# A Probability Problem in Heredity – Part 3

In my previous two posts I showed worked solutions to problems 2.5 and 11.7 in Bulmer’s Principles of Statistics, both of which involve the characteristics of self-fertilizing hybrid sweet peas. It turns out that problem 11.8 also involves this same topic, so why not work it as well for completeness. The problem asks us to assume that we were unable to find an explicit solution for the maximum likelihood equation in problem 11.7 and to solve it by using the following iterative method:

$$\theta_{1} = \theta_{0} + \frac{S(\theta_{0})}{I(\theta_{0})}$$

where $$S(\theta_{0})$$ is the value of $$\frac{d \log L}{d\theta}$$ evaluated at $$\theta_{0}$$ and $$I(\theta_{0})$$ is the value of $$-E(\frac{d^{2}\log L}{d\theta^{2}})$$ evaluated at $$\theta_{0}$$.

So we begin with $$\theta_{0}$$ and the iterative method returns $$\theta_{1}$$. Now we run the iterative method again starting with $$\theta_{1}$$ and get $$\theta_{2}$$:

$$\theta_{2} = \theta_{1} + \frac{S(\theta_{1})}{I(\theta_{1})}$$

We repeat this process until we converge upon a value. This is called the Newton-Raphson method. Naturally this is something we would like to have the computer do for us.

First, recall our formulas from problem 11.7:

$$\frac{d \log L}{d\theta} = \frac{1528}{2 + \theta} – \frac{223}{1 – \theta} + \frac{381}{\theta}$$
$$\frac{d^{2}\log L}{d \theta^{2}} = -\frac{1528}{(2 + \theta)^{2}} -\frac{223}{(1 – \theta)^{2}} -\frac{381}{\theta^{2}}$$

Let’s write functions for those in R:

# maximum likelihood score
mls <- function(x) {
1528/(2 + x) - 223/(1 - x) + 381/x
}
# the information
inf <- function(x) {
-1528/((2 + x)^2) - 223/((1 - x)^2) - 381/(x^2)
}



Now we can use those functions in another function that will run the iterative method starting at a trial value:

# newton-raphson using expected information matrix
nr <- function(th) {
prev <- th
repeat {
new <- prev + mls(prev)/-inf(prev)
if(abs(prev - new)/abs(new) <0.0001)
break
prev <- new
}
new
}


This function first takes its argument and names it "prev". Then it starts a repeating loop. The first thing the loop does it calculate the new value using the iterative formula. It then checks to see if the difference between the new and previous value - divided by the new value - is less than 0.0001. If it is, the loop breaks and the "new" value is printed to the console. If not, the loop repeats. Notice that each iteration is hopefully converging on a value. As it converges, the difference between the "prev" and "new" value will get smaller and smaller. So small that dividing the difference by the "new" value (or "prev" value for that matter) will begin to approach 0.

To run this function, we simply call it from the console. Let's start with a value of $$\theta_{0} = \frac{1}{4}$$, as the problem suggests:

nr(1/4)
[1] 0.7844304


There you go! We could make the function tell us a little more by outputting the iterative values and number of iterations. Here's a super quick and dirty way to do that:

# newton-raphson using expected information matrix
nr <- function(th) {
k <- 1 # number of iterations
v <- c() # iterative values
prev <- th
repeat {
new <- prev + mls(prev)/-inf(prev)
v[k] <- new
if(abs(prev - new)/abs(new) <0.0001)
break
prev <- new
k <- k + 1
}
print(new) # the value we converged on
print(v) # the iterative values
print(k) # number of iterations
}


Now when we run the function we get this:

nr(1/4)
[1] 0.7844304
[1] 0.5304977 0.8557780 0.8062570 0.7863259 0.7844441 0.7844304
[1] 6


We see it took 6 iterations to converge. And with that I think I've had my fill of heredity problems for a while.

# A Probability Problem in Heredity – Part 2

In my last post I solved a problem from chapter 2 of M.G. Bulmer’s Principles of Statistics. In this post I work through a problem in chapter 11 that is basically a continuation of the chapter 2 problem. If you take a look at the previous post, you’ll notice we were asked to find probability in terms of theta. I did it and that’s nice and all, but we can go further. We actually have data, so we can estimate theta. And that’s what the problem in chapter 11 asks us to do. If you’re wondering why it took 9 chapters to get from finding theta to estimating theta, that’s because the first problem involved basic probability and this one requires maximum likelihood. It’s a bit of a jump where statistical background is concerned.

The results of the last post were as follows:

 purple-flowered red-flowered long pollen $$\frac{1}{4}(\theta + 2)$$ $$\frac{1}{4}(1 – \theta)$$ round pollen $$\frac{1}{4}(1 – \theta)$$ $$\frac{1}{4}\theta$$

The table provides probabilities of four possible phenotypes when hybrid sweet peas are allowed to self-fertilize. For example, the probability of a self-fertilizing sweet pea producing a purple-flower with long pollen is $$\frac{1}{4}(1 – \theta)$$. In this post we’ll estimate theta from our data. Recall that $$\theta = (1 – \pi)^{2}$$, where $$\pi$$ is the probability of the dominant and recessive genes of a characteristic switching chromosomes.

Here’s the data:

 Purple-flowered Red-flowered Long pollen 1528 117 Round pollen 106 381

We see from the table there are 4 exclusive possibilities when the sweet pea self-fertilizes. If we think of each possibility having its own probability of occurrence, then we can think of this data as a sample from a multinomial distribution. Since chapter 11 covers maximum likelihood estimation, the problem therefore asks us to use the multinomial likelihood function to estimate theta.

Now the maximum likelihood estimator for each probability is $$\hat{p_{i}} = \frac{x_{i}}{n}$$. But we can’t use that. That’s estimating four parameters. We need to estimate one parameter, theta. So we need to go back to the multinomial maximum likelihood function and define $$p_{i}$$ in terms of theta. And of course we’ll work with the log likelihood since it’s easier to work with sums than products.

$$\log L(\theta) = y_{1} \log p_{1} + y_{2} \log p_{2} + y_{3} \log p_{3} + y_{4} \log p_{4}$$

If you’re not sure how I got that, google “log likelihood multinomial distribution” for more PDF lecture notes than you can ever hope to read.

