Ex St 801Statistical Methods

Probability

and

Distributions

a graphic:

population sample

Gather data

Make inferences

parameters statistics

.,,,,, 2 etc.etc,b,p,SS,or

.,ˆ,ˆ,ˆ,ˆ,ˆ2

2

y

etc

some definitions.

Population: Set of all measurements of interest to the

investigator.

Parameter: Characteristic of the population; Greek letter, usually unknown.

Sample: Subset of measurements selected (usually at random) from the population.

Statistic: Characteristic of the sample; Latin letter or Greek letter with a hat ( ^ ).

Random Variable: The characteristic we measure, usually called ‘y’.

a graphic, again

population sample

Probability

Gather data

Make inferences

Statistics

Probability: Likelihood of occurrence; we know the population, and we predict the outcome or the sample.

Statistics: We observe the sample and use the statistics to describe the unknown population.

When we make an inference about the population, it is

desirable that we give a measure of ‘confidence’ in our

being correct (or incorrect). This is done giving a

statement of the ‘probability’ of being correct. Hence,

we need to discuss probability.

More terms:

So, Probability: more terms:

event: is an experiment

y: the random variable, is the outcome from one event

sample space: is the list of all possible outcomes

probability, often written P(Y=y), is the chance that Y will be a certain value ‘ y ’ and can be computed:

number of successes p = -------------------------. number of trials

8 things to say about probability:

1. 0 p 1 ; probabilities are between 0 and 1.

2. pi = 1 ; the sum of all the probabilities of all the possible outcomes is 1.

Event relations

3. Complement: If P(A=a) = p then A complement is

is P( A not a) = 1-p (= q sometimes).

Note: p + (1-p) = 1 (p+q=1).

A complement is also called (not the mean).A

8 things continued

4. Mutually exclusive: two events that cannot happen together. If P(AB)=0, then A and B are M.E.

5. Conditional: Probability of A given that B has alreadyhappened. P(A|B)

6. Independent: Event A has no influence on the outcomeof event B. If P(A|B) = P(A) orP(B|A) = P(B) then A and B are independent.

8 things continued again

two laws about events:

7. Multiplicative Law: (Intersection ; AND)

P(AB) = P(A B) = P(A and B) = P(A) * P(B|A) = P(B) * P(A|B)

if A and B are independent then P(AB) = P(A) * P(B).

8. Additive Law: (Union; OR)

P(A B) = P(A or B) = P(A) + P(B) - P(AB).

A BAB

The Venn diagram below can be used to explainthe 8 ‘things’.

An example: Consider the deck of 52 playing cards:

sample space:

(A, 2, 3, … , J, Q, K) spades (A, 2, 3, … , J, Q, K) diamonds (A, 2, 3, … , J, Q, K) hearts (A, 2, 3, … , J, Q, K) clubs

Now, consider the following events:

J= draw a J: P(J)= 4/52=1/13 F = draw a face card (J,Q,K): P(F)= 12/52=3/13H = draw a heart: P(H)= 13/52

An example.2: Compute the following:

1. P(F complement) = (52/52 - 12/52) = 40/52

2. Are J and F Mutually Exclusive ?

No: P(JF) = 4/52 is not 0.

3. Are J and F complement M.E. ?

Yes: P(J and ) = 0

4. Are J and H independent ?

Yes: P(J) = 13/52 = 1/13 = P(J|H)

F

An example.3: Compute the following:

5. Are J and F independent ?

No: P(J) = 4/52 but P(J|F) = 4/12

6. P(J and H) = P(J) * P(H|J) = 4/52*1/4 = 1/52

7. P(J or H) = P(J) + P(H) - P(JH)

= 4/52 + 13/42 - 1/52 = 16/52.

Two sorts of random variables are of interest:

DISCRETE: the number of outcomes is countable

CONTINUOUS: the number of outcomes in infinite(not countable).

Random variables are often described with probability

distribution functions. These are graphs, tables or

formula which allow for the computation of

probabilities.

A few common Discrete probability distributions are:

Uniform: P(Y=y) = 1/number of outcomes (all outcomes are equally likely)

Binomial P(Y=y) = nCy * py * (1-p)(n-y)

where: nCy is the combination of n things taken y at the time n is the number of trials

y is the number of successes p is the probability of succeeding in

one trial in each trial, the only outcomes are success and failure (0,1).

Poisson P(Y=y) = ;

y=0,1,2,… (for example number of people waiting in line at a teller) = the population mean of Y.

These are a few of many and are used when there areonly a few possible outcomes: number of defects on a circuit board, number of tumors in a mouse, pregnant or not, dead or alive, number of accidents at an intersection and so on.

!yey

PROBABILITY DEFINITIONS

• An EXPERIMENT is a process by which an observation is obtained.

• A SAMPLE SPACE (S) is the set of possible outcomes or results of an experiment.

• A SIMPLE EVENT (Ei) is the smallest possible

element of the sample space.

