical tests T-Tests ANOVA Regression

3. Organizing an ecological study What is the aim of the study? What is the main question being asked? What are your hypotheses? Collect data Summarize data in tables Present data graphically Statistically test your hypotheses Analyze the statistical results Present a conclusion to the proposed question

4. Basic sampling terminology Variables Populations Samples Parameters Statistics

5. What is a variable? Variable: any defined characteristic that varies from one biological entity to another. Examples: plant height, bird weight, human eye color, no. of tree species If an individual is selected randomly from a population, it may display a particular height, weight, etc. If several individuals are selected, their characteristics may be very similar or very different.

6. What is a population? Population: the entire collection of measurements of a variable of interest. Example: if we are interested in the heights of pine trees in Everglades National Park (Plant height is our variable) then our population would consist of all the pine trees in Everglades National Park.

7. What is a sample? Sample: smaller groups or subsets of the population which are measured and used to estimate the distribution of the variable within the true population Example: the heights of 100 pine trees in Everglades National Park may be used to estimate the heights of trees within the entire population (which actually consists of thousands of trees)

8. What is a parameter? populationParameter: any calculated measure used to describe or characterize a population Example: the average height of pine trees in Everglades National Park

9. What is a statistic? Statistic: an estimate of any population parameter Example: the average height of a sample of 100 pine trees in Everglades National Park

10. Why use statistics? It is not always possible to obtain measures and calculate parameters of variables for the entire population of interest Statistics allow us to estimate these values for the entire population based on multiple, random samples of the variable of interest The larger the number of samples, the closer the estimated measure is to the true population measure Statistics also allow us to efficiently compare populations to determine differences among them Statistics allow us to determine relationships between variables

11. Statistical analysis of data Measures of central tendency Measures of dispersion and variability Site 1Site 2 54 72 38 83 67 Heights of pine trees at 2 sites in Everglades National Park

12. Where is the center of the distribution? mean (  or μ ): arithmetic mean…… median: the value in the middle of the ordered data set mode: the most commonly occurring value Example data set : 1, 2, 2, 2, 3, 5, 6, 7, 8, 9, 10 Mean = (1 + 2 + 2 + 2+ 3 + 5 + 6 + 7 + 8 + 9 + 10)/11 = 55/11 = 5 Median = 1, 2, 2, 2, 3, 5, 6, 7, 8, 9,10 = 5 1, 2, 2, 2, 3, 5, 6, 7, 8, 9,10,11 = (5+6)/2 = 5.5 Mode = 1, 2, 2, 2, 3, 5, 6, 7, 8, 9, 10 = 2 Measures of central tendency

13. How widely is the data distributed? range: largest value minus smallest value variance (s 2 or σ 2 ) ………….…………. standard deviation (s or σ)………………… Large spread Small spread Measures of dispersion and variability

14. Example data set: 0, 1, 3, 3, 5, 5, 5, 7, 7, 9, 10 Variance = 9.8 Standard Deviation = 3.13 Range = 10 Example data set: 0, 10, 30, 30, 50, 50, 50, 70, 70, 90, 100 Variance = 980 Standard Deviation = 31.30 Range = 100 Measures of dispersion and variability

15. Normal distribution of data A data set in which most values are around the mean, with fewer observations towards the extremes of the range of values The distribution is symmetrical about the mean

16. Proportions of a Normal Distribution A normal population of 1000 body weights μ = 70kgσ = 10kg 500 weights are > 70kg 500 weights are < 70 kg

17. Proportions of a Normal Distribution How many bears have a weight > 80kg μ = 70kgσ = 10kgX = 80kg We use an equation to tell us how many standard deviations from the mean the X value is located: = = We then use a special table to tell us what proportion of a normal distribution lies beyond this Z value This proportion is equal to the probability of drawing at random a measurement (X) greater than 80kg Z = X – μ σ Z = 80 – 70 10 1

18. Z table Look for Z value on table (1.0) Find associated P value (0.1587) P value states there is a 15.87% ((0.1587/1)x100) chance that a bear selected from the population of 1000 bears measured will have a weight greater than 80kg

19. Probability distribution tables There are multiple probability tables for different types of statistical tests. e.g. Z-Table, t-Table, Χ 2 -Table Each allows you to associate a “critical value” with a “P value” This P value is used to determine the significance of statistical results

