probability
DESCRIPTION
Probability. Principles of probability calculations. Probability values range from 0 to 1. Adding all probabilities of the sample yields 1. The probability that an event A will not occur is 1 minus the probability of A. - PowerPoint PPT PresentationTRANSCRIPT
Probability
Probability values range from 0 to 1. Adding all probabilities of the sample yields 1. The probability that an event A will not occur is 1
minus the probability of A. If two events are independent, the probability that
one or the other event occurs is the sum of their individual probabilities.
Principles of probability calculations
Simple probability
P(A) = 1/6 = 0.1666
Sample space: 1,2,3,4,5,6
Joint probability
P(5,6) =
P(A,B) = P(A) P(B)
P(0.166) P(0.166) = 0.0277
Joint probability
-> V NP PP-> V [NP PP]
(1) keep the dogs on the beach
keep
VP VP → V NP XP [.15]
V NP PP
the dogs on the beach
keep: V NP XP [.81]
.15 x .81 = .12
Conditional probability
keep
VP VP → V NP XP [.15]
V
NP PP
the dogs on the beach
keep: V NP [.19]
.19 x .39 x 14 = .01
NP NP → NP PP [.14]
Conditional probability
Conditional probability
Conditional probability
In a corpus including 12.000 nouns and 3.500 adjectives, 2.000 adjectives precede a noun. What is the likelihood that a noun occurs after an adjective?
P(2000)
P(12000)P(ADJ|N) = 0.1666
Conditional probability
What is the likelihood that an adjective precedes a noun?
P(2000)
P(3500)P(N|ADJ) = 0.5714
Probability distribution
Discrete probability distribution Continuous probability distribution
Types of probability distributions
Binomial distribution
two possible outcomes on each trail
the outcomes are independent of each other
the probability ratio is constant across trails
Bernoulli trail:
Binomial distribution
TH
HH HT TH TT
Binomial distribution
0 heads = HH
1 head = HT + TH
2 heads = TT
Binomial distribution
HH
HT
TH
TT
0
1
2
Sample space Random variable
Binomial distribution
Cumulative outcome Probability
0 = 11 = 22 = 1
0.250.500.25
P(x) = 1
Binomial distribution
TH
HH HT TH TT
HHH HHT HTH HTT THH THT TTH TTT
Sample space: HHH TTTHHT TTHHTH THTTHH HTT
Random variables: 0 Head1 Head2 Heads3 Heads
0 head: 11 head: 32 heads: 33 heads: 1
/ 8 = 0.125/ 8 = 0.375/ 8 = 0.375/ 8 = 0.125
Binomial distribution
Poisson distribution
Normal distribution
The center of the curve represents the mean, median, and mode.
The curve is symmetrical around the mean. The tails meet the x-axis in infinity. The curve is bell-shaped. The total under the curve is equal to 1 (by definition).
Normal distribution
Normal distribution
Standard normal distribution
1.96
x1 – x
SD
z-scores
z-scores
Zwei Kandidaten haben an zwei unterschiedlichen Sprachtests teilgenommen. Kandidat A hat 121 Punkte erzielt, Kandidat B hat 177 Punkte erzielt. Im ersten Test (an dem Kandidat A teilgenommen hat) lag der Mittelwert bei 92 und die Standardabweichung bei 14; im zweiten Test (an dem Kandidat B teilgenommen hat) lag der Mittelwert bei 143 und die Standardabweichung bei 21. Welcher der beiden Kandidaten hat besser abgeschnitten (im Vergleich zu allen übrigen Kandidaten)?
ZA = 121 – 92 / 14 = 2.07
ZB = 177 – 143 / 21 = 1.62
Central limit theorem
Central limit theorem
6, 2, 5, 6, 2, 3, 1,
6, 1, 1, 4, 6, 6, 2,
2, 1, 1, 5, 1, 3 = 2.64
X1 X2 X3 X4 M
Sample 1 6 2 5 6 4.75
Central limit theorem
X1 X2 X3 X4 M
Sample 1 6 2 5 6 4.75
Sample 2 2 3 1 6 3
Central limit theorem
X1 X2 X3 X4 M
Sample 1 6 2 5 6 4.75
Sample 2 2 3 1 6 3
Sample 3 1 1 4 6 3
Central limit theorem
X1 X2 X3 X4 M
Sample 1 6 2 5 6 4.75
Sample 2 2 3 1 6 3
Sample 3 1 1 4 6 3
Sample 4 6 2 2 1 2.75
Central limit theorem
X1 X2 X3 X4 M
Sample 1 6 2 5 6 4.75
Sample 2 2 3 1 6 3
Sample 3 1 1 4 6 3
Sample 4 6 2 2 1 2.75
Sample 5 1 5 1 3 2.5
Central limit theorem
4.75 + 3.0 + 3.0 + 2.75 + 2.5 = 3.2
5
Mean of sample mean
The sample means are normally distributed (even if the phenomenon in the parent population is not normally distributed).
