It is a course on applied statistics.

Hands-on: we use R, an open-source statistics software environment.

Course notes will be jupyter notebooks.

We will start out with a review of introductory statistics to see

`R`

in action.Main topic is

*(linear) regression models*: these are the*bread and butter*of applied statistics.

A regression model is a model of the relationships between some
*covariates (predictors)* and an *outcome*.

Specifically, regression is a model of the *average* outcome *given or having fixed* the covariates.

We will consider the heights of mothers and daughters collected by Karl Pearson in the late 19th century.

One of our goals is to understand height of the daughter,

`D`

, knowing the height of the mother,`M`

.

A mathematical model might look like $$ D = f(M) + \varepsilon$$ where $f$ gives the average height of the daughter of a mother of height

`M`

and $\varepsilon$ is*error*: not*every*daughter has the same height.A statistical question: is there

*any*relationship between covariates and outcomes -- is $f$ just a constant?

Let's create a plot of the heights of the mother/daughter pairs. The data is in an `R`

package that can be downloaded
from CRAN with the command:

```
install.packages("alr3")
```

If the package is not installed, then you will get an error message when calling `library(alr3)`

.

In [ ]:

```
library(alr3)
data(heights)
M = heights$Mheight
D = heights$Dheight
plot(M, D, pch = 23, bg = "red", cex = 2)
```

In the first part of this course we'll talk about fitting a line to this data. Let's do that and remake the plot, including this "best fitting line".

In [ ]:

```
plot(M, D, pch = 23, bg = "red", cex = 2)
height.lm = lm(D ~ M)
abline(height.lm, lwd = 3, col = "yellow")
```

How do we find this line? With a model.

We might model the data as $$ D = \beta_0+ \beta_1 M + \varepsilon. $$

This model is

*linear*in $\beta_1$, the coefficient of`M`

(the mother's height), it is a*simple linear regression model*.Another model: $$ D = \beta_0 + \beta_1 M + \beta_2 M^2 + \beta_3 F + \varepsilon $$ where $F$ is the height of the daughter's father.

Also linear (in the coefficients of $M,M^2,F$).

Which model is better? We will need a tool to compare models... more to come later.

Our example here was rather simple: we only had one independent variable.

Independent variables are sometimes called

*features*or*covariates*.In practice, we often have many more than one independent variable.

This example considers the effect of right-to-work legislation (which varies by state) on various factors. A description of the data can be found here.

The variables are:

Income: income for a four-person family

COL: cost of living for a four-person family

PD: Population density

URate: rate of unionization in 1978

Pop: Population

Taxes: Property taxes in 1972

RTWL: right-to-work indicator

In a study like this, there are many possible questions of interest. Our focus will be on the
relationship between `RTWL`

and `Income`

. However, we should recognize that other variables
have an effect on `Income`

. Let's look at some of these relationships.

In [ ]:

```
url = "http://www1.aucegypt.edu/faculty/hadi/RABE4/Data4/P005.txt"
rtw.table <- read.table(url, header=TRUE, sep='\t')
print(head(rtw.table))
```

A graphical way to
visualize the relationship between `Income`

and `RTWL`

is the *boxplot*.

In [ ]:

```
attach(rtw.table) # makes variables accessible in top namespace
boxplot(Income ~ RTWL, col='orange', pch=23, bg='red')
```

One variable that may have an important effect on the relationship between
is the cost of living `COL`

. It also varies between right-to-work states.

In [ ]:

```
boxplot(COL ~ RTWL, col='orange', pch=23, bg='red')
```

We may want to include more than one plot in a given display. The first line of the code below achieves this.

In [ ]:

```
par(mfrow=c(2,2))
plot(URate, COL, pch=23, bg='red')
plot(URate, Income, pch=23, bg='red')
plot(URate, Pop, pch=23, bg='red')
plot(COL, Income, pch=23, bg='red')
```

`R`

has a builtin function that will try to display all pairwise relationships in a given dataset, the function `pairs`

.

In [ ]:

```
pairs(rtw.table, pch=23, bg='red')
```

In looking at all the pairwise relationships. There is a point that stands out from all the rest.
This data point is New York City, the 27th row of the table. (Note that `R`

uses 1-based instead of 0-based indexing for rows and columns of arrays.)

In [ ]:

```
print(rtw.table[27,])
pairs(rtw.table[-27,], pch=23, bg='red')
```

Some of the main goals of this course:

Build a statistical model describing the

*effect*of`RTWL`

on`Income`

.This model should recognize that other variables also affect

`Income`

.What sort of

*statistical confidence*do we have in our conclusion about`RTWL`

and`Income`

?Is the model adequate do describe this dataset?

