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 different Unknown

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 QuantileNormal distribution = straight line

Non-normal = non-straight line

Detecting differences from normality:

Shapiro-Wilk test

Shapiro-Wilk Test is used to test statisticallywhether a set of data comes from a normal distribition

Ho: The data come from a normal distributionHa: 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 = ln Y[ ]

Y Y' = ln[Y]

ln = “natural log”, base elog = “log”, base 10EITHER WORK

biomass ratio

ln[biomass ratio]

Biomassratio

ln[BiomassRatio]

1.34 0.301.96 0.672.49 0.911.27 0.241.19 0.181.15 0.141.29 0.26

Carry out the test on the transformed data!

The log transformation is often useful when:

• the variable is likely to be the result of multipli cation

of various components.

• the frequency distribution of the data is skewed to the

right

• the variance seems to increase as the mean gets larger

( in comparisons across groups).

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

Other transformations

Arc sine

€

′ p =arcsinp[ ]Sq uar e -r oo t

€

′ Y = Y+12Sq uar e

€

′ Y =Y2Re cipr o ca l

€

′ Y =1Y

An t il o g

€

′ Y =eY

Arcsine

Square-root

Square

Reciprocal

Antilog

Example: Confidence interval with log-transformed data

Data: 5 12 1024 12398ln data: 1.61 2.48 6.93 9.43

€

Y '= 5.11 slog Y[ ] = 3.70

Y '±t0.05 2( ),3

slog Y[ ]

n= 5.11± 3.18

3.70

4= 5.11± 5.88

−0.773 < μ log Y[ ] <10.99

0.46 < μ < 59278

ln[Y]

ln[Y]

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 = eX

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 onlyDo the populations have equal variances?

If so, what shouldWe do about it?

Comparing the variance of two groups

€

H0 :σ 12 = σ 2

2

HA :σ 12 ≠ σ 2

2

One possible method: the F test

The test statistic F

€

F =s1

2

s22

Put the larger s2 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 = ni -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

Polygamousspecies

Monogamousspecies

Mean -19.3 10.25

Samplevariance

243.9 2.27

Sample size 7 9

Example: Variation in insect genitalia

€

s12 = 243.9 s2

2 = 2.27

F =243.9

2.27=107.4

Degrees of freedom

€

df1 = 7 −1 = 6

df2 = 9 −1 = 8

F0.025,6,8 = 4.7

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 s2 in the numerator, we are forcing F to be greater than 1.

By the null hypothesis there is a 50:50 chance of either s2 being greater, so we want the higher tail to include just /2.

Critical value for F

Conclusion

The F= 107.4 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.

SampleNull hypothesis

The two populations have the same variance

1 2

2

F-test

Test statistic Null distributionF with n1-1, n2-1 dfcompare

How unusual is this test statistic?

P < 0.05 P > 0.05

Reject Ho Fail to reject Ho

€

F =s1

2

s22

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:

€

Y = 4.8

s = 3.26

€

Y = 0.51

s = 0.89

Hypotheses

H0: Owls catch the same number of dung beetles with or without extra dung (1 = 2)

HA: Owls do not catch the same number of dung beetles with or without extra dung (1 2)

Welch’s t

€

t =Y 1 − Y 2s1

2

n1

+s2

2

n2

€

df =

s12

n1

+s2

2

n2

⎛

⎝ ⎜

⎞

⎠ ⎟

2

s12 n1( )

2

n1 −1+

s22 n2( )

2

n2 −1

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

Round down df to nearest integer

Owls and dung beetles

€

t =Y 1 −Y 2s1

2

n1

+s2

2

n2

=4.8 − 0.51

3.262

10+

0.892

10

= 4.01

Degrees of freedom

€

df =

s12

n1

+s2

2

n2

⎛

⎝ ⎜

⎞

⎠ ⎟

2

s12 n1( )

2

n1 −1+

s22 n2( )

2

n2 −1

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

=

3.262

10+

0.892

10

⎛

⎝ ⎜

⎞

⎠ ⎟

2

3.262 10( )2

10 −1+

0.892 10( )2

10 −1

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

=10.33

Which we round down to df= 10

Reaching a conclusion

t0.05(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?

SampleNull hypothesis

The two populations have the same mean

1

2

Welch’s t-test

Test statistic Null distributiont with df from formulacompare

How unusual is this test statistic?

P < 0.05 P > 0.05

Reject Ho Fail to reject Ho

€

t =Y 1 − Y 2s1

2

n1

+s2

2

n2

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?