2,50 3,00 3,50 4,00 4,50 5,00
case
0
2
4
6
8
10
12
Häu
fig
keit
Mean = 3,352Std. Dev. = 0,44802N = 25
Central limit theorem
Der Mittelwert der individuellen Mittelwerte nähert sich dem Mittelwert in der wahren Population an.
Die Mittelwerte der Stichproben ist normalverteilt, selbst wenn das Phänomen, das wir untersuchen, in der wahren Population nicht normalverteilt ist.
Alle parametrischen Tests nutzen die Tatsache, dass die Mittelwerte der Stichproben (ab einer bestimmten Anzahl von Stichproben) normalverteilt sind.
Central limit theorem
population
population
sample
population
sample
mean of this sample
population
sample
mean of this sample
distribution of many sample means
How many samples do you need to assume that the mean of the sample means is normally distributed?
Are your data normally distributed?
The distribution in the parent population (normal, slightly skewed, heavily skewed).
The number of observations in the individual sample.
The total number of individual samples.
Are your data normally distributed?
Confidence intervals
Confidence intervals indicate a range within which the mean (or other parameters) of the true population is located given the values of your sample and assuming a particular degree of certainty.
Confidence intervals
The mean of the sample means
The SDs of the sample means, i.e. the standard error
The degree of certainty with which you want to state
the estimation
Confidence intervals
(xn – x)2
N- 1
Standard deviation
Samples Mean
12345
1.51.81.32.01.7
8.3 / 5= 1.66 (mean)
Standard error
Samples Mean Individual means – Mean of means
12345
1.51.81.32.01.7
1.5 – 1.661.8 – 1.664 – 1.669 – 1.6612 – 1.66
8.3 / 5= 1.66 (mean)
Standard error
Samples Mean Individual means – Mean of means
12345
1.51.81.32.01.7
1.5 – 1.661.8 – 1.664 – 1.669 – 1.6612 – 1.66
0.160.14– 0.36– 0.360.04
8.3 / 5= 1.66 (mean)
Standard error
Samples Mean Individual means – Mean of means
squared
12345
1.51.81.32.01.7
1.5 – 1.661.8 – 1.664 – 1.669 – 1.6612 – 1.66
0.160.14– 0.36– 0.360.04
0.02560.01960.12960.11560.0016
8.3 / 5= 1.66 (mean)
Standard error
Samples Mean Individual means – Mean of means
squared
12345
1.51.81.32.01.7
1.5 – 1.661.8 – 1.664 – 1.669 – 1.6612 – 1.66
0.160.14– 0.36– 0.360.04
0.02560.01960.12960.11560.0016
8.3 / 5= 1.66 (mean)
0.292
Standard error
0.292
5 - 1= 0.2701
Standard error
[degree of certainty] [standard error] = x
[sample mean] +/–x = confidence interval
Confidence intervals
95% degree of certainty = 1.96 [z-score]
Confidence interval of the first sample (mean = 1.5):
1.96 0.2701 = 0.53
1.5 +/- 0.53 = 0.97–2.03
We can be 95% certain that the population mean is located in the range between 0.97 and 2.03.
Confidence intervals
SD
N
Confidence intervals
What is the 95% confidence interval of the following
sample: 2, 5, 6, 7, 10, 12?
SD: (2-7)2 + (5-7)2 + (6-7)2 + (7-7)2 + (10-7)2 + (12-7)2
6 -1
Standard error: 3.58 / 6 = 1.46
Mean: 7
= 3.58
Confidence I.: 1.46 1.96 = 2.86
7 +/– 2.86 = 4.14 – 9.86