Are there other (simpler, more complicated) better models?

In [ ]:

```
X = c(1,3,5,7,8,12,19)
print(X)
print(mean(X))
print((X[1]+X[2]+X[3]+X[4]+X[5]+X[6]+X[7])/7)
print(sum(X)/length(X))
```

We'll also illustrate thes calculations with part of an example we consider below, on differences in blood pressure between two groups.

In [ ]:

```
url = 'http://lib.stat.cmu.edu/DASL/Datafiles/Calcium.html'
calcium.table = read.table(url, header=TRUE, skip=28, nrow=21)
attach(calcium.table)
treated = Decrease[(Treatment == 'Calcium')]
placebo = Decrease[(Treatment == 'Placebo')]
treated
mean(treated)
```

Given a sample of numbers $X=(X_1, \dots, X_n)$ the sample standard deviation $S_X$ is $$ S^2_X = \frac{1}{n-1} \sum_{i=1}^n (X_i-\overline{X})^2.$$

In [ ]:

```
S2 = sum((treated - mean(treated))^2) / (length(treated)-1)
print(sqrt(S2))
print(sd(treated))
```

Given a sample of numbers $X=(X_1, \dots, X_n)$ the sample median is
the `middle`

of the sample:
if $n$ is even, it is the average of the middle two points.
If $n$ is odd, it is the midpoint.

In [ ]:

```
X
print(X)
median(X)
```

In [ ]:

```
quantile(X,c(0.25, 0.75))
```

We've already seen a boxplot. Another common statistical summary is a histogram.

In [ ]:

```
hist(treated, main='Treated group', xlab='Decrease', col='orange')
```

Suppose we want to determine the efficacy of a new drug on blood pressure.

Our study design is: we will treat a large patient population (maybe not so large: budget constraints limit it $n=20$) with the drug and measure their blood pressure before and after taking the drug.

One way to conclude that the drug is effective if the blood pressure has decreased. That is, if the average difference between before and after is positive.

The

*null hypothesis*, $H_0$ is: *the average difference is less than zero.*The

*alternative hypothesis*, $H_a$, is: *the average difference is greater than zero.*Sometimes (actually, often), people will test the alternative, $H_a$:

*the average difference is not zero*vs. $H_0$:*the average difference is zero.*The test is performed by estimating the average difference and converting to standardized units.

Formally, could set up the above test as drawing from a box of

*differences in blood pressure*.A box model is a useful theoretical device that describes the experiment under consideration. In our example, we can think of the sample of decreases drawn 20 patients at random from a large population (box) containing all the possible decreases in blood pressure.

In our box model, we will assume that the decrease is an integer drawn at random from $-3$ to 6.

We will draw 20 random integers from -3 to 6 with replacement and test whether the mean of our "box" is 0 or not.

In [ ]:

```
mysample = sample(-3:6, 20, replace=TRUE)
mysample
```

The test is usually a $T$ test that has the form $$ T = \frac{\overline{X}-0}{S_X/\sqrt{n}} $$

The formula can be read in three parts:

estimating the mean: $\overline{X}$;

comparing to 0: subtracting 0 in the numerator;

converting difference to standardized units: dividing by $S_X/\sqrt{n}$ our estimate of the variability of $\overline{X}$.

In [ ]:

```
T = (mean(mysample) - 0) / (sd(mysample) / sqrt(20))
T
```

This $T$ value is compared to a table for the appropriate $T$ distribution (in this case there are 19 *degrees of freedom*) and the 5% cutoff is

In [ ]:

```
cutoff = qt(0.975, 19)
cutoff
```

The result of the test is

In [ ]:

```
result = (abs(T) > cutoff)
result
```

If `result`

is `TRUE`

, then we reject $H_0$ the mean is 0 at a level of 5%, while if it is `FALSE`

we do not reject. Of course, in this example we know the mean in our "box" is not 0, it is 1.5.

This rule can be visualized with the $T$ density. The total grey area is 0.05=5%, and the cutoff is chosen to be symmetric around zero and such that this area is exactly 5%.

For a test of size $\alpha$ we write this cutoff $t_{n-1,1-\alpha/2}$.