Now let’s define the probabilities in terms of one parameter, theta:

$$\log L(\theta) = y_{1} \log f_{1}(\theta) + y_{2} \log f_{2}(\theta) + y_{3} \log f_{3}(\theta) + y_{4} \log f_{4}(\theta)$$

Now take the derivative. Once we have that we can set equal to 0 and find a solution for theta. The solution will be the point at which theta obtains its maximum value:

$$\frac{d \log L(\theta)}{d\theta} = \frac{y_{1}}{f_{1}(\theta)} f’_{1}(\theta) + \frac{y_{1}}{f_{2}(\theta)} f’_{2}(\theta) + \frac{y_{1}}{f_{3}(\theta)} f’_{3}(\theta) + \frac{y_{1}}{f_{4}(\theta)} f’_{4}(\theta)$$

Time to go from the abstract to the applied with our values. The y’s are the data from our table and the functions of theta are the results from the previous problem.

$$\frac{d \log L(\theta)}{d\theta} = \frac{1528}{1/4(2 + \theta)} \frac{1}{4} – \frac{117}{1/4(1 – \theta)}\frac{1}{4} – \frac{106}{1/4(1 – \theta)}\frac{1}{4} + \frac{381}{1/4(\theta)} \frac{1}{4}$$
$$\frac{d \log L(\theta)}{d\theta} = \frac{1528}{2 + \theta} – \frac{117}{1 – \theta} – \frac{106}{1 – \theta} + \frac{381}{\theta}$$
$$\frac{d \log L(\theta)}{d\theta} = \frac{1528}{2 + \theta} – \frac{223}{1 – \theta} + \frac{381}{\theta}$$

Set equal to 0 and solve for theta. Beware, it gets messy.

$$\frac{[1528(1 – \theta)\theta] – [223(2 + \theta)\theta] + [381(2 + \theta)(1 – \theta)]}{(2 + \theta)(1- \theta)\theta} = 0$$

Yeesh. Fortunately the denominator cancels out when we start multiplying terms and solving for theta. So we’re left with this:

$$1528\theta – 1528\theta^{2} – 446\theta – 223\theta^{2} + 762 – 381\theta – 381\theta^{2} = 0$$

And that reduces to the following quadratic equation:

$$-2132\theta^{2} + 701\theta + 762 = 0$$

I propose we use an online calculator to solve this equation. Here’s a good one. Just plug in the coefficients and hit solve to find the roots. Our coefficients are a = -2132, b = 701, and c = 762. Since it’s a quadratic equation we get two answers:

$$x_{1} = -0.4556$$
$$x_{2} = 0.7844$$

The answer is in terms of a probability which is between 0 and 1, so we toss the negative answer and behold our maximum likelihood estimator for theta: 0.7844.

Remember that $$\theta = (1 – \pi)^{2}$$. If we solve for pi, we get $$\pi = 1 – \theta^{1/2}$$, which works out to be 0.1143. That is, we estimate the probability of characteristic genes switching chromosomes to be about 11%. Therefore we can think of theta as the probability of having two parents that experienced no gene switching.

Now point estimates are just estimates. We would like to know how good the estimate is. That’s where confidence intervals come in to play. Let’s calculate one for theta.

It turns out that we can estimate the variability of our maximum likelihood estimator as follows:

$$V(\theta) = \frac{1}{I(\theta)}$$, where $$I(\theta) = -E(\frac{d^{2}\log L}{d \theta^{2}})$$

We need to determine the second derivative of our log likelihood equation, take the expected value by plugging in our maximum likelihood estimator, multiply that by -1, and then take the reciprocal. The second derivative works out to be:

$$\frac{d^{2}\log L}{d \theta^{2}} = -\frac{1528}{(2 + \theta)^{2}} -\frac{223}{(1 – \theta)^{2}} -\frac{381}{\theta^{2}}$$

The negative expected value of the second derivative is obtained by plugging in our estimate of 0.7844 and multiplying by -1. Let’s head to the R console to calculate this:

th <- 0.7844 # our ML estimate
(I <- -1 * (-1528/((2+th)^2) - 223/((1-th)^2) - 381/(th^2))) # information
[1] 5613.731


Now take the reciprocal and we have our variance:

(v.th <- 1/I)
[1] 0.0001781347


We can take the square root of the variance to get the standard error and multiply by 1.96 to get the margin of error for our estimate. Then add/subtract the margin of error to our estimate for a confidence interval:

# confidence limits on theta
0.784+(1.96*sqrt(v.th)) # upper bound
[1] 0.8101596
0.784-(1.96*sqrt(v.th)) # lower bound
[1] 0.7578404


Finally we convert the confidence interval for theta to a confidence interval for pi:

# probability of switching over
th.ub <- 0.784+(1.96*sqrt(v.th))
th.lb <- 0.784-(1.96*sqrt(v.th))
1-sqrt(th.ub) # lower bound
[1] 0.09991136
1-sqrt(th.lb) # upper bound
[1] 0.1294597


The probability of genes switching chromosomes is most probably in the range of 10% to 13%.

# A Probability Problem in Heredity

Here’s a fun problem in heredity from the Dover classic Principles of Statistics by M.G. Bulmer. It’s from chapter 2, problem 2.5.

The results of Table 7 on p. 25 (see below) can be explained on the assumption that the genes for flower colour and pollen shape are on the same chromosome but that there is a probability π that one of the genes will be exchanged for the corresponding gene on the other chromosome. If we denote the genes for purple or red flowers by P and p, and the genes for long and round pollen by L and l, then the hybrids from the cross considered will all be of the genotype PL/pl, the notation indicating that the P and L genes are on one chromosome and the p and l genes on the other. When these hybrids are allowed to self-fertilise, there is a chance π that the L and l genes will interchange in one parent, giving Pl/pL; there are therefore really three mating types, PL/pl X PL/pl, Pl/pL X PL/pl and Pl/pL x Pl/pL, which occur with probabilities $$(1 – \pi)^{2}$$, $$2\pi(1 – \pi)$$ and $$\pi^{2}$$ respectively. Find the probabilities of the four possible phenotypes resulting from the experiment in terms of $$\theta = (1 – \pi)^{2}$$.

Here’s the table the problem refers to:

 Purple-flowered Red-flowered Total Long pollen 1528 117 1645 Round pollen 106 381 487 Total 1634 498 2132

The interesting thing about that table is that it violates Mendel’s law of independent assortment. If it obeyed that law, then the probabilities of the resulting phenotypes would be:

 Purple-flowered Red-flowered Long pollen $$\frac{9}{16}$$ $$\frac{3}{16}$$ Round pollen $$\frac{3}{16}$$ $$\frac{1}{16}$$

We’d expect the purple/long pollen flower to happen about 9/16 = 56% of the time. Instead we see it occurring about 1528/2132 = 72% of the time. As the problem explains this has to do with the way genes are carried on chromosomes. This means we can’t calculate probabilities as you normally would for a dihybrid cross. Therefore it asks us to calculate those probabilities conditional on whether or not the gene for pollen switched chromosomes. Our answer will take the form above, but instead of actual numbers, we’ll express our answer in terms of $$\theta = (1 – \pi)^{2}$$. In other words theta is the probability that the gene for pollen did not switch chromosomes.

