Medical Biometry I. ( Biostatistics 511) Week 7 Discussion Section Lisa Brown. T he Normal Distribution. Many “Real world” measurements, such as IQ and height can be modeled was normal random variables (RVs).
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Biostat 511
Biostat 511
P(Z<1.65)
Biostat 511
Total area of a PDF=1
So P(Z<1.65)=.95 is the area of the shaded
region.
Interpretation: 95/100 samples of Z will
be less than or equal to 1.65
Biostat 511
STATA:
dispnormal(1.65)
.95052853
Or use normal probability tables (e.g. back of Baldi and Moore)
Biostat 511
=

P(1.65<Z<1.65)=P(Z<1.65)  P(Z<1.65)
= .95.05=.90
For a standard normal RV, 90% of values fall between 1.65 and 1.65
Biostat 511
P[Z < 1.65] = 0.9505
P[Z > 0.5] = 1P[Z < 0.5] = 0.3085
P[1.96 < Z < 1.96] = P[Z < 1.96]  P[Z < 1.96] = .95
P[0.50 < Z < 2.0] = P[Z <2.0]  P[Z <0.50]
2.0
0.50
Why?
0.50
2.0
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Step 1. Draw picture of area corresponding to probability.
Step 2. Use probability rules and tables or STATA to find quantities in (1).
Step 3. Get the answer.
Biostat 511
Q: This solves the problem for the N(0,1) case. How do we do calculate normal probabilities when the mean is not 0 and the standard deviation is not equal to 1?
A:Any normal random variable can be transformed to N(0,1)
E(X ) = 0
V(X ) = V(X) = 2
V( (X )/ ) =(1/2)*V(X)=1
Linear transformations of normal random variables are still normal. So
Z = (Xm)/s ~ N ( 0 , 1 )
Biostat 511
Z = (Xm)/s is a rescaled and shifted version of X—like going from Fahrenheit to Celsius.
In other words, the probability that
X<2.822 is the same as the probability
Z<1.65, since (X2)/.5 ~N(0,1).
Biostat 511
Step 0. Draw picture of area corresponding to probability.
Step 1. Reexpress probability statement about X as statement about Z by standardizing.
Step 2. Use probability rules and tables or STATA to find quantities in (1).
Step 3. Get the answer.
Biostat 511
Suppose X~N(m=2,s=.5). What is P(1.5<X<2.75)?
(STATA)
disp normal(1.5)normal(1)=.77453754
Biostat 511
Biostat 511
Suppose a clinically accepted value for mean systolic blood pressure in females, aged 6574 is 133 mmHg and the standard deviation is 20 mmHg. If a 70yearold woman is selected at random from the population, what is the probability that her systolic blood pressure is equal to or less than 120 mmHg?
X = systolic BP in woman age 6574.
= 133
= 20
What is P(X< 133)?
Biostat 511
Example Suppose a clinically accepted value for mean systolic blood pressure in females, aged 6574 is 133 mmHg and the standard deviation is 20 mmHg. If a 70yearold woman is selected at random from the population, what is the probability that her systolic blood pressure is equal to or less than 120 mmHg?
STATA: display normal(0.65)
Biostat 511
P(Z<1.65)=.95
The .95 quantile of a standard normal RV,
z.95, is 1.65.
In general, P(Z<zp)=p
Biostat 511
Suppose Z~N(0,1).
What is the .8 quantile (or 80th percentile) of Z?
P(Z<z.80)=.8
STATA: display invnorm(.8)
.84162123
Interpretation: There is an 80% chance that a randomly chosen Z~N(0,1)
will fall below .84.
Biostat 511
P(Z<z.80)=.8…Find values of z with p closest to .8
From the table, P(Z<.84)=.7995 and P(Z<.85)=.8023
So the .8th quantile is approximately .845.
Biostat 511
What about finding quantiles when X~N(m,s)?
We use standarization method…in reverse.
X has the same distribution as Zs+m, where Z~N(0,1)
Why?
E(Z)=E(Z)s+m=0*s+m=m
sd(X)=sd(Z)s=s
What is the .8 quantile (or 80th percentile) of X?
P(Z<z.80)=.8
P(Zs+m<z.80s+m)=P(X<z.80s+m)=.8
Interpretation: There is an 80% chance that a randomly chosen X~N(m,s)
will fall below z.80s+m=.84*s+m.