In [ ]:

```
library(ggplot2)
alpha = 0.05
df = 19
xval = seq(-4,4,length=101)
q = qt(1-alpha/2, df)
rejection_region = function(dens, q_lower, q_upper, xval) {
fig = (ggplot(data.frame(x=xval), aes(x)) +
stat_function(fun=dens, geom='line') +
stat_function(fun=function(x) {ifelse(x > q_upper | x < q_lower, dens(x), NA)},
geom='area', fill='#CC7777') +
labs(subtitle=text, y='Density', x='T') +
theme_bw())
}
T19_fig = rejection_region(function(x) { dt(x, df)}, -q, q, xval) +
annotate('text', x=2.5, y=dt(2,df)+0.3, label='Two sided rejection region, df=19')
```

In [ ]:

```
T19_fig
```

Suppose $H_0$ was true -- say the mean of the box was zero.

For example, we might assume the difference is drawn at random from integers -5 to 5 inclusive.

In [ ]:

```
# Generate a sample from a box for which the null is true
draw_sample = function(n) {
return(sample(-5:5, n, replace=TRUE))
}
# Compute the T statistic
null_T = function(n) {
cur_sample = draw_sample(n)
(mean(cur_sample) - 0) / (sd(cur_sample) / sqrt(n))
}
```

When the null hypothesis is true, like in our simulation, we expect that the $T$ statistic will exceed the cutoff only about 5% of the time.

If we use the cutoff $t_{19,0.975}$ to decide in favor or against $H_0$, rejecting
$H_0$ when the absolute value is larger than this value, then we have a test whose
**Type I error** is about 5%.

It is exactly 5% if the sample were drawn from a box whose values follow a normal curve...

In [ ]:

```
results = numeric(10000)
for (i in 1:10000) {
results[i] = null_T(20)
}
mean(abs(results) >= qt(0.975, 19))
```

We use the $T$ curve (close to the normal curve) because when $H_0$ is true, the distribution of the T statistic is close to the $T$ curve

In [ ]:

```
plot(density(results), lwd=3)
xval = seq(-4,4,length=201)
lines(xval, dt(xval, 19), col='red', lwd=3) # T_19 density
lines(xval, dnorm(xval), col='blue', lwd=3) # Normal(0,1) density
```

`R`

will compute this $T$ statistic for you, and many other things.

In [ ]:

```
t.test(mysample)
T
```

As mentioned above, sometimes tests are one-sided. If the null hypothesis we tested was that the mean is less than 0, then we would reject this hypothesis if our observed mean was much smaller than 0. This corresponds to a positive $T$ value.

In [ ]:

```
cutoff = qt(0.95, 19)
T19_pos = rejection_region(function(x) { dt(x, df)}, -Inf, cutoff, xval) +
annotate('text', x=2.5, y=dt(2,df)+0.3, label='One sided rejection region, df=19')
```

In [ ]:

```
T19_pos
```

The rejection rules are affected by the degrees of freedom. Here is the rejection region when we only have 5 samples from our "box".

In [ ]:

```
df = 4
cutoff = qt(0.975, df)
T4_fig = rejection_region(function(x) { dt(x, df)}, -cutoff, cutoff, xval) +
annotate('text', x=2.5, y=dt(2,19)+0.3, label='Two sided rejection region, df=4')
```

In [ ]:

```
T4_fig
```

Instead of testing a particular hypothesis, we might be interested in coming up with a reasonable range for the mean of our "box".

Statistically, this is done via a

*confidence interval*.If the 5% cutoff is $q$ for our test, then the 95% confidence interval is $$ [\bar{X} - q S_X / \sqrt{n}, \bar{X} + q S_X / \sqrt{n}] $$ where we recall $q=t_{n-1,0.975}$ with $n=20$.

If we wanted 90% confidence interval, we would use $q=t_{19,0.95}$. Why?

In [ ]:

```
cutoff = qt(0.95, 19)
L = mean(mysample) - cutoff*sd(mysample)/sqrt(20)
U = mean(mysample) + cutoff*sd(mysample)/sqrt(20)
data.frame(L,U)
```

Note that the endpoints above depend on the data. Not every interval will cover the true mean of our "box" which is 1.5. Let's take a look at 100 intervals of size 90%. We would expect that roughly 90 of them cover 1.5.

In [ ]:

```
cutoff = qt(0.95, 19)
L = c()
U = c()
covered = c()
box = -3:6
for (i in 1:100) {
mysample = sample(box, 20, replace=TRUE)
l = mean(mysample) - cutoff*sd(mysample)/sqrt(20)
u = mean(mysample) + cutoff*sd(mysample)/sqrt(20)
L = c(L, l)
U = c(U, u)
covered = c(covered, (l < mean(box)) * (u > mean(box)))
}
sum(covered)
```

A useful picture is to plot all these intervals so we can see the randomness in the intervals, while the true mean of the box is unchanged.