The hard part of this problem is setting it up. First, we have to recognize that we don’t know which chromosome has the characteristics. The characteristics of mating type PL/pl X PL/pl could both be on the PL chromosomes, in which case the result would be a PL, or a purple/long pollen flower. Or one could be PL and the other pl, which would still be PL, a purple/long pollen flower, since purple and long pollen are dominant characteristics. If both are on the pl chromosome, then the result would be pl, a red/round pollen flower. All told, for the PL/pl X PL/pl mating type we have the following possibilities:

 PL pl PL PL PL pl PL pl

For this mating type we get a purple/long pollen flower (PL) three out of four times and a red/round pollen flower (pl) one out of four times. We need to construct similar tables for the other two mating types. But notice the Pl/pL X PL/pl mating type can actually happen in reverse order as PL/pl X Pl/pL, so it needs two tables. Therefore we really need three more tables:

 PL pl Pl PL Pl pL PL pL
 Pl pL PL PL PL pl Pl pL
 Pl pL Pl Pl PL pL PL pL

Have to give a quick shout out to http://truben.no/latex/table/ for helping me make those tables! Anyway…so we have 4 tables displaying 16 possible results:

• 9 out of 16 times you get PL, a purple/long pollen flower
• 3 out of 16 times you get Pl, a purple/round pollen flower
• 3 out of 16 times you get pL, a red/long pollen flower
• 1 out of 16 times you get pl, a red/round pollen flower

If the results were independent, we could just call those probabilities. But they’re not. That’s the whole point of the problem. We have to take into account the probability π of the exchange of genes from one chromosome to the other. Let’s reset the probabilities for the four mating types:

1. PL/pl X PL/pl = $$(1-\pi)^{2}$$
2. Pl/pL X PL/pl = $$\pi(1-\pi)$$
3. PL/pl X pl/pL = $$\pi(1-\pi)$$
4. Pl/pL X Pl/pL = $$\pi^{2}$$

So the probability of getting pl in the first mating type is $$\frac{1}{4}(1-\pi)^{2}$$. Recall the problem asks us to find these probabilities in terms of $$\theta = (1 – \pi)^{2}$$, so we can express this as $$\frac{1}{4}\theta$$. And there’s one of our answers since pl does not occur in any of the other mating types.

Now we just need to find the other probabilities. To make life easier, let’s go ahead and convert all the probabilities in terms of $$\theta$$:

• $$(1 – \pi)^2 = \theta$$
• $$(1 – \pi) = \theta^{1/2}$$
• $$\pi = 1 – \theta^{1/2}$$
• $$\pi^{2} = (1- \theta^{1/2})^{2}$$

The hardest one to find is PL:

$$PL = \frac{3}{4}\theta + \frac{2}{4}(1 – \theta^{1/2})\theta^{1/2} + \frac{2}{4}(1 – \theta^{1/2})\theta^{1/2} + \frac{2}{4}(1 – \theta^{1/2})^{2}$$
$$PL = \frac{3}{4}\theta + \frac{2}{4}[2(1 – \theta^{1/2})\theta^{1/2} + 1 – \theta^{1/2} – \theta^{1/2} + \theta]$$
$$PL = \frac{3}{4}\theta + \frac{2}{4}[2(\theta^{1/2} – \theta) + 1 – 2\theta^{1/2} + \theta]$$
$$PL = \frac{3}{4}\theta + \frac{2}{4}[2\theta^{1/2} – 2\theta + 1 – 2\theta^{1/2} + \theta]$$
$$PL = \frac{3}{4}\theta + \frac{2}{4}(1 – \theta)$$
$$PL = \frac{3}{4}\theta + \frac{2}{4} – \frac{2}{4}\theta$$
$$PL = \frac{1}{4}\theta + \frac{2}{4}$$
$$PL = \frac{1}{4}(\theta + 2)$$

Next up is pL:

$$pL = \frac{1}{4}(1 – \theta^{1/2})\theta^{1/2} + \frac{1}{4}(1 – \theta^{1/2})\theta^{1/2} + \frac{1}{4}(1 – \theta^{1/2})^{2}$$
$$pL = \frac{1}{4}(\theta^{1/2} – \theta + \theta^{1/2} – \theta + 1 – \theta^{1/2} – \theta^{1/2} + \theta)$$
$$pL = \frac{1}{4}(-2\theta + 1 + \theta)$$
$$pL = \frac{1}{4}(1 – \theta)$$

That just leaves Pl. But if you look at the tables above, you’ll notice Pl appears in the same tables with pL the same number of times. So if we set up an equation to find its probability, we’ll get the same equation we started with when we solved Pl. That means we’ll get the same answer. So we don’t need to solve it. We already know it: $$pL = \frac{1}{4}(1 – \theta)$$

And that finishes the problem. The probabilities of the four possible phenotypes are as follows:

 purple-flowered red-flowered long pollen $$\frac{1}{4}(\theta + 2)$$ $$\frac{1}{4}(1 – \theta)$$ round pollen $$\frac{1}{4}(1 – \theta)$$ $$\frac{1}{4}\theta$$

# Testing Composite Hypotheses about Fixed Effects (Ch 4 of ALDA)

I’m still on my ALDA kick here, this time posting about section 4.7 of Chapter 4. In my last post I talked deviance-based hypothesis tests in the context of model building. Recall you have to be aware of which method of estimation you used when you deploy the deviance statistic. Namely, if you used Restricted Maximum Likelihood (RML) to estimate model parameters, you can only use the deviance statistic to test hypotheses about variance components. This is an important point as many programs default to RML, such as the lme4 package in R and SAS PROC MIXED. But the deviance statistic is not the only tool for testing composite hypotheses.

That leads us to section 4.7 and the Wald statistic. The Wald statistic allows you to “test composite hypotheses about multiple effects regardless of the method of estimation used. This means that if you use restricted methods of estimation, which prevent you from using deviance-based tests to compare models with different fixed effects, you still have a means of testing composite hypotheses about sets of fixed effects.” (p. 122) Sounds good to me!

The authors give two examples, one of which I want to review in this post. As usual they don’t show you how to do the test using statistical software. Unfortunately either does the UCLA stats consulting page for ALDA. So I had to figure it out on my own.

