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Statistical inference: probability

distributions and confidence intervals

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We are now familiar with descriptivestatistics; but the main use of statisticalmethods is not description, but prediction

o i.e. we collect samples mostly to predict

characteristics of the whole population

The key instrument of extrapolation fromsample to population is the analysis ofprobability distributions:

o By assuming that our variables have a certaindistribution (normal, uniform, etc.), we can usesamples to infer population properties

In the following, well examine the concept

and uses of statistical distributions 2

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Most utilised statistical distribution is the

normal distribution (the Bell curve)

o also the most infamous due to certain misuses

o

http://crab.rutgers.edu/~goertzel/normalcurve.htm

However, there is nothing intrinsically wrong

with using probability distributions

o

well, anything in the wrong hands (from a breadknife to a fundamental law of nature proposed by

a pacifist) may become a weapon

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http://crab.rutgers.edu/~goertzel/normalcurve.htmhttp://crab.rutgers.edu/~goertzel/normalcurve.htmhttp://crab.rutgers.edu/~goertzel/normalcurve.htmhttp://crab.rutgers.edu/~goertzel/normalcurve.htmhttp://crab.rutgers.edu/~goertzel/normalcurve.htmhttp://crab.rutgers.edu/~goertzel/normalcurve.htmhttp://crab.rutgers.edu/~goertzel/normalcurve.htmhttp://crab.rutgers.edu/~goertzel/normalcurve.htm

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The first reason for popularity of the normal curve isdescriptive; i.e. we use it to model distribution of certaintraits that look bell-shaped

What traits are bell-shaped? Typically, traits that are

optimised or established by biological or social processes,and thus have a tendency to occur at anexpected/established/optimal value

o classic example: biological traits under natural selection

o but the only reason Darwin applied the principle of optimisationto nature is that this was a current concept in Victorian society;

human societies also define optimal values of certain features,with deviations (in both directions) being less common

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Normal curve can describe features of optimised traitso mean value should be the most likely or frequently observed

o the furthest from the mean, the less likely a value should be

o sum of all cases (or probabilities) should be 100% (thats thewhole sample)

What kind of curve/distribution produces this pattern?

Lets try an exponential:o > x y plot(x,y) # what does it look like?

Try others:o > y y y

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The normal distribution is just a modified version ofour exponential

The curve

N(0,1) =

is thestandard normal distribution with

mean=0

sd=1

sum of frequencies=1

Distribution N(0, 1) is possibly the most used instatistical analyses

Is says that for example:

the probability of being well above average (+3standard deviations above mean) is only 0.1%

probability of being one standard deviation

below average (-1 sd) is 0.1+2.1+13.6=15.8%(i.e. everything below -1) 6

-3 -2 -1 0 +1 +2 +3

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However, real traits (body height,income, schooling years,number of social mediaaccounts) may have a normaldistribution (bell shape), butrarely with mean=0 andstandard deviation=1

That is not a problem: we canstandardise variables, i.e.

transform them so thateverything you measure hasmean=0 and sd=1

How is this done? With z-scores7

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1) We take variablex and subtract the mean fromeach caseo If mean height is 180 cm, someone 170 cm tall now

measures 170-180=-10

2) We take all residuals (case minus mean) anddivide by standard deviation of sampleo If sd=10 and mean is 180cm, then someone measuring

190 cm deviates -10 cm/10 cm= -1 standard deviationbelow the mean

In summary, standardisation or calculation ofz-scores is simply convertingany measurements into

standard deviation units using

=

i.e., if your height, or age, or income, etc. are average,then on the z-scale all those things measure zero

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So: if in a population

o mean height = 180 cm

o standard deviation=10

and you are 170cm, then

o you measure 10 cm above the average

o you measure z = (170 180)/10 = -1

This means that the probability of beingshorter than 170 cm in this populationis

o 0.1 + 2.1 + 13.6 = 15.8%

The reason for standardising is clear: itis the theoretical step that allows theapplication of the normal distribution tomany quantifiable aspects of reality

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We are interested in intervals of the normal

curve, not points

Why? What does it mean to ask what is the

probability of being a millionaire in the UK?(or their frequency)

o it does not mean the probability of having

exactly 1 million (thats a point)

o

it means everyone havingover 1 million(and thats an interval)

Cumulative probabilityis the probability of an

interval of values

10

a lower interval

an upper interval

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It is easy to estimate cumulative probability of being

smaller than a value

o You provide the individual value, the mean, and

the sd, and R calculates z-score and the

probability of the interval defined by that value

Command pnorm(test value, mean, sd) calculates

cumulative probability from left to right, i.e. from to a valuex;

o (thats the blue area)

Example: if your height is 170 cm (and average is

180 cm, sd=10 cm), then

o > pnorm(170,180,10)

o [1] 0.1586553 11

a lower interval

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pnorm can estimate upper intervals too (i.e. the probability of

being over a given value:

Example:

o what is the probability of being at least (i.e. taller than)

190cm in the same population?

Probability of beingsmallerthan 190 cm is

> pnorm(190,180,10)

[1] 0.8413447

i.e. 0.841=84.1%

Therefore the probability of being over 190 is the rest. i.e.

> 1-pnorm(190,180,10)

[1] 0.1586553

i.e.: probability of being taller than 190 cm is 1 (100%) minus the

probability of being smaller than 190 cm 12

an upper interval

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Important: we can combine the two

things to calculate probability of extreme

values (i.e. too large or too small)

So what is the probability of being

shorter than 170cm OR taller than 190

cm, with N(180, 10)?

> 1pnorm(190, 180, 10)+pnorm(170, 180, 10)

(check why)

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Now the most important case (well see why):

What about probability ofnot being extreme, i.e. of being

between 170 cm and 190 cm? (This means less than 10 cm

off average of 180 cm)

o > pnorm(190, 180, 10) pnorm(170, 180, 10)

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Take the estimates of years at school by country (from the HDR2011

database); this is the variableschoolingyears:

How can we estimate the proportion of countries with children havinga) less than 3 years of schooling?

b) less than 5 years of schooling?

c) at least 7 years of schooling?

Hints:

-You need to use function pnorm

-To use pnorm you need the test value, the mean and the standard

deviation of variableschooling years 15