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Part III The General Linear Model Chapter 10 GLM. ANOVA.

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Part IIIThe General Linear ModelChapter 10GLM. ANOVA.

Chapter 10.1

Single Sample t-test

William Sealy Gosset

AKA Student

V

D

F

G

- Verbal
- Drug A increases time slept.

- Graphical model
- Define quantity of interest

T

=

TDrugA

–

TControl

V

D

F

G

- Formal
- Response:
- Explanatory:
- Formal:

?

- Straight line model ok?
- Errors homogeneous?
- Errors normal?
- Errors independent?

NA

- Straight line model ok?
- Errors homogeneous?
- Errors normal?
- Errors independent?

NA

- Straight line model ok?
- Errors homogeneous?
- Errors normal?
- Errors independent?

NA

NA

- Straight line model ok?
- Errors homogeneous?
- Errors normal?
- Errors independent?

NA

NA

- Straight line model ok?
- Errors homogeneous?
- Errors normal?
- Errors independent?

NA

Differences in time slept relative to control for all possible subjects

All possible differences in time slept relative to control, given the experimental protocol

Differences in time slept relative to control, given the experimental protocol

- Does Drug A affect hours of sleep?
- Don’t know if the answer is yes or no

6. State HA / Ho pair, with tolerance for Type I error

- HA:
- Ho:
- State test statistic:
- Distribution of test statistic:
- Tolerance for Type I error:

- Assumptions met so no need to recompute

9. Declare and report decision about model terms (compare p to α).

- p = 0.218 < α = 0.05, so reject HA: βo ≠ 0
- Report decision:
- There is no significant difference in extra time slept, for drug A (F1,9= 1.76, p = 0.218)
- But might Type II error be a problem here?

9. Declare and report decision about model terms (compare p to α).

- There may be a difference, but it is hidden in the variance
- Power analysis:
- Compute the minimum detectable difference
- T = 0.75 hours F = 1.76 p = 0.218
- T = 1.00 hours F = 3.12 p = 0.111
- T = 1.25 hours F = 4.88 p = 0.054
- T = 1.28 hours F = 5.12 p = 0.050

- Compute the minimum detectable difference

9. Declare and report decision about model terms (compare p to α).

- Another experiment, with more subjects, should be considered before concluding there is no evidence of an effect
- Power analysis:
- Compute sample size needed to detect a difference
- n = 10 F = 1.76 p = 0.218
- n = 20 F = 3.71 p = 0.0692
- n = 24 F = 4.49 p = 0.0451

- Compute sample size needed to detect a difference

10. Report and interpret parameters of biological interest.

- Parameters are not of interest there because there appears to be no difference
- BUT, with this sampling effort and variability, the study needs to be repeated to be conclusive.

Chapter 10.2

Two Sample t-test

William Sealy Gosset

AKA Student

V

D

F

G

- Verbal
- Extra time slept depends on drug

- Graphical model
- Formal model
- Response:
- Explanatory:
Measurement scale?

lm1 <- lm(diff~drug, data=drugs)

Parameter estimates

Based on output:

GLM routine:

lm1 <- lm(diff~drug, data=drugs)

Parameter estimates

Based on output:

GLM routine:

- Straight line model ok?
- Errors homogeneous?
- Errors normal?
- Errors independent?

NA

- Straight line model ok?
- Errors homogeneous?
- Errors normal?
- Errors independent?

NA

- Straight line model ok?
- Errors homogeneous?
- Errors normal?
- Errors independent?

NA

- Straight line model ok?
- Errors homogeneous?
- Errors normal?
- Errors independent?

NA

- Straight line model ok?
- Errors homogeneous?
- Errors normal?
- Errors independent?

X

NA

- Straight line model ok?
- Errors homogeneous?
- Errors normal?
- Errors independent?

X

- All possible differences in time slept relative between the two groups

5. Decide on mode of inference. Is hypothesis testing appropriate?

- Yes. The question is whether one drug is better than the other.
- It is not clear whether the greater hours of sleep due to the one drug is more than just chance.

6. State HA / Ho pair, with tolerance for Type I error

- HA:
- Ho:
- State test statistic:
- Distribution of test statistic:
- Tolerance for Type I error:

- When assumptions not met, recompute if:
- n small (n = 19, so _____)
- p near α (p =0.079, so _____)

- Colquhoun (1971) carried out a randomization test
- p = 0.0813 (976/12000)

9. Declare and report decision about model terms (compare p to α).

- p = 0.0813< α = 0.05, so reject HA
- Report decision:
- There is no significant difference in extra time slept for the two drugs (F1,18= 3.46, p = 0.081)
- Again, Type II error may be a problem
- Run a Power Analysis to guide future study

10. Report and interpret parameters of biological interest.

- Parameters are not of interest there because there appears to be no difference
- BUT, with this sampling effort and variability, the study needs to be repeated to be conclusive.
- The inclusion of 10 more samples may allow the detection of a significant difference

Chapter 10.3

One way ANOVA, Fixed Effects

- Pea section growth data, from Box 9.4 in Sokal and Rohlf (1995).
- Does growth depend on treatment (control versus 4 different sugars with auxin present)?

ε

+

=

- Verbal
- Pea section length in treated groups differ from the control (untreated) group.

- Graphical model
- Formal model
- Response:
- Explanatory:
- Fixed effect ↗
Measurement scale?

- Fixed effect ↗

lm1 <- lm(len~trt, data=peas)

Parameter estimates

- Straight line model ok?
- Errors homogeneous?
- Errors normal?
- Errors independent?

NA

- Straight line model ok?
- Errors homogeneous?
- Errors normal?
- Errors independent?

NA

- Straight line model ok?
- Errors homogeneous?
- Errors normal?
- Errors independent?