PROBABILITY DEFINITIONS

• A COMPOUND EVENT is a collection of two or more simple events.

• Two events are INDEPENDENT if the

occurrence of one event does not affect the

occurrence of the other event.

SAMPLE SPACE FOR THE COINS IN A JAR EXAMPLE

(P, N1, N2) (P, N1, D) (P, N1, Q) (P, N2, D) (P, N2, Q)

(P, D, Q) (N1, N2, D) (N1, N2, Q) (N1, D, Q) (N2,

D, Q)

INTERPRETATIONS OF PROBABILITY

Classical:

Relative Frequency:

PROPERTIES OF PROBABILITIES

RANDOM VARIABLES

• A RANDOM VARIABLE (r.v.) is a numerical valued function defined on a sample space

• A DISCRETE RANDOM VARIABLE is a random variable that can assume a countable number of values.

RANDOM VARIABLES (CONT.):

• A CONTINUOUS RANDOM VARIABLE is a random variable that can assume an infinitely large number of values corresponding to the points on a line interval.

A RANDOM VARIABLE FOR THE COINS IN A JAR EXAMPLE

Let Y be the amount of money taken out of the jar.

11 16 31 16 31

(P, N1, N2) (P, N1, D) (P, N1, Q) (P, N2, D) (P, N2, Q)

36 20 35 40 40

(P, D, Q) (N1, N2, D) (N1, N2, Q) (N1, D, Q) (N2,

D, Q)

PROBABILITY DISTRIBUTIONS

• A DISCRETE PROBABILITY DISTRIBUTION is a

formula, table or graph that shows the probability

associated with each value of the discrete random

variable.

• A CONTINUOUS PROBABILITY DISTRIBUTION

is given by an equation f (y) (probability density

function) that shows the density of probability as it

varies with the continuous random variable.

COINS IN A JAR EXAMPLE

PROBABILITY DISTRIBUTION

Y 11 16 20 31 35 36 40P(Y) .1 .2 .1 .2 .1 .1 .2

EXPECTED VALUE (MEAN)E(Y) = 27.6

VARIANCE VAR (Y) = 105.84

EXPECTATION AND VARIANCE FOR A

DISCRETE RANDOM VARIABLE

yall

yall

ypyyEyVar

ypyYE

)()(])[()(

)()(

222

EXPECTATION FORMULA

SUPPOSE E(Y) = µY AND E(X)= µX

• E(aY) = a E(Y) = a µY

• E(Y + X) = E(Y) +E(X) = µY + µX

• E(aY + bX) = aE(Y) +bE(X) = aµY + b µX

VARIANCE FORMULA

Suppose Var(Y) = Y2 and Var(X) = X

2

•Var(aY) = a2 Var(Y)

•If Y and X are independent, then

–Var(Y + X) = Var(Y) + Var(X)

–Var(aY + bX) = a2 Var(Y) + b2 Var(X)

THE NORMAL DISTRIBUTIONSome properties

1. The area under the entire

curve is always 1.

2. The distribution is

symmetric about the

mean 3. The mean and the median

are equal.

4. Probabilities may be found

by determining the

appropriate area under

the curve.

SAMPLING DISTRIBUTIONS

• The SAMPLING DISTRIBUTION of a statistic is the probability distribution for the values of the statistic that results when random samples of size n are repeatedly drawn from the population.

• The STANDARD ERROR is the standard deviation of the sampling distribution of a statistic.

DIAGRAM FOR OBTAINING A SAMPLING DISTRIBUTION

POP.

SAMPLE 1

SAMPLE 2

SAMPLE 499

SAMPLE 500

Y

Y1

Y2

Y499

Y500

THE CENTRAL LIMIT THEOREM

• If random samples of n observations are drawn from a population with a finite mean, , and a finite variance 2, then, when n is large (usually greater than 30), the SAMPLE MEAN, will be approximately normally distributed with mean and variance 2/n.

• The approximation becomes more accurate as n becomes large.

THE CENTRAL LIMIT THEOREM

nlargeforNYthen

Yif

n

2

2

,~

),(~

AN APPLICATION OF THE SAMPLING DISTRIBUTION

• The Ybar CONTROL CHART can be used to detect shifts in the mean of a process.

• The chart looks at a sequence of sample means and the process is assumed to be “IN CONTROL” as long as the sample means are within the control limits.

Y CONTROL CHART

UPPERCONTROL LIMIT

LOWERCONTROL LIMIT

CENTERLINE

SAMPLE

GLASS BOTTLE EXAMPLE

• A glass-bottle manufacturing company wants to maintain a mean bursting strength of 260 PSI.

• Past experience has shown that the standard deviation for the bursting strength is 36 PSI.

• The company periodically pulls 36 bottles off the production line to determine if the mean bursting strength has changed.

GLASS BOTTLE EXAMPLE (CONT)

• Construct a control chart so that 95% of the sample means will fall within the control limits when the process is “IN CONTROL.”

• The END