Let’s reset the example study motivating the work in this chapter. 82 adolescents were surveyed on alcohol use. Some of the variables collected included:

• alcuse, a rating-scale measure of alcohol use (the response)
• age_14, age of participant centered about 14 (14 = 0, 15 = 1, 16 = 2)
• coa, an indicator whether or not participant is a child of an alcoholic (1 = yes, 0 = no)
• cpeer, a rating-scale measure of alcohol use among peers centered on its sample mean of 1.018
• id, an arbitrary level to group persons

These variables are part of Model F, the model of interest in section 4.7, which aims to explain the variability in alcuse. (Models A through E precede Model F in the earlier model-building portion of the chapter.) Here’s Model F in its multilevel form:

level 1
$Y_{ij} = \pi_{0i} + \pi_{1i}*age14_{ij} + \epsilon_{ij}$

level 2
$\pi_{0i} = \gamma_{00} + \gamma_{01}*coa + \gamma_{02}*cpeer + \zeta_{0i}$
$\pi_{1i} = \gamma_{10} + \gamma_{12}*cpeer + \zeta_{1i}$

So this model posits that individuals have a liner trajectory over time (level 1), and that the parameters themselves of that linear trajectory differ between individuals based on coa and cpeer (level 2).

We can combine the two levels into one scary-looking composite representation of the model:
$Y_{ij} = \gamma_{00} + \gamma_{01}*coa + \gamma_{02}*peer + \gamma_{10}*age14 + \gamma_{12}*peer*age14 + \zeta_{0i} + \zeta_{1i}*age14 + \epsilon_{ij}$

Then we can estimate the parameters of that model in R with the following code:

alcohol1 <- read.table("http://www.ats.ucla.edu/stat/r/examples/alda/data/alcohol1_pp.txt",
attach(alcohol1)
library(lme)
model.f1 <- lmer(alcuse ~ coa + cpeer*age_14 + (age_14 | id), alcohol1, REML = FALSE)
summary(model.f1)


And now we are ready to test composite hypotheses about this model. The first example in the book asks whether the average child of non-alcoholic parents - with an average value of peer - drinks no alcohol at age 14 (intercept = 0) and remains abstinent over time (slope = 0). So we set coa and cpeer both equal to 0 in our composite model and we're left with this:

$Y_{ij} = \gamma_{00} + \gamma_{10}*age14 + \zeta_{0i} + \zeta_{1i}*age14 + \epsilon_{ij}$

Thus our question is essentially asking if the slope and intercept in this model are 0. Or to state it formally, our composite null hypothesis is as follows:

$H_{0}: \gamma_{00} = 0 \: and \: \gamma_{10} = 0$

Now to carry out this test, we need to express this hypothesis as a general linear hypothesis in matrix notation. First let's restate the hypothesis using our fixed effects:

$1\gamma_{00} + 0\gamma_{01} + 0\gamma_{02} + 0\gamma_{10} + 0\gamma_{12} = 0$
$0\gamma_{00} + 0\gamma_{01} + 0\gamma_{02} + 1\gamma_{10} + 0\gamma_{12} = 0$

We have weighted our coefficients so that the only two we're interested in are viable. Now we create a matrix of the weights. This is easier and faster to show in R than trying to use LaTeX, which I'm not even sure I can pull off with the Word Press plugin I'm using.

C <- matrix(c(1,0,0,0,0,0,0,0,1,0), nrow=2, byrow=TRUE)
C
[,1] [,2] [,3] [,4] [,5]
[1,]    1    0    0    0    0
[2,]    0    0    0    1    0


Now using the matrix we just created we can conduct a linear hypothesis test using the linearHypothesis function available in the car package, like so:

library(car)
linearHypothesis(model.f1, C)


This returns a Wald statistic of 51.03 on 2 degrees of freedom, almost matching the book which reports 51.01. The p-value is practically 0, which means we reject this composite hypothesis.

Now it's nice to know there's an R function that will calculate the Wald statistic, but what is it? How can we find out? The following code reveals the source of linearHypothesis:

print(getAnywhere(linearHypothesis.lme))


In it we see the following calculation:

SSH <- as.vector(t(L %*% b - rhs) %*% solve(L %*% V %*% t(L)) %*%
(L %*% b - rhs))


That's it. So we have some matrix multiplication happening. In this calculation L is our hypothesis matrix, b is our fixed effects, rhs means "right hand side" (which in our example is 0) and V is the variance-covariance matrix of the model parameters.

If we wanted to calculate the Wald statistic by hand, we could do the following:

# extract the model coefficients
b <- matrix(summary(model.f1)@coefs[,1],nrow=5)

# create the "right-hand side"
q <- matrix(c(0,0),nrow=2)

# extract the variance-covariance matrix
V <- vcov(model.f1)

# calculate the Wald statistic
W <- t(C%*%b - q) %*% solve(C%*% V %*%t(C)) %*% (C%*%b - q)


To calculate the p-value, use the pchisq function:

pchisq(as.numeric(W),2,lower.tail=FALSE)


# Comparing Multilevel Models using Deviance Statistics (Ch 4 of ALDA)

The tour of Applied Longitudinal Data Analysis (ALDA) by Singer and Willett continues today with section 4.6, Comparing Models using Deviance Statistics. In the section prior to this they walk through building a model by way of examining hypothesis tests for fixed effects and variance components. While the former will be familiar to those who’ve done classical linear regression, the latter is likely something new. And perhaps not advised. As I mentioned in my previous post (last paragraph), they state in section 3.6.2 that “statisticians disagree as to the nature, form, and effectiveness of these tests.” They also “suggest you examine them only with extreme caution.” Therefore I decided not to blog about that particular tactic and instead focus on “a superior method for testing hypotheses about variance components.” (Also their words.) Of course I refer to the title of this post: Deviance Statistics.

As I’ve done in my other ALDA posts, I’m going to forgo exposition and get down to business. This post is meant to serve as a reference guide for my future self and maybe yours as well.

• The Deviance Statistic is used to test the hypothesis that additional model predictors do not improve the fit of the model. The null hypothesis is that the coefficients of the additional predictors are 0.
• To use the Deviance Statistic, one model must be nested in the other. That is, the smaller model can be derived from the bigger model by setting certain coefficients in the bigger model equal to 0.
• Deviance = -2 * (Log Likelihood (LL) of model)
• Deviance Statistic = -2 * (LL of model nested in bigger model – LL of bigger model)
• Smaller Deviance is better. If adding more predictors to a model reduces deviance, that may be a good thing. The hypothesis test using the Deviance Statistic helps us determine whether or not the reduction in deviance is significant. A large p-value tells us no, it is not significant and that our model is not improved by the additional predictors. A small p-value tells us to reject the null and keep the extra predictors.
• The distribution of the deviance statistic is chi-square with DF equal to the number of extra parameters in the bigger model.
• Deviance obtained under Restricted Maximum Likelihood (REML) should only be used if the two models compared have the same fixed effects and differ only in their random effects. If this is not the case, the deviance obtained using Full ML should be used instead.

