Statistic in Health Care Management: Assignment Week 3
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chapter5.pdf
Chapter 5
The Role of Probability
Learning Objectives
• Define the terms “equally likely” and “at random”
• Compute and interpret unconditional and conditional probabilities
• Evaluate and interpret independence of events
• Explain the key features of the binomial distribution model
• Calculate probabilities using the binomial formula
Learning Objectives
• Explain the key features of the normal distribution model
• Calculate probabilities using the standard normal distribution table
• Compute and interpret percentiles of the normal distribution
• Define and interpret the standard error
• Explain sampling variability
• Apply and interpret the results of the Central Limit Theorem
Two Areas of Biostatistics
Goal: Statistical Inference
POPULATION
SAMPLE
= ? n, X
Descriptive Statistics
Sampling from a Population
Population N
n
n
n
n
n
n
n
n
n
n
SAMPLES
Sampling:
Population Size=N, Sample Size=n
• Simple random sample
– Enumerate all members of population N (sampling frame), select n individuals at random (each has same probability of being selected)
• Systematic sample
– Start with sampling frame; determine sampling interval (N/n); select first person at random from first (N/n) and every (N/n) thereafter
Sampling:
Population Size=N, Sample Size=n
• Stratified sample
– Organize population into mutually exclusive strata; select individuals at random within each stratum
• Convenience sample
– Non-probability sample (not for inference)
• Quota sample
– Select a pre-determined number of individuals into sample from groups of interest
Basics
• Probability reflects the likelihood that outcome will
occur
• 0 < Probability < 1
N
outcomewithNumber yProbabilit
Example 5.1.
Basic Probability
Age 5 6 7 8 9 10 Total
Boys 432 379 501 410 420 418 2560
Girls 408 513 412 436 461 500 2730
Total 840 892 913 846 881 918 5290
P(Select any child) = 1/5290 = 0.0002
Example 5.1.
Basic Probability
P(Select a boy) = 2560/5290 = 0.484
P(Select boy age 10) = 418/5290
= 0.079
P(Select child at least 8 years of age)
= (846+881+918)/5290
= 2645/5290 = 0.500
Conditional Probability
• Probability of outcome in a specific sub-
population
• Example 5.1,
P(Select 9 year old from among girls) =
P(Select 9 year old|girl)
= 461/2730 = 0.169
P(Select boy|6 years of age)
= 379/892=0.425
Example 5.2.
Conditional Probability
Prostate Cancer
No Prostate Cancer
Total
Low PSA 3 61 64
Moderate PSA 13 28 41
High PSA 12 3 15
Total 28 92 120
Example 5.2.
Conditional Probability
P(Prostate Cancer|Low PSA)
= 3/64 = 0.047
P(Prostate Cancer|Moderate PSA)
= 13/41 = 0.317
P(Prostate Cancer|High PSA)
= 12/15 = 0.80
Sensitivity and Specificity
Sensitivity = true positive fraction
= P(test +|disease)
Specificity = true negative fraction
= P(test -|disease free)
False negative fraction=P(test -|disease)
False positive fraction=P(test +|disease free)
Example 5.4.
Sensitivity and Specificity
Affected Fetus
Unaffected Fetus
Total
Positive Screen
9 351 360
Negative Screen
1 4449 4450
Total 10 4800 4810
Sensitivity and Specificity
Sensitivity = P(test +|disease) =9/10=0.90
Specificity = P(test -|disease free)
= 4449/4800 = 0.927
False negative fraction= P(test -|disease)
= 1/10 = 0.10
False positive fraction=P(test +|disease free)
= 351/4800 = 0.073
Independence
• Two events, A and B, are independent if P(A|B) = P(A) or if P(B|A) = P(B)
• Example 5.2. Is screening test independent of prostate cancer diagnosis?