Order Family Multiple matinggroup

Numberof

species

Singlemating group

Numberof

speciesBeetles Anobiidae Ernobius 53 Xestobium 10

Dermestidae Dermestes 73 Trogoderma 120Elateridae Agriotes 228 Selatosomus 74

Flies Muscidae Coenosia 353 Delia 289Cecidomyiidae Rhopalomyia 157 Mayetiola 30Chironomidae Chironomus 300 Pontomyia 4Chironomidae Stictochironomus 34 Clunio 18Drosophilidaeand Culicidae

Drosophilidae 3,400 Culicidae 3,500

Dryomyzidaeand

Calliphoridae

Dryomyzidae 20 Calliphoridae 1,000

Tephritidae Anastrepha 196 Bactrocera 486Sciaridae and

BibionidaeSciaridae 1,750 Bibionidae 660

Scatophagidae Scatophaga 55 Musca 63Mayflies Siphlonuridae Siphlonurus 37 Caenis 115

Homoptera Psyllidae Cacopsylla 100 Aonidiella 30Butterfliesand moths

Noctuidae andPsychidae

Noctuidae 21,000 Psychidae 600

Tortricidae Choristoneura 37 Epiphyas 40Nymphalidae Eueides

(aliphera clade)7 Eueides

(vibiliaclade)

5

Nymphalidae Heliconius(silvaniform

clade)

15 Heliconius(sarasapho

clade)

7

Nymphalidae Polygonia / 18 Nymphalis 6

Etc....

The differences are not normal

-5000 0 5000 10000 20000

43 -47 154 64 127 296 16-100 -980 -290 1090 -8 -78 70

20940 -3 2 8 12 227 161 1 79 78

Hypotheses

H0: The median difference in number of species between singly-mating and multiply-mating insect groups is 0.

HA: The median difference in number of species between these groups is not 0.

7 out of 25 comparisons are

negative43 -47 154 64 127 296 16

-100 -980 -290 1090 -8 -78 7020940 -3 2 8 12 227 1

61 1 79 78

€

Pr X ≤ 7[ ] =25

i

⎛

⎝ ⎜

⎞

⎠ ⎟

i =1

7

∑ 0.5( )i

0.5( )25−i

= 0.02164

P = 2 (0.02164) = 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 tendenciesof 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 --> R1 and R2

Calculating the test statistic, U

€

U1 = n1n2 +n1 n1 +1( )

2− R1

€

U2 = n1n2 − U1

Example: Garter snake resistance to newt

toxin

Rough-skinned newt

Comparing snake resistance to TTX (tetrodotoxin)

Locality ResistanceBenton 0.29Benton 0.77Benton 0.96Benton 0.64Benton 0.70Benton 0.99Benton 0.34

Warrenton 0.17Warrenton 0.28Warrenton 0.20Warrenton 0.20Warrenton 0.37

This variable is known to be not normally distributed within populations.

Hypotheses

H0: The TTX resistance for snakes from Benton is the same as for snakes from Warrenton.

HA: The TTX resistance for snakes from Benton is different from snakes from Warrenton.

Calculating the ranksLocality Resistance RankBenton 0.29 5Benton 0.77 10Benton 0.96 11Benton 0.64 8Benton 0.70 9Benton 0.99 12Benton 0.34 6

Warrenton 0.17 1Warrenton 0.28 4Warrenton 0.20 2.5Warrenton 0.20 2.5Warrenton 0.37 7

Rank sum for Warrenton: R1=1+4+2.5+2.5+7=17

Calculating U1 and U2

€

U1 = n1n2 +n1 n1 +1( )

2− R1 = 5 7( ) +

5 6( )2

−17 = 33

€

U2 = n1n2 −U1 = 5 7( ) − 33 = 2

For a two-tailed test, we pick the larger of U1 or U2:

U=U1=33

Compare U to the U table

• Critical value for U for n1 =5 and n2=7 is 30

• 33 >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

Group Y Rank2 12 12 14 21 17 31 19 4.52 19 4.51 24 62 27 71 28 8

Mann-Whitney: Large sample approximation

For n1 and n2 both greater than 10, use

€

Z =2U − n1n2

n1n2 n1 + n2 +1( ) / 3

Compare this Z to the standard normal distribution

Example:

U1=245 U2=80n1=13 n2=25

€

Z =2U − n1n2

n1n2 n1 + n2 +1( ) /3

=2 245( ) −13 25( )

13 25( ) 13+ 25 +1( ) /3

= 2.54

Z0.05(2)=1.96, Z>1.96, so we could reject the null hypothesis

Assumption of Mann-Whitney U test

Both samples are random samples.