Example

The example in Chapter 4 of ALDA involves alcohol use by adolescents. 82 were surveyed over time (3 waves). Some of the data collected include:

alcuse, a continuous measure of alcohol use based on a rating scale (the response)
age_14, age of participant centered about 14 (14 = 0, 15 = 1, 16 = 2)
coa, an indicator whether or not participant is a child of an alcoholic (1 = yes, 0 = no)
id, an arbitrary level to group persons

The model building process is reproduced in R on the UCLA stats consulting site. They use the nlme package. I will use the lme4 package below to demonstrate the use of the deviance statistic.

# read in and attach the data
attach(alcohol1)
library(lme4)


We're going to fit a model that only has age_14 as a predictor. Then we're going to build a model that has age_14 and coa as predictors. Notice the first model is "nested" in the second. In other words we can get the first model from the second model by setting the coa coefficients to 0.

FIRST MODEL
$alcuse = \gamma_{00} + \gamma_{10}*age14 + \zeta + \zeta*age14 + \epsilon$

SECOND MODEL
$alcuse = \gamma_{00} + \gamma_{10}*age14 + \gamma_{01}*coa + \gamma_{11}*age14*coa + \zeta + \zeta_{1i}*age14 + \epsilon$

Is the second model better than the first? The null hypothesis is no, it is not better.

$H_{0}: \gamma_{01} = \gamma_{11} = 0$

The second model has two additional fixed effects and no change in the random effects. Therefore to carry out this test, both models need to be fitted using Full Maximum Likelihood. (Note the argument "REML = FALSE" in the calls to lmer() below.)

# FIRST MODEL
model.b1 <- lmer(alcuse ~ age_14 + (age_14 | id), alcohol1, REML = FALSE)
summary(model.b1)

# SECOND MODEL
model.c1 <- lmer(alcuse ~ age_14*coa + (age_14 | id), alcohol1, REML = FALSE)
summary(model.c1)


Now we're ready to carry out the test. We can access the deviance of each model from the summary object, like so:

summary(model.b1)@AICtab$deviance [1] 636.6111 summary(model.c1)@AICtab$deviance
[1] 621.2026


Notice the deviance of the bigger model is smaller than the deviance of the nested model. Is the reduction in deviance significant? To carry out the test we take the deviance of the smaller nested model and subtract from it the deviance of the bigger model. The difference is then compared to a chi-square distribution for significance. In this case, we'll compare the difference to a chi-square distribution with 2 degrees of freedom since the bigger model has two extra coefficients.

dev <- summary(model.b1)@AICtab$deviance - summary(model.c1)@AICtab$deviance
dev
[1] 15.40846
1 - pchisq(dev,2)
[1] 0.0004509163


Now that's a small p-value. That's the probability we would see a difference this large (or larger) in deviance if the two coefficients really were 0. We reject the null hypothesis and conclude our model is improved by including the two coefficients associated with the coa predictor. If we were planning to do several such tests, we could write a function to make the process go a little faster.

# function to calculate deviance statistic and return p-value
# a = nested model object, b = bigger model object, df = degrees of freedom
dev <- function(a,b,df){
return(1 - pchisq(
summary(a)@AICtab$deviance - summary(b)@AICtab$deviance,
df))
}

dev(model.b1,model.c1,2)
[1] 0.0004509163


# Unconditional Multilevel Models for Change (Ch 4 of ALDA)

In Chapter 4 (section 4.4) of Applied Longitudinal Data Analysis (ALDA), Singer and Willett recommend fitting two simple unconditional models before you begin multilevel model building in earnest. These two models “allow you to establish: (1) whether there is systematic variation in your outcome that is worth exploring; and (2) where that variation resides (within or between people).” (p. 92) This is a great idea. Why start building models if there is no variation to explain? In this post I want to summarize these two models for reference purposes.

Model 1: The Unconditional Means Model

• The keyword here is “means”. This model has one fixed effect that estimates the grand mean of the response across all occasions and individuals.
• The main reason to fit this model is to examine the random effects (i.e., the within-person and between-person variance components). This tells us the amount of variation that exists at the within-person level and the between-person level.
• Model specification: $$Y_{ij} = \gamma_{00} + \zeta_{0i} + \epsilon_{ij}$$
• $$\gamma_{00}$$ = grand mean (fixed effect)
• $$\zeta_{0i}$$ = the amount person i’s mean deviates from the population mean (between-person)
• $$\epsilon_{ij}$$ = the amount the response on occasion j deviates from person i’s mean (within-person)
• $$\epsilon_{ij} \sim N(0,\sigma_{\epsilon}^{2})$$
• $$\zeta_{0i} \sim N(0, \sigma_{0}^{2})$$
• Use the intraclass correlation coefficient to describe the proportion of the total outcome variation that lies “between” people: $$\rho = \sigma_{0}^{2} / (\sigma_{0}^{2} + \sigma_{\epsilon}^{2})$$
• In the unconditional means model the intraclass correlation coefficient is also the “error autocorrelation coefficient”, which estimates the average correlation between any pair of composite residuals: $$\zeta_{0i} + \epsilon_{ij}$$
• Sample R code for fitting the unconditional means model (where “id” = person-level grouping indicator):
library(nlme)
lme(response ~ 1, data=dataset, random= ~ 1 | id)


Or this:

library(lme4)
lmer(response ~ 1 + (1 | id), dataset)


To replicate the Unconditional Means Model example in ALDA, the UCLA stats page suggests the following:

alcohol1 <- read.table("http://www.ats.ucla.edu/stat/r/examples/alda/data/alcohol1_pp.txt",
library(nlme)
model.a <- lme(alcuse~ 1, alcohol1, random= ~1 |id)
summary(model.a)


This works OK, but returns slightly different results because it fits the model using REML (Restricted Maximum Likelihood) instead of ML (Maximum Likelihood). It also does not return the estimated between-person variance $$\sigma_{0}^{2}$$. We can "fix" the first issue by including the argument method="ML". There doesn't appear to be anything we can do about the second. However, the lmer() function allows us to replicate the example and obtain the same results presented in the book, as follows (notice we have to specify ML implicitly with the argument REML = FALSE):

model.a1 <- lmer(alcuse ~ 1 + (1 | id), alcohol1, REML = FALSE)
summary(model.a1)