– P(Prostate Cancer) = 28/120 = 0.023
– P(Prostate Cancer|Low PSA) = 0.047
– P(Prostate Cancer|Moderate PSA) = 0.317
– P(Prostate Cancer|High PSA) = 0.80
Bayes Theorem
• Using Bayes Theorem we revise or update a
probability based on additional information
– Prior probability is an initial probability
– Posterior probability is a probability that is revised
or updated based on additional information
Bayes Theorem
P(B)
A)P(A)|P(B B)|P(A
)A'|)P(BP(A' A)|P(A)P(B
A)P(A)|P(B B)|P(A
Example
• In Boston, 51% of adults are male
• One adult is randomly selected to participate in
a study
Prior probability of selecting a male= 0.51
Example
• Selected participant is a smoker
• 9.5% of males in Boston smoke as compared
to 1.7% of females
• Find the probability that we selected a male
given he is a smoker
Example - Find P(M|S)
• P(M)=0.51 P(M’)=0.49
P(S|M)=0.095 P(S|M’) = 0.017
• Bayes Theorem
Knowing the participant smokes – increases P(M)
)M'|)P(SP(M' M)|P(M)P(S
M)P(M)|P(S S)|P(M
853.0 )0.49(0.017 )0.51(0.095
)0.095(0.51 S)|P(M
Example 5.8.
Bayes Theorem
P(disease) = 0.002
Sensitivity = 0.85 = P(test +|disease)
P(test +)=0.08 and P(test -) = 0.92
What is P(disease|test +)?
Example 5.8.
Bayes Theorem
What is P(disease|test +)?
P(disease) = 0.002
Sensitivity = 0.85 = P(test +|disease)
P(test +)=0.08 and P(test -) = 0.92
)P(test
disease)disease)P(|P(test )test|P(disease
Example 5.8.
Bayes Theorem
P(disease) = 0.002
Sensitivity = 0.85 = P(test +|disease)
P(test +)=0.08 and P(test -) = 0.92
021.0 08.0
)002.0(85.0
)P(test
disease)disease)P(|P(test
Binomial Distribution
• Model for discrete outcome
• Process or experiment has 2 possible
outcomes: success and failure
• Replications of process are independent
• P(success) is constant for each replication
Binomial Distribution
Notation:
n=number of times process is replicated,
p=P(success),
x=number of successes of interest
0< x<n
xnx p)(1p
x)!(nx!
n! successes)P(x
Example 5.9.
Binomial Distribution
Medication for allergies is effective in reducing symptoms
in 80% of patients. If medication is given to 10 patients,
what is the probability it is effective in 7?
7-107 0.8)(10.8
7)!-(107!
10! successes)P(7
= 120(0.2097)(0.008) = 0.2013
Binomial Distribution
Antibiotic is claimed to be effective in 70% of the patients it is
given to. If antibiotic is given to 5 patients, what is the
probability it is effective on exactly three?
success = antibiotic is effective: n=5, p=0.7, x=3
3-53 0.7)(10.7
3)!-(53!
5! 3)P(X
= 10(0.343)(0.09) = 0.3087
Binomial Distribution
What is the probability that the antibiotic is
effective on all 5 ?
1681.0)1)(1681.0(1
0.7)(10.7 5)!-(55!
5! 5)P(X
5-55
Binomial Distribution
What is the probability that the antibiotic is
effective on at least 3 ?
P(X > 3) = P(3) + P(4) + P(5)
= 0.3087 + 0.3601 + 0.1681 = 0.8369
Binomial Distribution
Mean and Variance of the Binomial Distribution
= np s2 = n p ( 1 - p)
For Example, the mean (or expected) number of patients in whom the antibiotic
is effective is 5*0.7 = 3.5
Normal Distribution
• Model for continuous outcome
• Mean=median=mode
Normal Distribution
Notation: =mean and s=standard deviation
3s 2s s s 2s 3s
Normal Distribution
Properties of Normal Distribution
I) The normal distribution is symmetric about the mean
(i.e., P(X > ) = P(X < ) = 0.5).
ii) The mean and variance, and s2, completely characterize the normal distribution.
iii) The mean = the median = the mode.
P( - s < X < + s) = 0.68,
P( - 2s < X < + 2s) = 0.95,
P( - 3s < X < + 3s) = 0.99
iv) P(a < X < b) = the area under the normal curve from a to b.
Example 5.11.
Normal Distribution
Body mass index (BMI) for men age 60 is normally
distributed with a mean of 29 and standard deviation
of 6.
What is the probability that a male has BMI less than
29?
Example 5.11.
Normal Distribution
11 17 23 29 35 41 47
P(X<29)=?
Example 5.11.
Normal Distribution
11 17 23 29 35 41 47
P(X<29)=0.5
0.5 0.5
Example 5.11.