In [ ]:

```
mu = 1.5
plot(c(1, 100), c(-2.5+mu,2.5+mu), type='n', ylab='Confidence Intervals', xlab='Sample')
for (i in 1:100) {
if (covered[i] == TRUE) {
lines(c(i,i), c(L[i],U[i]), col='green', lwd=2)
}
else {
lines(c(i,i), c(L[i],U[i]), col='red', lwd=2)
}
}
abline(h=mu, lty=2, lwd=4)
```

A study was conducted to study the effect of calcium supplements on blood pressure.

More detailed data description can be found here.

We had loaded the data above, storing the two samples in the variables

`treated`

and`placebo`

.Some questions might be:

- What is the mean decrease in BP in the treated group? placebo group?
- What is the median decrease in BP in the treated group? placebo group?
- What is the standard deviation of decrease in BP in the treated group? placebo group?
- Is there a difference between the two groups? Did BP decrease more in the treated group?

In [ ]:

```
summary(treated)
summary(placebo)
boxplot(Decrease ~ Treatment, col='orange', pch=23, bg='red')
```

In our setting, we have two groups that we have reason to believe are different.

We have two samples:

- $(X_1, \dots, X_{10})$ (
`treated`

) - $(Z_1, \dots, Z_{11})$ (
`placebo`

)

- $(X_1, \dots, X_{10})$ (
We can answer this statistically by testing the null hypothesis $$H_0:\mu_X = \mu_Z.$$

If variances are equal, the

*pooled $t$-test*is appropriate.

The test statistic is $$ T = \frac{\overline{X} - \overline{Z} - 0}{S_P \sqrt{\frac{1}{10} + \frac{1}{11}}}, \qquad S^2_P = \frac{9 \cdot S^2_X + 10 \cdot S^2_Z}{19}.$$

For two-sided test at level $\alpha=0.05$, reject if $|T| > t_{19, 0.975}$.

Confidence interval: for example, a $90\%$ confidence interval for $\mu_X-\mu_Z$ is $$ \overline{X}-\overline{Z} \pm S_P \sqrt{\frac{1}{10} + \frac{1}{11}} \cdot t_{19,0.95}.$$

T statistic has the same form as before!

In [ ]:

```
sdP = sqrt((9*sd(treated)^2 + 10*sd(placebo)^2)/19)
T = (mean(treated)-mean(placebo)) / (sdP * sqrt(1/10+1/11))
c(T, cutoff)
```

`R`

has a builtin function to perform such $t$-tests.

In [ ]:

```
t.test(treated, placebo, var.equal=TRUE)
```

If we don't make the assumption of equal variance, `R`

will give a slightly different result.

In [ ]:

```
t.test(Decrease ~ Treatment)
```

The rule for the $SD$ of differences is $$ SD(\overline{X}-\overline{Z}) = \sqrt{SD(\overline{X})^2+SD(\overline{Z})^2}$$

By this rule, we might take our estimate to be $$ \widehat{SD(\overline{X}-\overline{Z})} = \sqrt{\frac{S^2_X}{10} + \frac{S^2_Z}{11}}. $$

The pooled estimate assumes $\mathbb{E}(S^2_X)=\mathbb{E}(S^2_Z)=\sigma^2$ and replaces the $S^2$'s above with $S^2_P$, a better estimate of $\sigma^2$ than either $S^2_X$ or $S^2_Z$.

Well, the $X$ sample has $10-1=9$ degrees of freedom to estimate $\sigma^2$ while the $Z$ sample has $11-1=10$ degrees of freedom.

Therefore, the total degrees of freedom is $9+10=19$.

We can put the two samples together: $$Y=(X_1,\dots, X_{10}, Z_1, \dots, Z_{11}).$$

Under the same assumptions as the pooled $t$-test: $$ \begin{aligned} Y_i &\sim N(\mu_i, \sigma^2)\\ \mu_i &= \begin{cases} \mu_X & 1 \leq i \leq 10 \\ \mu_Z & 11 \leq i \leq 21. \end{cases} \end{aligned} $$

This is a (regression) model for the sample $Y$. The (qualitative) variable

`Treatment`

is called a*covariate*or*predictor*.The decrease in BP is the

*outcome*.We assume that the relationship between treatment and average decrease in BP is simple: it depends only on which group a subject is in.

This relationship is

*modelled*through the mean vector $\mu=(\mu_1, \dots, \mu_{21})$.

In [ ]:

```
print(summary(lm(Decrease ~ Treatment)))
print(sdP*sqrt(1/10+1/11))
print(sdP)
```