Tenerife gecko

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Tenerife gecko

• Distribution Canary Islands
• It has been suggested that individuals of this
  species vary in size between islands
• Use body length to assess body size difference
• Plan take a random sample of 30 individuals per
  island
• H0 length does not differ between islands
• H1 length does differ between islands

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The circled islands were included in the study
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Gecko example

• H0
• H1
• H0
• H1
• .
• .
• .
• H0
• H1

From lecture 1 two sample hypothesis In this
case we need to test all pairwises difference
among all sampled populations. Since we have 5
populations we need to make 10 simultaneous
t-tests
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For each of those two sample hypothesis tests

• For each separate test we have a
• significance level of
• The probability of making a type I error
  rejecting a
• true null hypothesis
• The probability of NOT making a type I error

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The real significance level for all 10
simultaneous tests

• The probability of not making a type I error in
• any of the tests is
• The probability of making a type I error in at
• least one of the 10 tests is

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The real significance level

• The probability of not making a type I error in
• any of the tests is
• The probability of making a type I error in at
• least one of the 10 tests is
• HUGE!

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• So why do we not just make smaller?
• For example if we solve
• We get
• However, remember from last lecture If you
  decrease
• the Type I error you increase the type II error,
• meaning that you are reducing the statistical
• power of each test. If other analyses are
  available with
• a higher power choose that.

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• We are considering if there is a difference in
  the body lengths
• of the geckos. We therefore want to write our
  hypothesis in
• the form that there is NO difference among the
  islands.
• Denote the expected mean for each island
  population i as
• i1,..,5. Our hypothesis is then
• H0
• H1 the body length is not the same on all
  islands

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We make two important assumptions

• Body length has a normal distribution within each
• island.
• This variable has the same variance on all
  islands

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• Body length is the variable we are testing
• Island is a factor that we want to make
• inferences about
• Each island is a level of the factor
• In this example there is one factor with five
  levels

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• Our null hypothesis states that all islands have
• the same expected mean. We made the
• assumption that body length has a normal
• distribution. We also assumed that all
• populations had the same variance.
• So the null hypothesis basically says that we
• should consider all islands as one big
• population with the same expected mean and
• variance.

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Say that the null hypothesis is true

• And that expected mean is 22.6 while the
• variance is 5.
• Even if the five islands have exactly the same
• distribution, it does not mean that they are
• going to produce exactly the same
• observations.

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These 5 samples (each with 1000 individuals) have
been sampled from exactly the same normal
distribution with expected mean 22.6 and variance
5.
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These 5 samples (each with 15 individuals) have
been sampled from exactly the same normal
distribution Conclusion even if the null
hypothesis is true we do not expect to find
exactly the same mean and variance for each
island. There is going to be some differences
just by chancesampling error.
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• 1. If null hypothesis is true
• A) There is going to be some differences among
• populations that has to do with sampling from a
  probability
• distribution
• 2. If the null hypothesis is not true there
• A) There is going to be some differences among
• populations that has to do with sampling from a
  probability
• distribution
• B) There is going to some differences among
  populations that
• has to do with actual difference between the
  means among the
• islands

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Means Island 1 25 Island 2 20 Island 3
18 Island 4 28 Island 5 22
These islands are sampled from populations with
variance (4) .
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Means Island 1 25 Island 2 20 Island 3
18 Island 4 28 Island 5 22
These islands are sampled from populations that
have variance (100). Compare the means with the
previous slide, they are the same but the outcome
look very different due to difference in variance!
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• In the first case we had a small variance
• within each population, in the second the
• variance is much larger. The means were exactly
• the same in the two examples.
• We therefore need to construct a test that
• compares the difference between the islands
• and at the same time takes the within island
  variation
• into account!

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ANOVAANALYSIS OF VARIANCE

• Idea Partition the total variation into
  variation
• within islands and variation among islands
• SSTSSISLANDSSE
• These sums of squares are then used to estimate
  the MSISLAND
• and MSE (MSmean square) which we use in the
  actual test

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Sum of squares total SST

• To get the variation for the whole data set
• 1. Calculate the average for all
  observationsgrand mean
• 2. Obtain the difference between an observation
  and the grand mean
• 3. Square the difference
• 4. Repeat for each observation (we have 530150)
• 5. Add up all squared differences and you have
  SST

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Sum of squares error SSE

• To get the variation within each island
• 1. Calculate the mean for the island
• 2. Obtain the difference between an island
  observation and the island mean
• 3. Square the difference
• 4. Repeat for each island observation
• 5. Add up all squared differences
• Then add up all islands and you have the SSE

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To get then get the Mean Square Error term MSE

• Divide the SSE with degrees of freedom for
• the error
• So what is this?
• This an estimate of
• the within population variance

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Sum of squares islands SSISLAND

• To get the variation among islands
• 1. Calculate the mean of each island
• 2. Calculate the grand mean, the mean of all
  observations in study
• 3. Obtain the difference between the an island
  mean and the grand mean
• 4. Square the difference
• 5. Repeat for each island mean
• 6. Add up all squared differences
• This is the SSISLAND

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To get the Mean Square Island term MSISLAND

• Divide the SSISLAND with degrees of freedom
• for the island
• (degrees of freedom islands 5-1 4)

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• If the null hypothesis is true then we would
• expect the variation between islands to be
• small compared to the variation within islands
• So should be small. How small?
• If the null hypothesis is true then
• have a F-distribution with dfisland, dferror

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How do I get the degrees of freedom for the error
term?