The output provides the values discussed in the book in the "Random effects" section under the variance column:

> summary(model.a1)
Linear mixed model fit by maximum likelihood
Formula: alcuse ~ 1 + (1 | id)
Data: alcohol1
AIC   BIC logLik deviance REMLdev
676.2 686.7 -335.1    670.2     673
Random effects:
Groups   Name        Variance Std.Dev.
id       (Intercept) 0.56386  0.75091
Residual             0.56175  0.74950
Number of obs: 246, groups: id, 82

Fixed effects:
Estimate Std. Error t value
(Intercept)   0.9220     0.0957   9.633


The "Random effect" id has variance = 0.564. That's the between-person variance. The "Random effect" Residual has variance = 0.562. That's the within-person variance. We can access these values using "summary(model.a1)@REmat" and calculate the intraclass correlation coefficient like so:

icc_n <- as.numeric(summary(model.a1)@REmat[1,3])
icc_d <- as.numeric(summary(model.a1)@REmat[1,3]) +
as.numeric(summary(model.a1)@REmat[2,3])
icc_n / icc_d
[1] 0.5009373


Model 2: The Unconditional Growth Model

• This model partitions and quantifies variance across people and time.
• The fixed effects estimate the starting point and slope of the population average change trajectory.
• Model specification: $$Y_{ij} = \gamma_{00} + \gamma_{10}*time_{ij} + \zeta_{0i} + \zeta_{1i}*time_{ij} + \epsilon_{ij}$$
• $$\gamma_{00}$$ = average intercept (fixed effect)
• $$\gamma_{10}$$ = average slope (fixed effect)
• $$\zeta_{0i}$$ = the amount person i's intercept deviates from the population intercept
• $$\zeta_{1i}$$ = the amount person i's slope deviates from the population slope
• $$\epsilon_{ij}$$ = the amount the response on occasion j deviates from person i's true change trajectory
• $$\epsilon_{ij} \sim N(0,\sigma_{\epsilon}^{2})$$
• $$\zeta_{0i} \sim N(0, \sigma_{0}^{2})$$
• $$\zeta_{1i} \sim N(0, \sigma_{1}^{2})$$
• $$\zeta_{0i}$$ and $$\zeta_{1i}$$ have covariance $$\sigma_{1}^{2}$$
• The residual variance $$\sigma_{\epsilon}^{2}$$ summarizes the average scatter of an individual's observed outcome values around his/her own true change trajectory. Compare this to the same value in the unconditional means model to see if within-person variation is systematically associated with linear time.
• The level-2 variance components, $$\sigma_{0}^{2}$$ and $$\sigma_{1}^{2}$$ quantify the unpredicted variability in the intercept and slope of individuals. That is, they assess the scatter of a person's intercept and slope about the population average change trajectory. DO NOT compare to the same values in the unconditional means model since they have a different interpretation.
• The level-2 covariance $$\sigma_{01}$$ quantifies the population covariance between the true initial status (intercept) and true change (slope). Interpretation is easier if we re-express the covariance as a correlation coefficient: $$\hat{\rho}_{01} = \hat{\sigma}_{01} / \sqrt{\hat{\sigma}_{0}^{2}\hat{\sigma}_{1}^{2}}$$
• Sample R code for fitting the unconditional growth model (where "id" = person-level grouping indicator):
lme(response ~ time , data=dataset, random= ~ time | id)


Or this:

lmer(alcuse ~ time + (time | id), dataset)


To replicate the Unconditional Growth Model example in ALDA, the UCLA stats page suggests the following:

alcohol1 <- read.table("http://www.ats.ucla.edu/stat/r/examples/alda/data/alcohol1_pp.txt",
library(nlme)
model.b <- lme(alcuse ~ age_14 , data=alcohol1, random= ~ age_14 | id, method="ML")
summary(model.b)


However I think the following is better as it gives the same values in the book:

model.b1 <- lmer(alcuse ~ age_14 + (age_14 | id), alcohol1, REML = FALSE)
summary(model.b1)


For instance it provides variance values instead of standard deviation values. It doesn't really matter in the long run, but it makes it easier to quickly check your work against the book. Here's the output:

> summary(model.b1)
Linear mixed model fit by maximum likelihood
Formula: alcuse ~ age_14 + (age_14 | id)
Data: alcohol1
AIC   BIC logLik deviance REMLdev
648.6 669.6 -318.3    636.6   643.2
Random effects:
Groups   Name        Variance Std.Dev. Corr
id       (Intercept) 0.62436  0.79017
age_14      0.15120  0.38885  -0.223
Residual             0.33729  0.58077
Number of obs: 246, groups: id, 82

Fixed effects:
Estimate Std. Error t value
(Intercept)  0.65130    0.10508   6.198
age_14       0.27065    0.06246   4.334

Correlation of Fixed Effects:
(Intr)
age_14 -0.441


Again the main section to review is the "Random effects". The Residual variance (within-person) has decreased to 0.337 from 0.562. We can conclude that $$(0.562 - 0.337)/0.562 = 0.40$$ (i.e., 40%) of the within-person variation in the response is systematically associated with linear time. We also see the negative correlation (-0.223) between the true initial status (intercept) and true change (slope). However, the book notes this correlation is not statistically significant. As you can see this is not something the output of the lmer object reports. The book mentions in chapter 3 (p. 73) that statisticians disagree about the effectiveness of such significance tests on variance components, and I can only assume the authors of the lme4 package question their use. Finally, we notice the level-2 variance components: 0.624 and 0.151. These provide a benchmark for predictors' effects as the authors proceed to build models.

# The Multilevel Model for Change (Ch 3 of ALDA) – revisited

In my previous post I talked about how to replicate the example in Chapter 3 of ALDA. It was mostly about finding the dataset and modifying the R code on the UCLA web site so it works. In this post I want to talk a little more about the statistics.