Biostat 511
Suppose a clinically accepted value for mean systolic blood pressure in females, aged 6574 is 133 mmHg and the standard deviation is 20 mmHg.
Between what two blood pressure readings will 80% of all systolic blood pressures for 6574yearold women lie?
We want the .1 and .9 quantiles of X,
since 80% of all values lie in this range.
P(z.1<Z<z.9)=.80
P(20z.1+133<20Z+133<20z.9+133)=.8
P(20z.1+133<X<20z.9+133)=.8
P(20*(1.2816)+133<X<20*1.2816+133)
So 80% of BP readings will fall between
107.4 and 158.6.
Biostat 511
X~Binomial(n,p)
Goal: What is the P(X<c) or P(X>c)?
Tail probabilities using the binomial distribution can be tedious to compute, especially by hand!
If np and n(1p) are large enough (>10), then approximately
Biostat 511
If np and n(1p) are large enough (>10), then approximately
X~Binomial(n=200, p=.4).
What is P(X<70)?
200*.4>10 and 200*.6>10, so, approximately
Exact calculation P(X<70)=.0843 STATA dispbinomial(20,12,.5)
Biostat 511
What happens if np and n(1p) are not large enough? The normal approximation can be terrible!
X~Binomial(n=10, p=.1).
What is P(X<1)?
Does not meet “rule of thumb” for normal approx: np=1, n(1p)=9.
If we assume it anyway,
Exact calculation P(X<1)= .74
STATA: display binomial(10,1,.1)
Biostat 511
Assume that X1, X2,...,Xn are an independent, identically distributed sample of RVs
from a distribution with mean m and variance s2 (sds) .
The sample mean is another random variable
So as n gets, bigger, the standard deviation of the sample mean goes down.
If sd(X) =10, what is the sd of the the sample mean when n=100?
Biostat 511
Assume that X1, X2,...,Xn are an independent, identically distributed sample of RVs from a distribution with mean m and variance s2 (sds) .
Remarkably, regardless of the distribution of Xi, as the sample size n gets large,
Or, for large sample sizes, approximately
Biostat 511
The CLT is very powerful: no matter how skewed the distribution of X, the distribution of a sample mean will approach normality with increasing n.
How large does n need to be for the normal approximation to be good?
It depends on the distribution of X.
Distribution of sample mean for different N
Biostat 511
One goal of statistical inference is to estimate population means.
We use the sample mean, as a point estimate.
This estimate is better for larger n, since is less variable and
closer to m with increasing n.
Confidence intervals allow us to express the uncertainty about our estimate
of the mean, by citing a range of values rather than a single point.
We construct a “ppercent” confidence interval for mu as follows:
Biostat 511
We need to find the standard normal quantile, z*, such that the shaded area P(Z<z*)=p.
This corresponds to the 1(1p)/2 quantile (see picture)!
For a 90% confidence interval, 1(1p)/2=1.10/2=.95, so z* is z.95.=1.645
What about 95% confidence? z*=z.975=1.96.
That is, each of the tail regions have area
(1p)/2. So z* corresponds to the
1(1p)/2 quantile of the standard normal
Distribution.
p
Right tail probability
(1p)/2
Biostat 511
For a given sample, the (for example) 95% confidence interval
either contains the population mean m or it doesn’t!!!
So it doesn’t make sense to to say that there is “a 95% probability that this interval contains m.”
Rather, with repeated samples, a 95% confidence interval constructed with this method will contain m 95% of the time.
Biostat 511
Your goal is to estimate the mean of systolic BP in a population of women 6575. You collect a sample of 100 women. Suppose you know that the standard deviation for systolic BP in the population is 20. The mean BP in your sample is 125.
Construct and interpret a 95% confidence interval for the population mean BP.
For a 95% CI, the critical value z*=1.96
95% Confidence interval: [1251.96*20/10, 125+1.96*20/10]=
[121.08, 128.92].
Interpretation: with repeated samples, 95% of intervals formed with this method would contain the true mean BP.
Biostat 511
What affects the width of the confidence interval?
Confidence intervals depend on the CLT and normal approximation for the sample mean’s distribution. For small n, is this still a good approach?
Biostat 511