Statistical Methods Notes

Math 222 - Spring 2026

• Examples
• Index
• Notes
• Software & Tables

Week 1 Notes

Mon, Jan 12

Today we went over the course syllabus and talked about making R-markdown files in Rstudio. We started the following lab in class, I recommend finishing the second half on your own. I also recommend installing Rstudio on your own laptop (it’s free).

• Lab: Working with R and Rstudio

Wed, Jan 14

Today we reviewed populations and samples. We started with a famous example of a bad sample.

• Slides: Literary Digest election poll 1936

Then we reviewed population parameters, sample statistics, and sampling frames. The difference between a sample statistic and a population parameter is called the sample error.
There are two sources of sample error:

1. Bias. Can be caused by a non-representative sample (sample bias) or by measurement errors, non-response, or biased questions (non-sample bias). The only way to avoid sample bias is a simple random sample (SRS) from the whole population.

2. Random error. This is non-systematic error. It tends to get smaller with larger samples.

To summarize: $$\text{Statistics} = \text{Parameters} + \underbrace{\text{Bias } + \text{ Random error}}_\text{Sample error}.$$

We finished with this workshop.

• Workshop: Sampling

Fri, Jan 16

If you find an association between an explanatory variable and a response variable in an observational study, then you can’t say for sure that the explanatory variable is the cause. We say that correlation is not causation because there might be lurking variables that are confounders, that is, they are associated with both the explanatory and response variables and so you can tell what is the true cause.

It turns out that randomized experiments can prove cause and effect because random assignment to treatment groups controls all lurking variables. We also talked about blocking and double-blind experiments.

• Example: 1954 polio vaccine trials

• Workshop: Experiments

We finished by simulating the results of the polio vaccine trials to see if they might just be a random fluke. We wrote this R code in class:

results = c()
trials <- 1000
for (x in 1:trials) {
  simulated.result <- sample(c(0,1), size = 244, replace = TRUE)
  percent <- sum(simulated.result) / 244
  results <- c(results, percent)
}
hist(results)
sum(results < 0.336) / trials

Week 2 Notes

Wed, Jan 21

Today we did a lab about using R to visualize data.

• Lab: High Bridge half marathon (Rmd, html)

You should be able to open this file in your browser, then hit CTRL-A and CTRL-C to select it and copy it so that you can paste it into Rstudio as an R-markdown document.

We had a little trouble with R-markdown on the lab computers.

Fri, Jan 23

Last time we talked about how to visualize data with R. Here are two quick summaries of how to make plots in R:

• Example: Basic plots with R
• Example: Facier plots with ggplot2

After that, we started talking about probability. We review some of the basic rules.

The notation $P(B \, |\, A)$ means “the probability of B given that A happened”. Two events $A$ and $B$ are independent if the probability of $A$ does not depend on whether or not $B$ happens. We did the following examples.

1. If you shuffle a deck of 52 playing cards and then draw two, what is the probability that the second card is an ace if the first card is?

We also talked about tree diagrams (see subsection 3.2.7 from the book) and how to use them to compute probabilities.

2. Based on a study of women in the United States and Germany, there is an 0.8% chance that a woman in her forties has breast cancer. Mammograms are 90% accurate at detecting breast cancer if someone has it. They are also 93% accurate at not detecting cancer in people who don’t have it. If a woman in her forties tests positive for cancer on a mammogram screening, what is the probability that she actually has breast cancer?

3. 5% of men are color blind, but only 0.25% of women are. Find $P( \text{male} \, | \, \text{color-blind})$.

Week 3 Notes

Wed, Jan 28

Class was canceled today because I had a doctor’s appointment. But I recommended that everyone watch the following video and then complete a workshop about the R functions pnorm, qnorm, and rnorm.

• Video: https://youtu.be/qLBmYfAVUdg
• Workshop: Normal distribution calculations with R

Fri, Jan 30

Today we talked about random variables and probability distributions. We talked about some example probability distributions:

• Flip a coin until you get a tail. Let $X$ represent the number of flips needed. (geometric distribution)

• About 1 meteorite bigger than 1000 kg hits the Earth every year. The time until the next meteorite hits the Earth has probability density function $f(t) = e^{-t}$. (exponential distribution)

We talked about the difference between continuous and discrete probability distributions. Then we introduced expected value.

We did the following example.

1. In roulette, if you bet $1 on black, there is an 18/38 probability that you win $2, and a 20/38 chance that you lose (and win nothing). What is the expected amount of money you will win?

We finished by talking about what we mean when we say something is “expected”.

Week 4 Notes

Mon, Feb 2

We did these examples in class.

1. Find the variance and standard deviation when you roll a 6-sided die.

2. Exercise 3.34(a)

Here is an extra example from Kahn academy that we did not do in class.

3. Suppose a random variable $X$ has the following probability model.

   $X$	0	1	2	3	4
   $P(X)$	0.1	0.15	0.4	0.25	0.1

   1. Find the expected value of $X$. (https://youtu.be/qafPcWNUiM8)
   2. Find the variance and standard deviation of $X$. (https://youtu.be/2egl_5c8i-g)

4. A single six-sided die has expected value $\mu = 3.5$ and standard deviation $\sigma = 1.7078$. What is the mean and standard deviation if you roll two dice and add them?

5. Exercise 3.34(b)

Wed, Feb 4

1. If you play roulette and bet $1 on black, you have an 18/38 chance of winning $2. If you bet on a number like 7, then you have a 1/38 chance of winning $36. Both bets have the same expected value μ = $0.947. What are the variances for both bets?