Recall the example involves two groups of infants: one assigned to a program to enhance cognitive functioning and the other acting as a control. Cognitive measurements were taken at three different time periods for both groups. Did the group assigned to the program perform differently than the control group? To answer this question the authors postulate a linear model where cognitive test results are explained by time, as follows:

$Y = \beta_{0} + \beta_{1}*time$

But the intercept and slope coefficients in that model are modeled as follows:

$\beta_{0} = \gamma_{00} + \gamma_{01}*program$
$\beta_{1} = \gamma_{10} + \gamma_{11}*program$

So we have two levels of modeling happening at the same time. The first level concerns within-person change while the second level concerns between-person differences in change. We can consolidate the two levels into one formula, like this:

$Y = \gamma_{00} + \gamma_{10}*time + \gamma_{01}*program + \gamma_{11}*time*program$

So we have an intercept and three coefficients. When we fit the model to the data, we get:

$Y = 107.84 - 21.13*time + 6.85*program + 5.27*program*time$

All of which are significant coefficients. When program = 0, our linear model is $Y = 107.84 - 21.13*time$. When program = 1, our linear model is $Y = 114.69 - 15.86*time$. The intercept in these models is interpreted as the cognitive score at the first measurement (when the infants were 12 months old). We can see that infants in the program had a higher performance at the first measurement than those not in the program: 114.69 – 107.84 = 6.85. The slope tells us the rate of decline of cognitive performance (decline?). We see the infants in the program had a slower rate of decline over time: -15.86 versus -21.13. Or put another way: -21.13 – -15.86 = -5.27, which is the coefficient of the interaction. That is, the difference in slopes between the two groups is -5.27.

Now it’s interesting to note that the model does not appear to make use of the fact that each subject contributed three waves of data. Our dataset has 309 records. But those 309 records are in 103 groups. Those 103 groups are the 103 infants. Each infant contributed three cognitive test scores over time. The dataset has a variable, ID, to indicate those groups. But ID is not in our model. Shouldn’t we make use of that information? Well, in fact we did. Have a look at the R code:

lme(cog ~ time*program, data=ch3, random = ~ time | id, method="ML",
control=list(opt = "optim"))


Notice how “id” indicates the grouping structure in the “random” argument. Time is specified as the random effect and “| id” indicates it is grouped by “id” (i.e., the 103 infants). In so many words, this allows us to capture the variability in each infant’s own change trajectory. We can think of plotting the cognitive test score for one infant over time and fitting a line to those three points. There will be some error in that line. But not as much error than if we fit a line to all infants in a group over the three times. In this latter scenario we’re not accounting for the grouping of the measurements by infant. We can actually see what happens if we don’t account for this grouping by doing an Analysis of Covariance (ANCOVA).

With ANCOVA, we’re basically doing regression with continuous and categorical variables. The usual approach to ANCOVA is to think of doing a regular ANOVA analysis but blocking on a continuous variable. For example, comparing cholesterol levels (y) between a treated group and a reference group adjusted for age (x, in years). We’re interested in the treatment effect but we want to account for the effect of age.

We can naively do ANCOVA with the Chapter 3 example from ALDA as follows:

lm(cog ~ program + time + time:program, data=ch3)


Look at the results:

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)   107.841      1.773  60.822  < 2e-16 ***
program         6.855      2.363   2.901  0.00399 **
time          -21.133      2.747  -7.694 1.99e-13 ***
program:time    5.271      3.660   1.440  0.15087


Now compare those to the multilevel modelling results, obtained from the call to lme() above:

                 Value Std.Error  DF   t-value p-value
(Intercept)  107.84074  2.048799 204  52.63608  0.0000
time         -21.13333  1.903664 204 -11.10140  0.0000
program        6.85466  2.730259 101   2.51063  0.0136
time:program   5.27126  2.536850 204   2.07788  0.0390


Notice the similarities? That's right, both return the same model coefficients! But compare the difference in standard errors. The most dramatic is the interaction between time and program. In the ANCOVA analysis the interaction appears to be insignificant (SE = 3.7; p = 0.15). But in the multilevel model it's significant at the 5% level (SE = 2.5; p = 0.03). We see that the ANCOVA model does not take into account the change trajectories at the individual level, and is thus not sensitive enough to detect the significant difference in rates of cognitive decline between the infants in the program and those in the control group.

# The Multilevel Model for Change (Ch 3 of ALDA)

ALDA stands for the book, Applied Longitudinal Data Analysis, by Singer and Willett. I’ve had this lying around for a while and decided to give it another go. If you know a little about regression and want to get started with data analysis using multilevel models, this is probably a good place to start. It’s light on the math and heavy on application. Well, maybe not that heavy. They don’t actually provide anything in the way of computer code when it comes time to fit a model. They just do it and produce a table of results. Fortunately, our friends at the UCLA Statistical Consulting Group have generously provided programs to reproduce the examples in the book using R, SAS, and Stata, among others. So that’s where I turned when I wanted to replicate the example running through chapter 3.

The example involves 103 African-American infants born into low-income families. When the children were 6 months old, about half were randomly assigned to an early intervention program designed to enhance cognitive functioning. The other half received no intervention and acted as a control. Cognitive performance was measured at ages 12, 18 and 24 months by way of a test. The researches wanted to examine the effects of program participation on the test results. There’s actually a lot more to this study. Here’s the abstract. But the authors scaled it back and took a portion of the data to create a friendly introductory example to multilevel modeling. And I have to say, they succeeded. It is indeed an easy-to-follow example. The chapter can be read in one sitting and gives you a sense of what multilevel modeling is all about. In this case we want to model individual change in cognitive ability over time with a straight line,

$y = \beta_{0} + \beta_{1}(time)$

but then also model the intercept and slope of that line with respect to program:

$\beta_{0} = \gamma_{00} + \gamma_{01}(program)$
$\beta_{1} = \gamma_{10} + \gamma_{11}(program)$

So we have two levels of modeling going on. Hence, “multilevel” modeling.

Naturally I wanted to replicate the Chapter 3 results in R. That’s why I’m reading the book, so I can learn how to do this stuff. So I hit up the UCLA site to get the necessary R code. And there at the top the page, before I can even get started, is a roadblock:

Please note that the “early_int” data file (which is used in Chapter 3) is not included among the data files. This was done at the request of the researchers who contributed this data file to ensure the privacy of the participants in the study. Although the web page shows how to obtain the results with this data file, we regret that visitors do not have access to this file to be able to replicate the results for themselves.