20. Using Excel Program used to organize data Produce tables Perform calculations Make graphs Perform statistical tests

21. Organizing data in tables Allows you to arrange data in a format that is best for analysis The following are the steps you would use:

25. Performing calculations Allows you to perform several calculations Sum, Average, Variance, Standard deviation Basic subtraction, addition, multiplication More complex formulas

40. Bar Charts……. Scatter Plots…………………. Making graphs

55. Bar Charts……. Scatter Plots…………………. Making graphs

64. Analyzing Data in Excel Statistical tests can be done to determine: Whether or not there is a significant difference between two data sets (Student’s t-test) Whether or not there is a significant difference between more than two data sets (ANOVA) Whether or not there is a significant relationship between two variables (Regression analysis)

65. Analyzing Data in Excel The following steps must be followed: 1.Choose an appropriate statistical test 2.State H 0 and H A 3.Run test to produce Test Statistic 4.Examine P-value 5.Decide to accept or reject H 0

66. Normally, you would have to calculate the critical value and look up the P value on a table All tests done in Excel provide the P value for you This P value is used to determine the significance of statistical results This P value must be compared to an α value α value is usually 0.05 or less (e.g. 0.01) Less than 5% chance that the null hypothesis is true The lower the α value the more certain we about rejecting the null Hypothesis First thing you must do is select which statistical test you want to perform This is how it is done…….. Analyzing Data in Excel

71. t-Tests Used to compare the means of two populations and answer the question: Is there a significant difference between the two populations? Example: Is there a significant difference between the average height of pine trees from 2 sites in Everglades National Park? You cannot use this test to compare two different types of data (e.g. water depth data and soil depth data). It can only compare two sets of data based on the same data type (e.g. water depth data from two different sites) The two data sets that are being compared must be presented in the same units. (e.g. you can compare two sets of data if both are recorded in days. You cannot compare data recorded in units of days with data recorded in units of months)

72. Your Null Hypothesis is always: There is no significant difference between the two compared populations (μ1= μ2) Your Alternative Hypothesis is always: There is a difference between the two compared populations (μ1 ≠ μ2) 1. Choose an appropriate statistical test 2. State H 0 and H A 3. Run test to produce Test Statistic 4. Examine P-value 5. Decide to accept or reject H0

79. t-Tests 1. Choose an appropriate statistical test 2. State H 0 and H A 3. Run test to produce Test Statistic 4. Examine P-value 5. Decide to accept or reject H0 When you run the test, look for the p-value If p > 0.05 then fail to reject your Null Hypothesis and state that “there is no significant difference between the two compared populations” If p < 0.05 then reject your Null Hypothesis and state that “there is a significant difference between the two compared populations”

80. t-Tests 1. Choose an appropriate statistical test 2. State H 0 and H A 3. Run test to produce Test Statistic 4. Examine P-value 5. Decide to accept or reject H0 When you run the test, look for the p-value Our results show P = 0.09903 Therefore P > 0.05 (This means that there is greater than a 5% chance that our null hypothesis is true) So we must fail to reject the Null Hypothesis and state that “there is no significant difference between the two compared populations”

81. ANOVA Used to compare the means of more than two populations and answer the question: Is there a significant difference between the populations? Example: Is there a significant difference between the average height of pine trees from 4 sites in Everglades National Park? For comparing a particular feature of two or more populations, use a Single Factor ANOVA For comparing a particular feature of two or more populations, subdivided into two groups, use a Two Factor ANOVA

82. Your Null Hypothesis is always: There is no significant difference between the compared populations (μ1 = μ2 = μ3 = μ4 …..) Your Alternative Hypothesis is always: There is a difference between the compared populations (μ1 ≠ μ2 ≠ μ3 ≠ μ4 …..) 1. Choose an appropriate statistical test 2. State H 0 and H A 3. Run test to produce Test Statistic 4. Examine P-value 5. Decide to accept or reject H0

87. ANOVA 1. Choose an appropriate statistical test 2. State H 0 and H A 3. Run test to produce Test Statistic 4. Examine P-value 5. Decide to accept or reject H0 When you run the test, look for the p-value If p > 0.05 then fail to reject your Null Hypothesis and state that “there is no significant difference between the compared populations” If p < 0.05 then reject your Null Hypothesis and state that “there is a significant difference between at least two of the compared populations”

88. ANOVA 1. Choose an appropriate statistical test 2. State H 0 and H A 3. Run test to produce Test Statistic 4. Examine P-value 5. Decide to accept or reject H0 When you run the test, look for the p-value Our results show P = 0.002197 Therefore P < 0.05 (This means that there is less than a 5% chance that our null hypothesis is true) So we must reject your Null Hypothesis and state that “there is a significant difference between at least two of the compared populations”

89. ANOVA Remember: The ANOVA result will only tell you that i)None of the data sets are significantly different from each other OR ii)At least two of the data sets among the data sets being compared are significantly different If there is a significant difference between at least two data sets, it will not tell you which two.