NA

- Straight line model ok?
- Errors homogeneous?
- Errors normal?
- Errors independent?

NA

- Straight line model ok?
- Errors homogeneous?
- Errors normal?
- Errors independent?

NA

- Straight line model ok?
- Errors homogeneous?
- Errors normal?
- Errors independent?

NA

- Straight line model ok?
- Errors homogeneous?
- Errors normal?
- Errors independent?

- Population is all possible measurements, given the method of applying treatments and the protocol for taking measurements.
- It is taken to be representative (i.e. not biased)

5. Decide on mode of inference. Is hypothesis testing appropriate?

- Yes. We want to know if there are any differences between treatments
- ANOVA tells us if there are ANY differences in variance among groups

Research Hypothesis (HA)

Null Hypothesis (Ho)

Treatment effects do not differ

μC= μG= μF= μG+F= μS

OR

var(βTrtTrt) = 0

Treatment effects differ

μC≠ μG≠ μF≠ μG+F≠ μS

OR

var(βTrtTrt) > 0

- 5·5 = 25 possible comparisons

- Based on these, we can undertake planned or a priori comparisons
- Examples:
- HA: βC ≠ (1/4)(βG+ βF+ βG+F + βS) [Control vs. Treatment]
- HA: βG+F≠ (1/3)(βG+ βF+βS)[Mixed vs. Pure]
- HA: βS≠ (1/2)(βG+ βF)[Poly vsMono]

- State test statistic :F-ratio
- Distribution of test statistic: F-distribution
- Tolerance for Type I error: 5%
- BUT we need to adjust α for planned comparisons
- 5% = 1 in 20; hence, we would likely reject one true HA if we did 20 comparisons
- Adjust α level for three comparisons (use Dunn-Sidak method)
- αexpwise= 1– (1 – α)k = 1 – (1 – 0.05)3 = 0.017

n = 50

- Assumptions met, skip

9. Declare decision about model terms.

- p < 0.001
- p < 0.05 so accept HA That μC≠ μG≠ μF≠ μG+F≠ μS

- F4,45 = 49.37, p < 0.0001

10. Report and interpret parameters of biological interest.

- Where are the differences, among the 5 groups?
- Two approaches: A priori
A posteriori

- Two approaches: A priori
- First planned comparison: Growth in treated media differs from that in untreated

10. Report and interpret parameters of biological interest.

- Run t-test on Control vs. Treatment

lm2 <- lm(len~comp1, data=peas)

anova(lm2)

Analysis of Variance Table

Response: len

Df Sum Sq Mean Sq F value Pr(>F)

comp1 1 832.32 832.32 81.45 6.516e-12 ***

Residuals 48 490.50 10.22

10. Report and interpret parameters of biological interest.

- First work out UCL ≤ mean ≤ LCL
Control: 67.6 ≤ 70.1 ≤ 72.6 units (n=40)

Treatment: 58.9 ≤59.9 ≤ 60.9 units (n=40)

- Then work out degree of difference
(μCon – μTrt)/μCon

(70.1 - 59.9)/70.1 = 15%

- Sugar supressed growth by 15%

10. Report and interpret parameters of biological interest.

- Conclusions from the 3 planned comparisons:
- A 2% sugar solution reduces growth
- F1,45 = 152.564, α= 0.017 > p < 0.0001

- Mixed glucose + fructose reduces growth relative to pure sugars
- F1,45 = 8.82, α= 0.017 > p = 0.00476

- The monosaccharides (fructose, glucose) suppress growth more than the polysaccharide (sucrose)
- F = 34.98, α= 0.017 > p = 0.00000417

- A 2% sugar solution reduces growth
- Conclusions reaffirmed using CL

Chapter 10.4

One way ANOVA, Random Effects

- Example. Box 9.1 of Sokal and Rohlf 1995, p. 210.
- Does tick size, as measured by scutum width, differ among hosts (rabbits)?
- Random effects example – contrast with fixed effects

- Verbal
- Scutum width depends on host identity

- Graphical model NEXT
- Formal model
- Response:
- Explanatory: (Random effect)

βH = [+12.55, -5.33, -4.4, +1.6]

βo = 359.7

SSTotal

βH = [+12.55, -5.33, -4.4, +1.6]

βo = 359.7

SSH

βH = [+12.55, -5.33, -4.4, +1.6]

βo = 359.7

SSRes

- Execute analysis.
- Evaluate model.
- State the population and whether the sample is representative.
- Fixed vs. Random
- Fixed: All possible measurement of scutum widths from ticks on these four rabbits only
- A fixed factor has levels that are the only ones of interest (e.g. different sugar treatments).

- Random: All possible measurement of scutum widths from ticks found on all possible rabbits
- A random factor has levels that are considered a sample from some larger population of levels (such as rabbits)

- Fixed: All possible measurement of scutum widths from ticks on these four rabbits only

- Fixed vs. Random

Depends on context

Because we want to infer to a larger population level

Because we’re usually interested in specific contrasts

- Decide on mode of inference. Is hypothesis testing appropriate?
- State HA / Ho pair, test statistic, distribution, tolerance for Type I error.
- Q: Is additional variation in size due to their host?
- HA: Var(βH·H) > 0
- Ho: Var(βH·H) = 0

- ANOVA - Compute and partition the df in the response variable according to the model

0.004

Recommendation: Cross-check your workings with the computer generated ANOVA table.

This ensures the computer did what you wanted it to do!

- Recomputep-value by randomization if necessary.
- Declare and report statistical decision, with evidence
- The variance among hosts exceeds variance within hosts (F3,33= 5.26, p = 0.004)

- Report and interpret parameters of biological interest.
- In this example the interest was in whether there was variance among the hosts.
- There was no stated interest in which hosts differed, or by how much.