Seriously? C’mon. Did the authors not think about this when writing the book? Couldn’t they come up with a different example that can be replicated? Very frustrating. So I did what any honorable statistician would do and gave up, turned off the computer and went outside to enjoy nature Googled the name of the dataset used in the R code on the UCLA web site (earlyint_pp.txt). The very first result takes you to this Harvard web page, where the data in question is available in SAS and Stata formats. Now the authors of ALDA are both Harvard professors. And here is the data – that I’m not supposed to have access to – available on a Harvard web page for an old statistics class taught by one of the authors of the book. I guess the researchers changed their mind? Anyway, I now had the data and could try to replicate the results. And you’ll be happy to know the data contained no identifying information. It had four variables: (1) an ID number for the infant, (2) the test score, (3) age at test, and (4) an indicator for program participation.

library(foreign)


With the data loaded, I was ready to start replicating the results. This required me to copy and paste the R code from the UCLA web site. I don’t usually go for copying-and-pasting, but in this case I was OK with it. It was mostly exploratory analysis. And it worked like a charm. That is until I got to the part where you actually fit a multilevel model to the data. Running the following R code produced an error:

library(nlme)
lme(cog ~ time*program, data=early.int, random = ~ time | id, , method="ML")

Error in lme.formula(cog ~ time * program, data = early.int, random = ~time |  :
nlminb problem, convergence error code = 1
message = iteration limit reached without convergence (10)


Hmm. Seemed to work for the UCLA person. What to do? Long story short, I needed to specify a different optimizer. The default optimizer for lme is “nlminb”. I needed to specify “optim”, like so:

model1 <- lme(cog ~ time*program, data=early.int, random = ~ time | id,
method="ML", control=list(opt = "optim"))
summary(model1)


That worked! Or at least it gave me some output. Upon closer inspection the random effects were slightly different, but everything else looked about the same. Why does “optim” work and not “nlminb”? I have no idea. Obviously they go about optimization in different ways. But I’m not too worried at the moment since I’m just getting my feet wet with multilevel modeling.

In researching the error above I discovered I could also fit the multilevel model using the following code:

library(lme4)
mod1 <- lmer(cog ~ time*program + (time | id), early.int)
summary(mod1)


The output from this is also slightly different from the book as well as the output from the lme call above using “optim”. The coefficients of the fixed effects are all the same, but the random effects are slightly different.

So what does all this mean? How do you interpret the output? What is multilevel modeling? Well, that wasn’t quite the point of this post. I wanted to document for posterity how to replicate the results of ALDA Chapter 3 using R despite the message to the contrary on the UCLA site. Besides, if you’ve read this far then you're probably also working through ALDA and don't need me to repeat what's in the book. However, if you insist, I point you to this solid PowerPoint presentation that boils Chapter 3 of ALDA to its essence. It tells you everything you need to know about modeling this example and interpreting the results.

# Using a Bootstrap to Estimate Power and Significance Level

I’ve been reading Common Errors in Statistics (and How to Avoid Them) by Phillip Good and James Hardin. It’s a good bathroom/bedtime book. You can pick it up and put it down as you please. Each chapter is self-contained and contains bite-size, easy-to-read sections. I’m really happy with it so far.

Anyway, chapter 3 had a section on computing Power and sample size that inspired me to hop on the computer:

If the data do not come from one of the preceding distributions, then we might use a bootstrap to estimate the power and signiﬁcance level.

In preliminary trials of a new device, the following test results were observed: 7.0 in 11 out of 12 cases and 3.3 in 1 out of 12 cases. Industry guidelines speciﬁed that any population with a mean test result greater than 5 would be acceptable. A worst-case or boundary-value scenario would include one in which the test result was 7.0 3/7th of the time, 3.3 3/7th of the time, and 4.1 1/7th of the time. i.e., $$(7 \times \frac{3}{7}) + (3.3 \times \frac{3}{7}) + (4.1 \times \frac{1}{7}) = 5$$

The statistical procedure required us to reject if the sample mean of the test results were less than 6. To determine the probability of this event for various sample sizes, we took repeated samples with replacement from the two sets of test results.

If you want to try your hand at duplicating these results, simply take the test values in the proportions observed, stick them in a hat, draw out bootstrap samples with replacement several hundred times, compute the sample means, and record the results.

Well of course I want to try my hand at duplicating the results. Who wouldn’t?

The idea here is to bootstrap from two samples: (1) the one they drew in the preliminary trial with mean = 6.69, and (2) the hypothetical worst-case boundary example with mean = 5. We bootstrap from each and calculate the proportion of samples with mean less than 6. The proportion of results with mean less than 6 from the first population (where true mean = 6.69) can serve as a proxy for Type I error or the significance level. This is proportion of times we make the wrong decision. We conclude the mean is less than 6 when in fact it’s really 6.69. The proportion of results with mean less than 6 from the second population (where true mean = 5) can serve as a proxy for Power. This is proportion of times we make the correct decision. We conclude the mean is less than 6 when in fact it’s really 5.

In the book they show the following table of results:

We see they have computed the significance level (Type I error) and power for three different sample sizes. Here’s me doing the same thing in R:

# starting sample of test results (mean = 6.69)
el1 <- c(7.0,3.3)
prob1 <- c(11/12, 1/12)

# hypothetical worst-case population (mean = 5)
el2 <- c(7, 3.3, 4.1)
prob2 <- c(3/7, 3/7, 1/7)

n <- 1000
for (j in 3:5){ # loop through sample sizes
m1 <- double(n)
m2 <- double(n)
for (i in 1:n) {
m1[i] <- mean(sample(el1,j, replace=TRUE,prob=prob1)) # test results
m2[i] <- mean(sample(el2,j, replace=TRUE,prob=prob2)) # worst-case
}
print(paste("Type I error for sample size =",j,"is",sum(m1 < 6.0)/n))
print(paste("Power for sample size =",j,"is",sum(m2 < 6.0)/n))
}



To begin I define vectors containing the values and their probability of occurrence. Next I set n = 1000 because I want to do 1000 bootstrap samples. Then I start the first of two for loops. The first is for my sample sizes (3 - 5) and the next is for my bootstrap samples. Each time I begin a new sample size loop I need to create two empty vectors to store the means from each bootstrap sample. I calls these m1 and m2. As I loop through my 1000 bootstrap samples, I take the mean of each sample and assign to the ith element of the m1 and m2 vectors. m1 holds the sample means from the test results and m2 holds the sample means from the worst-case boundary scenario. Finally I print the results using the paste function. Notice how I calculate the proportion. I create a logical vector by calling mx < 6.0. This returns a vector of 0s and 1s, where 0 is false and 1 is true. I then sum this vector to get the number of times the mean was less than 6. Dividing that by n (1000) gives me the proportion. Here are my results:

[1] "Type I error for sample size = 3 is 0.244"
[1] "Power for sample size = 3 is 0.845"
[1] "Type I error for sample size = 4 is 0.04"
[1] "Power for sample size = 4 is 0.793"
[1] "Type I error for sample size = 5 is 0.067"
[1] "Power for sample size = 5 is 0.886"


Pretty much the same thing! I guess I could have used the boot function in the boot package to do this. That’s probably more efficient. But this was a clear and easy way to duplicate their results.