90. Regression analysis Used to determine whether or not there is a linear relationship between two variables and answer the question: Is there a significant linear relationship between two variables? Example: Is there a significant relationship between the average height of pine trees and soil depth in Everglades National Park? It basically creates an equation (or line) that best predicts Y values based on X values. You cannot use this test to compare populations. It only compares variables. You are looking at two different variables (e.g. water depth (cm) and plant abundance (no. of individuals), so the data sets do not have to be presented in the same units

91. Your Null Hypothesis is always: There is no significant linear relationship between the two variables Your Alternative Hypothesis is always: There is a significant linear relationship between the two variables 1. Choose an appropriate statistical test 2. State H 0 and H A 3. Run test to produce Test Statistic 4. Examine P-value 5. Decide to accept or reject H0

92. Example: R square value of 0.04 The regression line does not fit the data well Many of the points lie far from the line, so there is not a defined linear relationship between the two variables “x” cannot be used to predict “y” Example: R square value of 0.94 The regression line fits the data well The points all lie fairly close to the line, so there is a defined linear relationship between the two variables “x” can be used to predict “y” R squared: how well “y” can be predicted by “x”, i.e. how strong the linear relationship is between the two variables. The closer R square is to 0, the less well it fits the data. The closer R square is to 1, more it fits the data.

93. 1. Choose an appropriate statistical test 2. State H 0 and H A 3. Run test to produce Test Statistic 4. Examine P-value 5. Decide to accept or reject H0

100. Regression analysis 1. Choose an appropriate statistical test 2. State H 0 and H A 3. Run test to produce Test Statistic 4. Examine P-value 5. Decide to accept or reject H0 When you run the test, look for the Significance F or Sample p-value If p > 0.05 then fail to reject your Null Hypothesis and state that “There is no significant linear relationship between the two variables” If p < 0.05 then reject your Null Hypothesis and state that “There is a significant linear relationship between the two variables”

101. 1. Choose an appropriate statistical test 2. State H 0 and H A 3. Run test to produce Test Statistic 4. Examine P-value 5. Decide to accept or reject H0 When you run the test, look for the p-value Our results show Significance F or Sample p-value = 1.65E08 = 0.0000000165 Therefore P < 0.05 (This means that there is less than a 5% chance that our null hypothesis is true) So we must reject your Null Hypothesis and state that “There is a significant linear relationship between the two variables” Next look at the R squared value Our results show R squared = 0.975 Therefore the line fits the data well “x” can be used to predict “y” Regression analysis

102. Ecological study What is the aim of the study? What is the main question being asked? What are your hypotheses? Collect data Summarize data in tables Present data graphically Statistically test your hypotheses Analyze the statistical results Present a conclusion to the proposed question

103. Aim: To determine whether or not there are changes in heights of Pine trees with distance from the edge of a forest trail in Everglades National Park. Hypotheses: HO: There is no significant relationship between distance from the edge of the trail and Pine tree height HA: There is a significant relationship between distance from the edge of the trail and Pine tree height Results: Discussion/Conclusion: The gap created by the trail may be adversely affecting Pine trees, such that they are shorter near the trail and become taller with distance from the trail. Distance from trail (m)Plant heights (m) 02.1 52.7 102.9 153.1 203.4 253.7 303.8 354.5 404.6 454.8 505.6 SUM41.2 AVERAGE3.74 STANDARD DEVIATION1.04 P = 1.65E-08 Since P < 0.05, reject Ho Therefore, there is a significant relationship between distance from the edge of the trail and Pine tree height R Square = 0.97, so there is a strong positive linear relationship between distance from the trail and plant height Average tree height of pine trees along transect from forest trail to interior forest at ENP

104. Three questions: 1.T-test 2.Single factor ANOVA 3.Regression analysis Assignment – Worksheet 1