Lecture: Forensic Evidence and Probability Characteristics of evidence Class characteristics...

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Lecture: Forensic Evidence and Lecture: Forensic Evidence and Probability Probability Characteristics of evidence Characteristics of evidence Class characteristics Individual characteristics features that place the item into a specific category features that distinguish one item from another of the same type

Transcript of Lecture: Forensic Evidence and Probability Characteristics of evidence Class characteristics...

Page 1: Lecture: Forensic Evidence and Probability Characteristics of evidence Class characteristics Individual characteristics  features that place the item.

Lecture: Forensic Evidence and Lecture: Forensic Evidence and ProbabilityProbability

Characteristics of evidenceCharacteristics of evidence

• Class characteristics

• Individual characteristics

features that place the

item into a specific

category

features that distinguish one item from another of the same type

Page 2: Lecture: Forensic Evidence and Probability Characteristics of evidence Class characteristics Individual characteristics  features that place the item.

The standard deviation (SD) quantifies variability. If the data follow a bell-shaped Gaussian distribution, then 68% of the values lie within one SD of the mean (on either side) and 95% of the values lie within two SD of the mean. The SD is expressed in the same units as your data.

The arithmetic mean is the "standard" average, often simply called the "mean"

Page 3: Lecture: Forensic Evidence and Probability Characteristics of evidence Class characteristics Individual characteristics  features that place the item.

What is a P value?

Assume that you've collected data from two samples of animals treated with different drugs. You've measured an enzyme in each animal's plasma, and the means are different. You want to know whether that difference is due to an effect of the drug - whether the two populations have different means. Observing different sample means is not enough to persuade you to conclude that the populations have different means. It is possible that the populations have the same mean (the drugs have no effect on the enzyme you are measuring), and that the difference you observed is simply a coincidence. There is no way you can ever be sure if the difference you observed reflects a true difference or if it is just a coincidence of random sampling. All you can do is calculate probabilities.

Statistical calculations can answer this question: If the populations really have the same mean, what is the probability of observing such a large difference (or larger) between sample means in an experiment of this size? The answer to this question is called the P value.

The P value is a probability, with a value ranging from zero to one. If the P value is small, you'll conclude that the difference between sample means is unlikely to be a coincidence. Instead, you'll conclude that the populations have different means.

P<0.05 "significant"

P>0.05, "not significant"

Page 4: Lecture: Forensic Evidence and Probability Characteristics of evidence Class characteristics Individual characteristics  features that place the item.

What is a null hypothesis?

When statisticians discuss P values, they use the term null hypothesis. The null hypothesis simply states that there is no difference between the groups. Using that term, you can define the P value to be the probability of observing a difference as large or larger than you observed if the null hypothesis were true.

Page 5: Lecture: Forensic Evidence and Probability Characteristics of evidence Class characteristics Individual characteristics  features that place the item.

1% of women at age forty who participate in routine screening have breast cancer.  80% of women with breast cancer will get positive mammographies.  9.6% of women without breast cancer will also get positive mammographies.  A woman in this age group had a positive mammography in a routine screening.  What is the probability that she actually has breast cancer?

STATISTICAL SOLUTIONTo put it another way, before the mammography screening, the 10,000 women can be divided into two groups:•Group 1:  100 women with breast cancer.•Group 2:  9,900 women without breast cancer.

  After the mammography, one gets:*  80 women with breast cancer, and a positive mammography. i.e. 80% of 100* 950 women without  breast cancer, and a positive mammography. i.e. 9.6% of 9900

The probability that a patient with a positive mammogram has breast cancer is: # (breast cancer + positive mammorgraphy) / #(positive mammorgraphy ) = 80/(80+950) = 7.8%

Page 6: Lecture: Forensic Evidence and Probability Characteristics of evidence Class characteristics Individual characteristics  features that place the item.

1% of women at age forty who participate in routine screening have breast cancer.  80% of women with breast cancer will get positive mammographies.  9.6% of women without breast cancer will also get positive mammographies.  A woman in this age group had a positive mammography in a routine screening.  What is the probability that she actually has breast cancer?

BAYESIAN SOLUTION

The original proportion of patients with breast cancer is known as the prior probability:  P(C) = 1% and P(~C) = 99%The chance of a patient having a positive mammography given that she has cancer, and the chance that of a patient having a positive mammography given that she does not have cancer, are known as the two conditional probabilities.  Collectively information is often termed the liklehood ratio: P(M | C) = 80% i.e probability of +ve mammogram given that she has cancerP(M | ~C) = 9.6% i.e probability of +ve mammogram given that she does not

have cancerThe final answer - the estimated probability that a patient has breast cancer given that we know she has a positive result on her mammography - is known as the revised probability or the posterior probability.

Page 7: Lecture: Forensic Evidence and Probability Characteristics of evidence Class characteristics Individual characteristics  features that place the item.

1% of women at age forty who participate in routine screening have breast cancer.  80% of women with breast cancer will get positive mammographies.  9.6% of women without breast cancer will also get positive mammographies.  A woman in this age group had a positive mammography in a routine screening.  What is the probability that she actually has breast cancer?

prior probability x conditional probability = posterior probability

P(C) . P(M | C) = P(C | M)P(~C) P(M | ~C) P(~C | M)

0.01 . 0.8 = 0.008 = 800.99 0.096 0.095 950

the estimated odds that a patient has breast cancer given that we know she has a positive result on her mammography are 80 to 950

the estimated probability that a patient has breast cancer given that we know she has a positive result on her mammography is 80 / (80+950) = 7.8%