Normal Distribution
Body mass index (BMI) for men age 60 is normally
distributed with a mean of 29 and standard deviation
of 6.
What is the probability that a male has BMI less than
35?
Example 5.11.
Normal Distribution
11 17 23 29 35 41 47
P(X<35)=?
Example 5.11.
Normal Distribution
11 17 23 29 35 41 47
P(X<35)=0.5 + 0.34 = 0.84
0.5 0.34
Standard Normal Distribution Z
Normal distribution with =0 and s=1
-3 -2 -1 0 1 2 3
Example 5.11.
Normal Distribution
σ
μx Z
11 17 23 29 35 41 47
P(X<35)= P(Z<1) = ?
Example 5.11.
Normal Distribution
P(X<35) = P(Z<1).
Using Table 1, P(Z<1.00) = 0.8413
Table 1. Probabilities of Z
Table entries represent P(Z < Zi)
Zi .00 .01 .02 .03 .04 …
0.0 0.5000 0.5040 0.5080 0.5120 0.5160 …
0.1 0.5398 0.5438 0.5478 0.5517 0.5557 …
.
.
1.0 0.8413 0.8438 0.8461 0.8485 0.8508 …
Example 5.11.
Normal Distribution
11 17 23 29 35 41 47
P(X<30)=?
What is the probability that a male has BMI less than 30?
Example 5.11.
Normal Distribution
0.17 6
2930
σ
μx Z
P(X<30)= P(Z<0.17) = 0.5675
Percentiles of the Normal Distribution
The kth percentile is defined as the score that holds k percent of the scores below it.
Eg., 90th percentile is the score that holds 90% of the scores below it.
Q1 = 25th percentile, median = 50th percentile, Q3 = 75th
percentile
Percentiles
For the normal distribution, the following is used to compute
percentiles:
X = + Z s
where
= mean of the random variable X,
s = standard deviation, and
Z = value from the standard normal distribution for the desired
percentile (See Table 1A).
Percentiles
Percentiles of the Standard Normal Distribution
(Table 1A) Percentile Z
1st -2.326
2.5th -1.960
5th -1.645
10th -1.282
50th 0
90th 1.282
95th 1.645
97.5th 1.960
99th 2.326
0 -4 -3 -2 -1 0 1 2 3 4
1.645
0.05
0.95
Example 5.12.
Percentiles of the Normal Distribution
BMI in men follows a normal distribution with =29, s=6. BMI in women follows a normal distribution with =28, s=7.
The 90th percentile of BMI for men:
X = 29 + 1.282 (6) = 36.69.
The 90th percentile of BMI for women:
X = 28 + 1.282 (7) = 36.97.
Central Limit Theorem
Suppose we have a population with known
mean and standard deviation s. If we take
simple random samples of size n with
replacement, then for large n, the sampling
distribution of the sample means is
approximately normal with mean and
standard deviation μμ
X
n
σ σ
X
Application
• Non-normal population
• Take samples of size n – as long as n is sufficiently
large (usually n > 30 suffices)
• The distribution of the sample mean is approximately
normal, therefore can use Z to compute probabilities
nσ
μx Z
Example 5.18.
Central Limit Theorem
HDL cholesterol has a mean of 54 and
standard deviation of 17 in patients over 50. A
physician has 40 patients over age 50 and
wants to know the probability that their mean
cholesterol is above 60.
?60)XP(
Example 5.18.
Central Limit Theorem
2.22 4017
5460
nσ
μX Z
0.0132 0.9868-1 2.22)P(Z60)XP(
Example
Suppose we wish to estimate the mean of a population () whose standard deviation is known and equal to 12. Suppose a simple random sample of 100 individuals is selected from the population.
Find the probability that the sample mean is no more than 2 units from the population mean.
Sampling Distribution of Sample
Mean
- 2
+ 2
Central Limit Theorem
P( -2 < < + 2) = ??
Z = { ( - 2) - } / 12/100 = -2/1.2 = -1.67
Z = { ( + 2) - } / 12/100 = 2/1.2 = 1.67
Then: P(-1.67 < Z < 1.67) = 0.9525 – 0.0475 = 0.905
The probability that the sample mean is no more than 2 units from the population mean is 0.905, or 90.5%.
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