The Normal Distribution

1
The Normal Distribution

• Forensic Statistics

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The Normal Distribution

• A theoretically appealing model to explain many
  forms of natural continuous variation.
• Many discrete distributions may be approximated
  by the normal distribution for large samples
• Most continuous variables, especially from the
  biological sciences, are distributed normally,
  e.g. the length of the femur (thigh bone) in
  adult humans, the length of the tibia (shin bone)
  in adult humans.

3
Probability density functions (PDF) for adult
human femurs and tibias. The mean and standard
deviation of the femur PDF are arrowed
4
Comparison between the PDFs for femurs and tibias

• Tibia is usually shorter than the femur, but some
  people have tibias which are longer than other
  peoples femurs.
• The mean of tibia lengths is shorter than the
  mean of the femur lengths.
• Femurs are more spread out than the tibias,
  leading to the PDF for tibias being more peaked.
• Each normal distribution has two parameters which
  control what it looks like (1) the mean, the
  location about which the distribution is
  centered (2) the standard deviation, the
  dispersion parameter.

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The standard deviation is the square root of the
sample variance
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The standard error of the mean

• Another measure is the standard error of the mean
  (or simply standard error) which arises from the
  fact that bar x is only an estimate of the
  population mean. Were we to repeatedly to sample
  from the same population the means of the samples
  would be unlikely to be the same in each case.
  This means that our estimates of the population
  mean would also vary. This variability is called
  the standard error of the mean (SEM)
• SEM(x) SD / vn

7
Empirical Probability Density for the ?9-THC
content of sample marijuana from 1986
8
Probability Density Function superimposed on the
histogram of ?9-THC in marijuana seizures from
1986.
9
Percentage Points of the Normal Distribution

• Use the table on slide 7 to determine the
  probability that the ?9-THC content of marijuana
  from 1986 is 8.
• This is done by summing the probabilities of
  finding 6.0 ? 6.5, 6.5 ? 7.0, 7.0 ? 7.5, and 7.5
  ? 8.0 ?9-THC.
• Very much the same thing can be done with the
  normal distribution, only the summation has to be
  calculated using a mathematical process called
  integration.
• As integration is a difficult process,
  statisticians have calculated tables for a
  standardised normal distribution which can be
  rescaled to fit any particular normal
  distribution (see appx. B).
• The standard normal distribution has a mean of 0
  and a standard deviation of 1.

10
Standard normal distribution with mean 0 and
standard deviation 1. The shaded area covers
the area -8 (minus infinity) to 2 standard
deviations.
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Area under the curve (1)

• The shaded area under the standard normal curve
  extends from -8 to 2 standard deviations.
• 8 is notation for infinity. The normal
  distribution is asymptotic in that it goes from
  -8 to 8, so values occurring at very large ve or
  ve numbers of standard deviations are very small
  but never zero.
• If we wish to find the area under this portion of
  the curve we simply look down the rows of appx B
  until we find the row labelled z 2.0. For the
  third decimal place the appropriate column is
  selected.
• Standardised variables obtained by subtracting
  the mean and dividing the result by the standard
  deviation are sometimes called z-scores, or z for
  short.
• For z 2.00, the shaded area is 0.9772. The
  total area under the curve is 1, so we can see
  that 97.22 of the total area lies between -8 and
  2 standard deviations.

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The distribution of slide 10 rescaled to the
normal distribution underlying the ?9-THC content
sample from 1986.
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Area under the curve (2)

• On the previous slide, the mean is 8.59 and the
  standard deviation is 1.09.
• The upper limit at the mean 2 standard
  deviations is at 8.59 (2 1.09) 10.75
• As we know from slide 10 that 97.72 of the total
  area lies in the shaded zone, we can say that
  97.72 of marijuana consignments seized in 1986
  will have a ?9-THC content of less than 10.75.

14
The normal distribution for ?9-THC content from
marijuana contents seized in 1986 showing the
mean and the 95 symmetric area about the mean
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Area under the curve (3)

• Find the shortest interval in which 95 of the
  samples fall.
• By symmetry, this is centred about the mean.
• The shaded area contains 95 of the area, leaving
  5 over both tails, which means 2.5 in each
  tail. Thus the appropriate percentage point in
  appx B is 100 2.5 97.5.
• We see that 97.5 lies at 1.96 standard
  deviations from the mean.
• Therefore the interval which contains 95 of the
  area goes between the mean minus 1.96 standard
  deviations and mean plus 1.96 standard
  deviations, which is 8.59 (1.96 1.09) 8.59
  2.13 6.45 ? 10.73
• This means that there is a 95 probability that
  the ?9-THC content of marijuana seized in 1986 is
  between 6.45 and 10.73.

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The t-distribution and the standard error of the
mean (SEM)

• We previously defined the SEM as the standard
  deviation divided by the square root of the
  sample size.
• It is a measure of the spread of confidence for
  the population mean when it has been estimated by
  a sample taken from the population.
• For example, the ?9-THC concentration in
  marijuana from 1986 has a standard deviation of
  1.09, so the standard error for a sample of 20
  for the estimate for the mean is 1.09 / 4.47
  0.244.
• A mean, however, is not in itself distributed
  with a normal distribution. If we took repeated
  samples, the means we found would not always be
  the same, but be distributed with a
  t-distribution.
• The t-distribution has only one parameter
  defining its shape. This is called the degrees of
  freedom (df), and is based on the sample size.

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t-distribution (df 4) with a standard normal
superimposed. The tails of the t-distribution are
shaded and contain 2.5 of the area in each.
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t-testing between two independent samples

• A widespread use for the t-distribution is
  testing between the means of two samples to
  examine the hypothesis that the means of the
  populations from which the two samples were drawn
  are equal (the null hypothesis).