We used this binomial distribution plotting tool to compare the distributions if you make these two bets 100 times. In one case we get something that looks roughly like a bell curve, in the other case we get something that is definitely skewed to the right.

2. In the 1954 polio vaccine trials there were 244 polio cases, but only 82 actually had polio. Use the binomial distribution to compute the probability that 82 or fewer fair coin tosses out of 244 come up heads. This can model the p-value under the hypothesis that the polio vaccine does not work. Use the pbinom(x, n, p) function in R.

Sometimes the assumption that the trials are independent is not justified.

3. Suppose you want to find the percent of Hampden-Sydney students who are left handed. So you interview a random sample of 50 students (out of the population of about 900 HSC students). Why aren’t these observations independent?

The correct probability distribution to model the example above is called the hypergeometric distribution. As long as the population is much larger than the sample, we typically do not need to worry about the trials not being independent.

We finished by discussing the normal approximation of a binomial distribution. When $n$ is large enough so that both $np \ge 10$ and $n(1-p) \ge 10$, then $\operatorname{Binom}(n, p)$ is approximately normal with mean $np$ and standard deviation $\sqrt{n p (1-p)}$.

Fri, Feb 6

1. If each $X_i$ has mean $\mu$ and standard deviation $\sigma$, then what is the mean and the standard deviation of the total?

2. In Dungeons and Dragons, you calculate the damage from a fireball spell by rolling 8 six-sided dice and adding up the results. This has an approximately normal distribution. What is the mean and standard deviation of this distribution. (Recall that the mean and standard deviation of a single six-sided die is $\mu = 3.5$ and $\sigma = 1.7078$).

We looked at a graph of the distribution from the previous example to see that it is indeed approximately normal.

3. Use the normal approximation to estimate the probability of doing more than 35 points of damage with a fireball spell.

When you use a normal approximation to estimate discrete probabilities, it is recommended to use a continuity correction (see Section 4.3.3). To estimate $P(X \le k)$, calculate $P(X < k + 0.5)$ using the normal approximation (and likewise, to estimate $P(X \ge k)$, compute $P(X > k - 0.5)$ using the normal approximation).

An important special case of the central limit theorem is the normal approximation of the binomial distribution, which has mean $\mu = np$ and standard deviation $\sigma = \sqrt{np(1-p)}$.

4. About 7% of the US population has type O-negative blood (universal donors). If a hospital has 700 patients, use the normal approximation to estimate the probability that more than 60 have type O-negative blood. Compare your answer with the result if you use the pbinom(x, n, p) function.

We finished by talking about the difference between the distribution of the total versus the distribution of the proportion of patients who are O-negative. The standard deviation of the sample proportion $\hat{p}$ is $$\sigma_{\hat{p}} = \sqrt{ \frac{p (1-p)}{n} }.$$

Week 5 Notes

Mon, Feb 9

Today we talked about confidence intervals for a proportion.

In practice, we usually don’t know the population proportion $p$. Instead we can use the sample proportion $\hat{p}$ to calculate the standard error of $\hat{p}$: $$SE_{\hat{p}} = \sqrt{\frac{\hat{p} (1-\hat{p})}{n}}.$$

1. This year, 34 out of 72 students in my statistics classes were born in Virginia. What is the sample proportion $\hat{p}$ and the standard error?

If the sample size is large enough, then there is a 95% chance that $\hat{p}$ will be within about two standard deviations of $p$. So if we know $\hat{p}$ and we assume that the standard error is close to the standard deviation for $\hat{p}$, then we can make a confidence interval for the location of the parameter $p$.

You can use the R command qnorm((1 - p) / 2) to find the critical z-value ($z^*$) when you want a specific confidence level $p$.

2. Use the data above to make a 95% confidence interval for the proportion of all HSC students born in VA.

After that, we talked about the prop.test() function in R which can make a confidence interval (among other things).

• Example: Confidence intervals for proportions with R (Rmd, html)

Notice that the prop.test() confidence interval is not the same as what we got using the formula above. Instead of using the formula above, R uses something called a Wilson score confidence interval with continuity correction. The idea is to solve for the two points $p$ where $$p - \hat{p} = \pm z^* \sqrt{\frac{p(1-p)}{n}}.$$ If you add in the continuity correction, this pretty much guarantees that there is at least a 95% chance (or whatever other confidence level you want) that the interval contains the true population parameter. The Wilson method confidence intervals are fairly trustworthy even with relatively small samples and small numbers of successes/failures.

Wed, Feb 11

Today we went over the midterm 1 review problems (the solutions are also available now). We also did some additional practice problems including these.

1. If you draw a random card from a deck of 52 playing cards, what is the probability that you draw an ace or a heart?

2. Suppose you need knee surgery. There is an 11% that the surgery fails. There is a 4% chance of getting an infection. And there is a 3% chance of both infection and the surgery failing. What is the probability that the surgery succeeds without infection?

3. In the Wimbledon tennis tournament, serving players are more likely to win a point. A server has two chances to serve the ball. There is a 59% chance that the first serve is in, and if it is, then the server has a 73% chance of winning the point. If the first serve is out, then they have an 86% of getting the second serve in, and in that case they have a 59% chance of winning the point. But if the second serve is out, then the server automatically loses the point.

   1. Make a tree diagram for this situation.
   2. Find the probability that the serving player wins the point.
   3. Find the conditional probability that 1st serve was in, if the server wins the point.

Week 6 Notes

Week 7 Notes

Week 8 Notes

Week 9 Notes

Week 10 Notes

Week 11 Notes

Week 12 Notes

Week 13 Notes

Week 14 Notes