# ML Wiki

## Confidence Intervals for Means

We want to build a Confidence Interval for a Point Estimate of the population mean

### Problem

• let $\mu$ be the true mean parameter
• $X_1, ..., X_n$ our sample of size $n$
• $\bar{X}$ - the average value for the sample $\bar{X} = \cfrac{1}{n} \sum_{i=1}^n X_i$
• we want to estimate $\mu$ using $\bar{X}$ and a Confidence Interval around it

## Normal Model

With sufficiently large sample and no violations of the assumptions, we can use Normal Distribution to model the Sampling Distribution of mean

• note that it's better to use the $t$ statistics described below

### Normal Approximation

Normal approximation is crucial for this - because we use Normal Distribution to find percentiles

### Assumptions

• sample observations are independent
• the distribution is not strongly skewed and there are few outliers
• sample size is sufficiently large (e.g. $\geqslant 30$)
• the larger the sample, the more tolerant we can be to the skews (thanks to the C.L.T)
• sampling distribution is symmetric, unimodal, no outliers - approximately normal

If these conditions are met, can use Normal Model to find the confidence intervals

#### Example

We have this data set that contains data about the whole population

• Suppose we take 10k samples

• and for each sample we calculate the mean
• and then draw the histogram of this data - thus we'll get the sampling distribution
• • we see that it's normal, but can also try to draw the Normal Probability Plot to see that it's indeed the case
• R code to reproduce the experiment
load(url('http://s3.amazonaws.com/assets.datacamp.com/course/dasi/ames.RData'))
population = ames$Gr.Liv.Area oldpar = par(no.readonly=TRUE) # fig=c(x1, x2, y1, y2) par(fig=c(0, 1, 0, 1)) par(mar=c(6, 2, 2, 1)) h = hist(population, col='grey', probability=T, axes=F, xlab='', main='Histogram') dens = density(population, adjust=2) lines(dens, col="black", lwd=2) axis(side=1, pos=c(-max(h$density)/4,0))
axis(side=2)

par(fig=c(0, 1, 0.16, 0.41), new=TRUE)
par(mar=c(0, 2, 0, 1))
boxplot(population, horizontal=TRUE, axes=F, col='grey')

par(oldpar)

set.seed(1237)
n = 50
samp.mean = replicate(10000, mean(sample(population, n)))

plot(x=NA, y=NA, xlim=c(1250, 1750), ylim=c(0, 0.006), axes=F,
xlab='Estimate of mean', ylab='Frequency',
main='Sampling Distribution of mean')

m = mean(samp.mean)
s = sd(samp.mean)

par(xpd=FALSE)

rect(xleft=m-3*s, xright=m+3*s, ybottom=-1, ytop=1,
rect(xleft=m-2*s, xright=m+2*s, ybottom=-1, ytop=1,
rect(xleft=m-s, xright=m+s, ybottom=-1, ytop=1,

axis(side = 1)

dens = dnorm(1200:1800, mean=m, sd=s)
lines(1200:1800, dens, col="blue", lwd=2)

qqline(samp.mean)


In this case all the assumptions hold - can use the Normal Approximation to calculate the confidence intervals

### Model

• $E[\bar{X}] = \mu$, it's an unbiased estimate of mean
• Standard Error: $\text{var}(\bar{X}) = \cfrac{\sigma^2}{n}$
• by C.L.T. have $\bar{X} \approx N\left(\mu, \cfrac{\sigma^2}{n}\right)$
• therefore
• $\cfrac{\bar{X} - \mu}{\sqrt{\sigma^2 / n}} \approx N(0, 1)$

So the 95% CI with $z$-score is:

• $\left[ \bar{X} - 1.96 \sqrt{\sigma^2 / n}; \bar{X} + 1.96 \sqrt{\sigma^2 / n} \right]$

### Estimating $\sigma$

To compute a CI we need to know $\sigma^2$, but it's a parameter - we need to estimate it

• We know that
• $\text{Var}(X) = \cfrac{1}{n - 1} \sum (x_i - \bar{X})^2$
• $s = \text{sd}(X) = \sqrt{\text{Var}(X)}$
• $\sigma^2$ is unknown. Can we substitute it by $s^2$?
• $s^2$ is unbiased estimate of $\sigma^2$
• $E[s^2] = \sigma^2$ (this is the reason for $n - 1$ instead of $n$)
• so yes, we can, but it's better to use the $t$-distribution (described below) instead and use $s^2$

## Using $t$ Statistic

To use normal approximation we need a sufficiently large sample

• what if we don't have it?
• use $t$-statistics that follows $t$-distribution
• note that with high degrees of freedom $t$-distribution very closely resembles $N(0,1)$

### $t$-distribution

$t$-distribution

• We say that value follows $t$-distribution with $n - 1$ degrees of freedom
• This distribution is similar to normal, but not quite: it's little wider and allows for more uncertainty

Use the $t$- distribution rather than the normal distribution when

• the variance is not known and
• has to be estimated from sample data.

Shape of $t$ vs Normal:

• When the sample size is large, e.g. $\geqslant$ 100, the $t$ is very similar to $N(0,1)$
• on smaller sizes, $t$ is leptokurtic - it has relatively more scores in its tails than the normal distribution
• $\Rightarrow$ you have to extend farther from the mean to span a given proportion of the area.
• for $N(0,1)$ 95% of data is within 1.96 $\sigma$ from the mean
• for $t$, if you a sample size is only 5, 95% of the area is within 2.78 std from the mean
• $\Rightarrow$ the SE of the mean would be multiplied by 2.78 rather than 1.96.

### $t$-Statistic Confidence Intervals

Thus

• we replace $\sigma^2 = s^2$ and
• use the $t$-distribution with $n-1$ degrees of freedom
• i.e. replace $z$-score with $z = t_{n - 1}$

95% CI becomes

• $\left[\bar{X} - t \cdot \sqrt{s^2 / 2}; \bar{X} + t \cdot \sqrt{s^2 / 2}\right]$
• we we have $n = 400$ (therefore $df=399$), $t \approx 1.969$

### R-code

In R:

n = 60
xbar = mean(d)
v = var(d)
t = qt(0.025, df=n-1, lower.tail=F)
ME = t * sqrt(v / n)
xbar + c(-ME, ME)


or:

t.test(d, conf.int=0.95)$confint  The last chuck actually uses$t\$-test and returns its confidence interval

### Examples

Example1:

• The mean for 51 observations was 20
• The sample variance was 5
• Calculate the 99% CI for the mean
xbar = 20
v = 5
t = qt(0.005, df=50, lower.tail=F)
ME = t * sqrt(v / 50)
xbar + c(-ME, ME)
// [19.16, 20.84]