• dftotal is 530-1149
• dfisland is 5-14
• We know that dftotaldfislanddferror
• dferrordftotal-dfisland 149-4145

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The test itself

• Calculate the F
• If Fgt then reject the null
  hypothesis
• Note that we are only interested if MSISLAND is
  large in
• comparison to MSE, not the reversed situation,
  this is why we
• are using a one-tailed test

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Each observation can be described as a linear
statistical model
i1,,a j1,..,n Where where
is the population mean, Ai is the effect of
level i of factor A, and is the random
error component
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This looks much worse than it is!

• We can break it down in the following way

This is not true in a strict sense
This is how far away the observation is from the
mean of that specfic level
This is how far away the mean of that specfic
level is from the total mean
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Random versus fixed effects

• We are looking at the effect the factor island
• has on the body length of these geckos.
• Fixed effect the experimenter chooses the levels
  of the factor for whatever reason
• Random effect the experimenter randomly selects
  a set of all possible levels that can be studied

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• Fixed effect model
• you can only draw conclusions for the levels
  investigated
• Random effect model
• you can make inferences on the whole set of
  levels
• So how does this apply to our gecko example for
• Do not reject null hypothesis
• Reject null hypothesis

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Purple sandgrass (Triplasis purpurea)
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Some background

• In coastal ecosystems, two important abiotic
• factors which influence growth and
• reproduction of plants are
• Airborne saltwater spray
• Sand deposition
• It has been suggested that some species have
  evolved
• adaptation (tolerance) to airborne salt

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Research question

• What is the impact of both saltwater spray and
• partial sand burial of seedling on plant growth
• and seed production?
• Adapted from American Journal of Botany 86
  703-710. 1999.

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• Investigated variable shoot mass
• 135 hundred seedlings are randomly selected
• at a beach and returned to greenhouse and
• then planted into pots

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Two factors

• Factor A saltwater spray
• Levels
• No spray
• 2 sprays/week
• 6 sprays/week
• Factor B sand burial
• Levels
• 1) unburied
• 2) buried to 50 height
• 3) buried to 75 height

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This design allows us to set up the following
hypotheses

• H0 There is no difference between the different
  levels of burial.
• HA There is difference between the different
  levels of burial.
• H0 There is no difference between the different
  levels of saltwater
• spray.
• HA There is difference between the different
  levels of saltwater
• spray.
• H0 There is no interaction between saltwater
  spray and burial.
• HA There is interaction between saltwater spray
  and burial.

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observations
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• Linear model
• Partition the sums of squares just like for the
• one way ANOVA

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Conclusion Both Burial and Spray show
significant effect on shoot mass. But there is no
interaction. WHY? Do we get the whole story from
this analysis?
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• If you get significant results you do not know
• which levels are significantly different from
  each
• other. This can be done by pairwise testing for
• example Tukeys test and and other available
• comparisons (we will not cover this here)
• In our purple sandgrass example
• saltwater spray had a negative effect on the
  shoot
• mass while the more buried the larger the shoot
  mass
• Adaptation to sand deposition on seedlings?
• Adaptation to saltwater spray on seedlings?

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Barn Swallows in Ukraine

• Partial alibinism in barn swallows generally
• rare, less than 1
• In Chernobyl this frequency is 13-15
• Albinism is usually due to mutation, in Chernobyl
  it
• is probably elevated due to radioactive
• contamination (nuclear catastrophe in May 1986)

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• Research question Are albino barn swallows
• smaller than non-albino?
• This could then in turn have implications for
• the evolution of the barn swallows. If albinism
  is
• NOT decreasing due to natural selection, then
  barn
• Swallows is general would get smaller.
• If albinism is affected by natural selection then
  the
• frequency of albinism should decrease with time.

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• Hypothesis There is no difference in body size
  between albino birds and wildtype (non-albino)
  birds
• There is a problem Males and females may not
• have equal size, so variation in our error term
• may come from that. Include therefore sex as
• factor.
• Hypothesis There is no difference in body size
  between males and females

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Samples came from 3 different years 1991, 1996,
2000

• There might be an effect of the year of
• sampling on the body size of the barn
• Swallows
• Hypothesis There is no difference in size
• between years.
• In, addition, there could be interactions among
  the
• three different factors. These need also to be
  tested