21
Feb 17

The pearls of AP Statistics 37

Confidence interval: attach probability or not attach?

I am reading "5 Steps to a 5 AP Statistics, 2010-2011 Edition" by Duane Hinders (sorry, I don't have the latest edition). The tip at the bottom of p.200 says:

For the exam, be VERY, VERY clear on the discussion above. Many students
seem to think that we can attach a probability to our interpretation of a confidence
interval. We cannot.

This is one of those misconceptions that travel from book to book. Below I show how it may have arisen.

Confidence interval derivation

The intuition behind the confidence interval and the confidence interval derivation using z score have been given here. To make the discussion close to Duane Hinders, I show the confidence interval derivation using the t statistic. Let X_1,...,X_n be a sample of independent observations from a normal population, \mu the population mean and s the standard error. Skipping the intuition, let's go directly to the t statistic

(1) t=\frac{\bar{X}-\mu}{s/\sqrt{n}}.

At the 95% confidence level, from statistical tables find the critical value t_{cr,0.95} of the t statistic such that

P(-t_{cr,0.95}<t<t_{cr,0.95})=0.95.

Plug here (1) to get

(2) P(-t_{cr,0.95}<\frac{\bar{X}-\mu}{s/\sqrt{n}}<t_{cr,0.95})=0.95.

Using equivalent transformations of inequalities (multiplying them by s/\sqrt{n} and adding \mu to all sides) we rewrite (2) as

(3) P(\mu-t_{cr,0.95}\frac{s}{\sqrt{n}}<\bar{X}<\mu+t_{cr,0.95}\frac{s}{\sqrt{n}})=0.95.

Thus, we have proved

Statement 1. The interval \mu\pm t_{cr,0.95}\frac{s}{\sqrt{n}} contains the values of the sample mean with probability 95%.

The left-side inequality in (3) is equivalent to \mu<\bar{X}+t_{cr,0.95}\frac{s}{\sqrt{n}} and the right-side one is equivalent to \bar{X}-t_{cr,0.95}\frac{s}{\sqrt{n}}<\mu. Combining these two inequalities, we see that (3) can be equivalently written as

(4) P(\bar{X}-t_{cr,0.95}\frac{s}{\sqrt{n}}<\mu<\bar{X}+t_{cr,0.95}\frac{s}{\sqrt{n}})=0.95.

So, we have

Statement 2. The interval \bar{X}\pm t_{cr,0.95}\frac{s}{\sqrt{n}} contains the population mean with probability 95%.

Source of the misconception

In (3), the variable in the middle (\bar{X}) is random, and the statement that it belongs to some interval is naturally probabilistic. People not familiar with the above derivation don't understand how a statement that the population mean (which is a constant) belongs to some interval can be probabilistic. It's the interval ends that are random in (4) (the sample mean and standard error are both random), that's why there is probability! Statements 1 and 2 are equivalent!

In all statistical derivations random variables are ex-ante (before the event). No book says that but that's the way it is. An estimate is an ex-post (after the event) value of an estimator. An estimate is, of course, a number and not a random variable. Ex-ante, a confidence interval always has a probability. Ex-post, the fact that an estimate belongs to some interval is deterministic (has probability either 0 or 1) and it doesn't make sense to talk about 95%.

Since confidence levels are always strictly between 0 and 100%, students should keep in mind that we deal with ex-ante variables.
23
Sep 16

The pearls of AP Statistics 30

Where do the confidence interval and margin of error come from?

They say: A confidence interval is an interval containing the most believable values for a parameter.
The probability that this method produces an interval that contains the parameter is called the confidence level. This is a number chosen to be close to 1, most commonly 0.95... The key is the sampling distribution of the point estimate. This distribution tells us the probability that the point estimate will fall within any certain distance of the parameter (Agresti and Franklin, p.352)... The margin of error measures how accurate the point estimate is likely to be in estimating a parameter. It is a multiple of the standard deviation of the sampling distribution of the estimate, such as 1.96 x (standard deviation) when the sampling distribution is a normal distribution (p.353)

I say: Confidence intervals, invented by Jerzy Neyman, were an important contribution to the statistical science. The logic behind them is substantial. Some math is better to hide from students but not in this case. The authors keep in mind complex notions involving math and try to deliver them verbally. Instead of hoping that students will recreate mentally those notions, why not give them directly?

Motivation

I ask my students what kind of information they would prefer:

a) I predict the price S of Apple stock to be $114 tomorrow or

b) Tomorrow the price of Apple stock is expected to stay within $1 distance from $114 with probability 95%, that is P(113<S<115)=0.95.

Everybody says statement b) is better. A follow-up question: Do you want the probability in statement b) to be high or low? Unanimous answer: High. A series of definitions follows.

An interval (a,b) containing the values of a random variable S with high probability

(1) P(a<S<b)=p

is called a confidence interval. The value p which controls probability is called a confidence level and the number \alpha=1-p is called a level of significance. The interpretation of \alpha is that P(S\ falls\ outside\ of\ (a,b))=\alpha as follows from (1).

How to find a confidence interval

We want the confidence level to be close to 1 and the significance level to be close to zero. In applications, we choose them and we need to find the interval (a,b) from equation (1).

Step 1. Consider the standard normal. (1) becomes P(a<z<b)=p. Note that usually it is impossible to find two unknowns from one equation. Therefore we look for a symmetric interval, in which case we have to solve

(2) P(-a<z<a)=p

for a. The solution a=z_{cr} is called a critical value corresponding to the confidence level p or significance level \alpha=1-p. It is impossible to find it by hand, that's why people use statistical tables. In Mathematica, the critical value is given by

z_{cr}=Max[NormalCI[0, 1, ConfidenceLevel -> p]].

Geometrically, it is obvious that, as the confidence level approaches 1, the critical value goes to infinity, see the video or download the Mathematica file.

Step 2. In case of a general normal variable, plug its z-score in (2):

(3) P(-z_{cr}<\frac{X-\mu}{\sigma}<z_{cr})=p.

The event -z_{cr}<\frac{X-\mu}{\sigma}<z_{cr} is the same as -z_{cr}\sigma<X-\mu<z_{cr}\sigma which is the same as \mu-z_{cr}\sigma<X<\mu+z_{cr}\sigma. Hence, their probabilities are the same:

(4) P(-z_{cr}<\frac{X-\mu}{\sigma}<z_{cr})=P(\mu-z_{cr}\sigma<X<\mu+z_{cr}\sigma)=p.

We have found the confidence interval (\mu-z_{cr}\sigma,\mu+z_{cr}\sigma) for a normal variable. This explains where the margin of error z_{cr}\sigma comes from.

Step 3. In case of a random variable which is not necessarily normal we can use the central limit theorem. z-scores of sample means z=\frac{\bar{X}-E\bar{X}}{\sigma(\bar{X})}, for example, approach the standard normal. Instead of (3) we have an approximation

P(-z_{cr}<\frac{\bar{X}-E\bar{X}}{\sigma(\bar{X})}<z_{cr})\approx p.

Then instead of (4) we get

P(E\bar{X}-z_{cr}\sigma(\bar{X})<\bar{X}<E\bar{X}+z_{cr}\sigma(\bar{X}))\approx p.

 

In my classes, I insist that logically interconnected facts should be given in one place. To consolidate this information, I give my students the case of one-sided intervals as an exercise.