My assumptions are violated!!

• Homework #8 (Chap. 10 and 11) due this Wed at 5 pm
• Lab #7 (Cleaning Data in R) due April 12-13
• Projects, Part 1 and 2: discuss in lab this week
• Reading Assignment #14 for Wednesday: Whitlock & Schluter, Chapter 13: Handling violations of assumptions

Blood sugar levels

Experimental Design: Phlebotomist takes 15 samples from each of 10 patients, yielding a total of 150 measurements.

Discuss: What is the replicate and sample size in this situation? Why?

Antibiotics and bacterial growth rates

Experimental Design: Two agar plates: one with antibiotic, one without. Spread bacteria on both plates, let them grow for 24 hours, then measure diameter of 100 colonies on each plate?

Discuss: What is the replicate and sample size in this situation? Why?

Three things:

• Plan for precision (estimation)
• Plan for power (hypothesis testing)
• Plan for data loss

We'll use a two-sample \( t \)-test as the example in this section.

We would like to compute a 95% confidence interval for \( \mu_{1}-\mu_{2} \).

\[ \bar{Y}_{1}-\bar{Y}_{2} \pm \mathrm{margin \ of \ error}, \]

where “margin of error” is the half-width of the 95% confidence interval.

In this case, the following formula is an approximation to the number of samples needed to achieve the desired margin of error (assuming balanced design, i.e. \( n_{1}=n_{2}=n \)):

\[ n \approx 8\left(\frac{\mathrm{margin \ of \ error}}{\sigma}\right)^{-2} \]

Two-sample \( t \)-test:

\[ H_{0}: \mu_{1} - \mu_{2} = 0. \] \[ H_{A}: \mu_{1} - \mu_{2} \neq 0. \]

A conventional power to aim for is 0.80, i.e. we aim to prove \( H_{0} \) is false in 80% of experiments.

Assuming a significance level of 0.05, a quick approximation to the planned sample size \( n \) in each of two groups is

\[ n \approx 16\left(\frac{D}{\sigma}\right)^{-2}, \]

where \( D = |\mu_{1}-\mu_{2}| \) is the effect size.

library(pwr)

Two-sample \( t \)-test with significance level 0.05, 80% power, and relative effect size \( d = \frac{|\mu_{1}-\mu_{2}|}{\sigma} = 0.3 \).

pwr.t.test(d=0.3, power=0.8, type="two.sample")

     Two-sample t test power calculation 

              n = 175.3847
              d = 0.3
      sig.level = 0.05
          power = 0.8
    alternative = two.sided

NOTE: n is number in *each* group

• Whitlock & Schluter, Interleaf 2: Pseudoreplication (pp. 115-116)
• Whitlock & Schluter, Chapter 14: Designing experiments

Four options for handling violations of assumptions:

• Ignore the violations of assumptions
• Transform the data
• Use a nonparametric method
• Use a permutation test (computer-intensive methods)

Need to detect deviations first

To check for normality, first (as always) look at your data. Histograms work best here.

The following data come from a normal distribution:

They don't look normal, but they:

• …don't have outliers
• …aren't skewed

Examples of data from non-normal distributions:

Definition: The normal quantile plot compares each observation in the sample with its quantile expected from the standard normal distribution. Points should fall roughly along a straight line if the data come from a normal distribution.

1. Sort measurements (\( x \))
2. Compute percentiles of \( x \) (cumulative probabilities, \( p \))
3. Compute standard normal quantiles from percentiles (\( q \))
4. Plot measurements against computed quantiles (\( q \) vs \( x \))

x <- sort(rnorm(20))  # (1)
p <- (1:20)/21  # (2)
q <- qnorm(p, lower.tail = TRUE)  # (3)
plot(q ~ x, xlab="Measurements", ylab="Normal quantiles")  # (4)

x <- sort(rnorm(20))  # (1)
p <- (1:20)/21  # (2)
q <- qnorm(p, lower.tail = TRUE)  # (3)
plot(q ~ x, xlab="Measurements", ylab="Normal quantiles")  # (4)

Fast way (note: axes are flipped by default!)

qqnorm(x, datax = TRUE)

Question: Are marine reserves effective in preserving marine wildlife?

Experimental design

Halpern (2003) matched 32 marine reserves to a control location, which was either the site of the reserve before it became protected or a similar unprotected site nearby. They then evaluated the “biomass ratio,” which is the ratio of total masses of all marine plants and animals per unit area of reserve in the protected and matched unprotected areas.

Experimental design

Halpern (2003) matched 32 marine reserves to a control location, which was either the site of the reserve before it became protected or a similar unprotected site nearby. They then evaluated the “biomass ratio,” which is the ratio of total masses of all marine plants and animals per unit area of reserve in the protected and matched unprotected areas.

Discuss: Observational or experimental? Paired or unpaired? Interpret response measure in terms of effect of protection.

Answer: Observational. Paired (matching). Biomass ratio = 1 (no effect); > 1 (beneficial effect); < 1 (detrimental effect).

Practice Problem #4: Interpret the following normal quantile plots.