Homework Chapter 11: 13 Chapter 12: 1, 2, 14, 16

Assumptions in Statistics

The wrong way to make a comparison of two groups “Group 1 is significantly different from a constant, but Group 2 is not. Therefore Group 1 and Group 2 are different from each other.”

A more extreme case...

Interpreting Confidence Intervals Different Not differentUnknown

Assumptions Random sample(s) Populations are normally distributed Populations have equal variances –2-sample t only

Assumptions Random sample(s) Populations are normally distributed Populations have equal variances –2-sample t only What do we do when these are violated?

Assumptions in Statistics Are any of the assumptions violated? If so, what are we going to do about it?

Detecting deviations from normality Histograms Quantile plots Shapiro-Wilk test

Detecting deviations from normality: by histogram Biomass ratio Frequency

Detecting deviations from normality: by quantile plot Normal data Normal Quantile

Detecting deviations from normality: by quantile plot Biomass ratio Normal Quantile

Detecting deviations from normality: by quantile plot Biomass ratio Normal Quantile Normal distribution = straight line Non-normal = non-straight line

Detecting differences from normality: Shapiro-Wilk test Shapiro-Wilk Test is used to test statistically whether a set of data comes from a normal distribition Ho: The data come from a normal distribution Ha: The data come from some other distribution

What to do when the distribution is not normal If the sample sizes are large, sometimes the standard tests work OK anyway Transformations Non-parametric tests Randomization and resampling

The normal approximation Means of large samples are normally distributed So, the parametric tests on large samples work relatively well, even for non-normal data. Rule of thumb- if n > ~50, the normal approximations may work

Data transformations A data transformation changes each data point by some simple mathematical formula

Log-transformation Y Y' = ln[Y] ln = “natural log”, base e log = “log”, base 10 EITHER WORK

Carry out the test on the transformed data!

Variance and mean increase together --> try the log-transform

Other transformations Arcsine Square-root Square Reciprocal Antilog

Example: Confidence interval with log-transformed data Data: ln data: ln[Y]

Valid transformations... Require the same transformation be applied to each individual Must be backwards convertible to the original value, without ambiguity Have one-to-one correspondence to original values X = ln[Y] Y = e X

Choosing transformations Must transform each individual in the same way You CAN try different transformations until you find one that makes the data fit the assumptions You CANNOT keep trying transformations until P <0.05!!!

Assumptions Random sample(s) Populations are normally distributed Populations have equal variances –2-sample t only Do the populations have equal variances? If so, what should We do about it?

Comparing the variance of two groups One possible method: the F test

The test statistic F Put the larger s 2 on top in the numerator.

F... F has two different degrees of freedom, one for the numerator and one for the denominator. (Both are df = n i -1.) The numerator df is listed first, then the denominator df. The F test is very sensitive to its assumption that both distributions are normal.

Example: Variation in insect genitalia

Degrees of freedom For a 2-tailed test, we compare to F  /2,df1,df2 from Table A3.4

Why  /2 for the critical value? By putting the larger s 2 in the numerator, we are forcing F to be greater than 1. By the null hypothesis there is a 50:50 chance of either s 2 being greater, so we want the higher tail to include just  /2.

Critical value for F

Conclusion The F= from the data is greater than F (0.025), 6,8 =4.7, so we can reject the null hypothesis that the variances of the two groups are equal. The variance in insect genitalia is much greater for polygamous species than monogamous species.

Sample Null hypothesis The two populations have the same variance   1  2 2 F-test Test statistic Null distribution F with n 1 -1, n 2 -1 df compare How unusual is this test statistic? P < 0.05 P > 0.05 Reject H o Fail to reject H o

What if we have unequal variances? Welch’s t-test would work If sample sizes are equal and large, then even a ten-fold difference in variance is approximately OK

Comparing means when variances are not equal Welch’s t test compared the means of two normally distributed populations that have unequal variances

Burrowing owls and dung traps

Dung beetles

Experimental design 20 randomly chosen burrowing owl nests Randomly divided into two groups of 10 nests One group was given extra dung; the other not Measured the number of dung beetles on the owls’ diets

Number of beetles caught Dung added: No dung added:

Hypotheses H 0 : Owls catch the same number of dung beetles with or without extra dung (  1 =  2 ) H A : Owls do not catch the same number of dung beetles with or without extra dung (  1   2 )

Welch’s t Round down df to nearest integer

Owls and dung beetles

Degrees of freedom Which we round down to df= 10

Reaching a conclusion t 0.05(2), 10 = 2.23 t=4.01 > 2.23 So we can reject the null hypothesis with P<0.05. Extra dung near burrowing owl nests increases the number of dung beetles eaten.

Assumptions Random sample(s) Populations are normally distributed Populations have equal variances –2-sample t only What if you don’t want to make so many assumptions?

Sample Null hypothesis The two populations have the same mean  1  2 Welch’s t-test Test statistic Null distribution t with df from formula compare How unusual is this test statistic? P < 0.05 P > 0.05 Reject H o Fail to reject H o

Non-parametric methods Assume less about the underlying distributions Also called "distribution-free" "Parametric" methods assume a distribution or a parameter

Most non-parametric methods use RANKS Rank each data point in all samples from lowest to highest Lowest data point gets rank 1, next lowest gets rank 2,...

Sign test Non-parametric test Compares data from one sample to a constant Simple: for each data point, record whether individual is above (+) or below (-) the hypothesized constant. Use a binomial test to compare result to 1/2.

Example: Polygamy and the origin of species Is polygamy associated with higher or lower speciation rates?

Etc....

The differences are not normal

Hypotheses H 0 : The median difference in number of species between singly-mating and multiply-mating insect groups is 0. H A : The median difference in number of species between these groups is not 0.

7 out of 25 comparisons are negative P = 2 ( ) = 0.043

The sign test has very low power So it is quite likely to not reject a false null hypothesis.

Non-parametric test to compare 2 groups The Mann-Whitney U test compares the central tendencies of two groups using ranks

Performing a Mann-Whitney U test First, rank all individuals from both groups together in order (for example, smallest to largest) Sum the ranks for all individuals in each group --> R 1 and R 2

Calculating the test statistic, U

Example: Garter snake resistance to newt toxin Rough-skinned newt

Comparing snake resistance to TTX (tetrodotoxin) This variable is known to be not normally distributed within populations.

Hypotheses H 0 : The TTX resistance for snakes from Benton is the same as for snakes from Warrenton. H A : The TTX resistance for snakes from Benton is different from snakes from Warrenton.

Calculating the ranks Rank sum for Warrenton: R 1 = =17

Calculating U 1 and U 2 For a two-tailed test, we pick the larger of U 1 or U 2 : U=U 1 =33

Compare U to the U table Critical value for U for n 1 =5 and n 2 =7 is >30, so we can reject the null hypothesis Snakes from Benton have higher resistance to TTX.

How to deal with ties Determine the ranks that the values would have got if they were slightly different. Average these ranks, and assign that average to each tied individual Count all those individuals when deciding the rank of the next largest individual

Ties

Mann-Whitney: Large sample approximation For n 1 and n 2 both greater than 10, use Compare this Z to the standard normal distribution

Example: U 1 =245 U 2 =80 n 1 =13 n 2 =25 Z 0.05(2) =1.96, Z>1.96, so we could reject the null hypothesis

Assumption of Mann-Whitney U test Both samples are random samples.