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Normal models for two sub-samples of n10 for the
?9-THC from seizures during 1986
20
Summary statistics for the two sub-samples of
slide 19
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The object of a t-test

• From slide 19 even random sampling from the same
  set of data has led to a difference in the means
  of the two sub-samples of 0.76.
• The object of a t-test is to look at the
  differences in means and see whether the
  difference is due to chance selection, or some
  real populational difference.

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The null hypothesis and the alternative hypothesis

• Conventionally we start by erecting two
  hypotheses.
• The null hypothesis, H0, is one of no difference,
  or that the means of the two sub-samples can be
  regarded as belonging to the same population.
• The second hypothesis is complementary to the
  null hypothesis and is called the alternative
  hypothesis, H1. It states that the means from the
  two sub-samples are not equal, and that there are
  grounds for treating the two sub-samples as being
  drawn from different populations.

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Calculating t (1)

• The difference in the mean values of sub-sample 1
  and sub-sample 2 is 0.77, and this difference
  will have a distribution centred around 0.77
  with a dispersion se(x1 x2) given by
• Where x1 and x2 refer to the two sub-samples, n1
  is n for sub-sample 1, n2 is n for sub-sample 2.

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Calculating t (2)

• The term s in the equation is an estimate for
  the pooled variance. The reason that we need a
  pooled variance is that we are trying to estimate
  a distribution for the difference between two
  means.
• We cannot have a single unimodal distribution
  which has two variances, so we need a single
  estimate of variance.
• This is done by a form of weighted average of the
  two component variances given by the formula on
  the next slide

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Calculating t (3)
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Calculating t (4)

• s1² and s2² are the variances of the two
  sub-samples. Substituting the information from
  the table on slide 20 into the equation on the
  previous slide

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Calculating t (6)

• Taking s and substituting into the equation of
  slide 23

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Calculating t (7)

• The estimate for the standard error for the
  difference between the two sample means is 0.50.
  .
• From appx C 95 of the probability for the
  difference between the two sample means will lie
  within t 2.101 standard errors of the mean (for
  the t-test use df n1 n2 -2), so a confidence
  interval for the difference of 0.76 will be
  0.76 (0.50 2.101) 0.76 1.05 -0.29 ?
  1.81
• The 95 confidence interval contains 0 as a
  possible value for the difference in means
  between the two samples.
• Hence one would accept the hypothesis H0, and
  conclude that there is no significant difference
  in the means of the sub-samples at 95 confidence
  (or 5 significance), and these samples could
  have been taken from consignments with the same
  ?9-THC content.

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Comparing marijuana seizures from 1986 and 1987
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Summary statistics for the ?9-THC concentrations
in marijuana from 1986 and 1987 seizures
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Calculating t
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Interpreting t

• The difference in the two means is 8.58-7.79
  0.76 with standard error 0.31
• Df 20 15 2 33
• There is no specific row in appx C for df 33,
  so use three tenths of the way between df 30
  and df 40.
• The value of t is 2.036 standard errors.
• A 95 confidence interval for the difference in
  the two means is therefore 0.76 (2.036 0.31)
  0.13 ? 1.39.
• This interval does not include 0, and so we may
  act as if the alternative hypothesis H1 is true.
  It has been shown with 95 confidence that the
  ?THC concentrations in the two groups is
  different.

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Testing between paired observations

• Sometimes a question may arise concerning the
  differences between two means which may not be
  considered independent, e.g. where two treatments
  are applied to the same group of individuals such
  that each person receives treatment A then B,
  with some observations of interest being taken
  for each person under each treatment.
• If the effects of treatment A are equivalent to
  the effects of treatment B then the mean of the
  differences should 0.
• Because the observations are subject to
  uncertainty, it would be unusual for the
  differences to be exactly zero.
• So how large must the mean of differences be
  before we think that one treatment is better than
  the other?

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Number of cells recovered from swabs from men
under two different extraction regimes, water and
phosphate buffered solution.
35
Means and differences of numbers of cells from
three men on each of three occasions
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Calculating t for matched pairs

• The sum of the squared deviations S 18371500,
  so the variance is 18371500 / n-1 2296438.
• The standard deviation is the square root of the
  variance 1515.4
• The standard error of the mean (SEM) is the
  standard error divided by the square root of n
  1514.4 / v9 505.13
• A 95 confidence interval for the mean of -714
  cells can be calculated from the SEM and the
  tabulated value for t in appx C for n 1 8
  degrees of freedom, which is 2.306
• The 95 confidence interval is -714 (2.306
  505.13) -1878.84 ? 450.84
• This confidence interval includes zero, so at 95
  confidence 0 is a possible value for the mean
  difference.
• This gives us grounds for accepting H0, and
  acting as if there were no differences in the
  incidence of cell recovery between water and PBS.
  

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Confidence, significance, p-values (1)

• H0 is a hypothesis of no difference
• A Type 1 error is the rejection of H0 when H0 is
  true
• a is a pre-defined probability, called a
  significance level, at which making a Type I
  error is acceptable.
• A p-value is the probability of finding the
  observed values, or any values more extreme,
  given the truth of the null hypothesis.
• Confidence is the complement of significance,
  that is 1-a

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Confidence, significance, p-values (2)

• H1 is a hypothesis of difference.
• A Type II error is the error of not rejecting H0
  when H0 is false
• ß is the probability of making a type II error
• 1-ß is called the power of a test, and can be
  interpreted as the probability of detecting a
  difference if one exists.