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The normal distribution. Binomial distribution is discrete events, (infected, not infected) The normal distribution is a probability density function for a continuous variable, and is represented by a continuous curve. Density = relative frequency of varites on the Y (horizontal) axis.

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slide1

The normal distribution

Binomial distribution is discrete events, (infected, not infected)

The normal distribution is a probability density function for a continuous variable, and is represented by a continuous curve.

Density = relative frequency of varites on the Y (horizontal) axis.

slide3

Cannot evaluate the probability of the variable being exactly equal to some value (that area of the curve is soooo small)

Must estimate the frequency of observations falling between two limits

Won’t work

freq

2

3

Some variable

slide5

A normal curve is defined as:

1

e-(x-)2/22

Y=

22

*

Y is the height of the ordinate

μ is the mean

σ is the standard deviation

π is the constant 3.14159

e is the base of natural logarithms and is equal to 2.718282x can take on any value from -infinity to +infinity.

slide7

The shape of a given normal curve results from different values of  and 

The mean, , determines the midpoint

The standard deviation, , changes the shape, it affects the spread or the dispersion of scores

The larger the value of  the more dispersed the scores; the smaller the value, the less dispersed.

slide8

There is not one Normal Curve

The mean, , determines the midpoint

A smaller , means less dispersion

Bonus question. What’s wrong with this graph?

slide10

How to determine what proportion of a normal population lies above/below a certain level

If distribution of Hobbit heights is normal with mean = 120 cm, SD = 20

Half < 120 & half >120

What is probability of finding a Hobbit taller than 130 cm??

120 cm

The average Hobbit

slide11

Calculate the normal deviate

- Any point on normal curve

- Here, 130 cm

Mean

Xi - 

Z =

SD

- Normal deviate

- Test statistic

Z = (130-120)/20 = 0.5

slide12

Table B.2; Zar

Table A S & R

P (probability) (Xi >130 cm) = P (Z>0.50) = 0.3085 or 30.85%

slide13

In any normal population:

68.27% of measurements lie w/in (  1)

99.73% of measurements lie w/in (  3)

50% lie w/in (  0.67 )

95% lie w/in (  1.96 )

- hence the 95% confidence interval of a sample = X  1.96 * s

- the range within one is 95% confident that the true population mean, , is to be found

slide14

The normal distribution in biology

Binomial distribution (p + q)k

Imagine a trait is controlled by many factors, ex skin pigmentation.

When a factor is present, it contributes 1 unit of pigmentation

If 3 factors were present, the animal would have skin that was 3 units dark

Assume 0.5 probability of each factor being present: p (hence 0.5 probability of each factor being absent): q

slide15

If only one factor existed, (p + q)1; k=1

Half the animals would have it, half would not

expected proportion w 0 pigmentation unit=0.5

expected proportion w 1 pigmentation unit=0.5

If two factors existed, (p + q)2; k=2

There will now be 3 classes: pp, pq, qq

Frequency of pp = 0.25 or (0.5)2

Frequency of pq = 0.5 or 2[0.5*0.5]

Frequency of qq = 0.25 or (0.5)2

slide16

If k (number of independent factors) becomes large, the distribution produced by binomial expansion would come very close to the normal distribution

Many biological variables behave like this

When samples are large, this occurs even when the factors are not strictly independent, or not all equal in magnitude of effect.

slide17

Assessing Normality

I'm not an outlier; I just haven't found my distribution yet

slide18

Skewness: asymmetry, one tail is drawn out

Mean not equal to median

freq

2

3

Some variable

slide19

kurtosis: the proportion of a curve located in the center, shoulders and tails

How fat or thin the tails are

leptokurtic

no shoulders

platykurtic

wide shoulders

slide21

2

Xi -

X

X

X

Revisit variance and SD relative to normal curve

Xi

Xii

2

Xii -

Mean SS =variance

Total sum of squares

I side =SD

=

n-1

slide22

Distribution of means assuming normality

If you take repeated samples of size N from a normally distributed population, the distribution of the the means of those samples will be normal

If you take repeated samples of size N from a non-normally distributed population, the distribution of the the means of those samples will tend towards normality

Central Limit Theorem

slide23

Calculate the variance of the mean of the distribution of means

2

Variance of mean =

n

Square root of the variance of the mean is the SD of the mean, also called the standard error

But rarely know pop parameters, so…….

2

x

=

n

slide24

For a sample,

s2

sx

=

n

or

s

sx

=

n

 We’ll come back to this with more on testing differences between a mean and a value or between 2 means

slide25

Changes in s2, SD, & SE with increasing N

Variance, s2

Become more accurate approximations of “true”

SD, s2

Parameter units

SE becomes smaller with increased sampling

SE, SD/ n

Sample size (n)

slide26

What SD and SE mean

SD is a parameter of a natural population (even though real populations are constantly changing). Its size reflects real, natural variance. Big is not good, small is not good. SD just is. Natural dispersion of population

SE becomes smaller with increasing sample size, therefore reflects sampling effort. Accuracy of mean.

Both frequently reported (graphed) in ecological / biological literature. SE smaller, so often favored– but this is wrong reasoning!

Practically either OK, if you state which is shown and report n!