Lecture 4. For assignment 2, recode the smoking variable replace smoke=smoke-1.

Lecture 4

• For assignment 2, recode the smoking variable

• replace smoke=smoke-1

Recap from probability distributions

• The binomial distribution describes the probability of x successes in n independent trials, each with probability p of success

• The normal distribution may be used to describe cutoffs for some continuous random variables with mean µ and standard deviation – How do you know if your data are normally distributed?

• Histograms (stata: hist varname, normal)• QQ plots – next Biostat class• Other statistical tests

– What to do if my data are not normal?• Transformations – like taking the log, or the inverse 1/x …

• We will discuss an important property of the normal distribution today that applies to non-normally distributed data

Sampling• When we cannot measure the entire population we

take a sample• We estimate the population characteristics, i.e. the

mean and variance, using the sample• We use statistical inference to draw conclusions

about the how the estimates from the sample relate to the population values

Pagano and Gavreau, Chapter 8

• To do make inference from our sample to the population, our sample must be representative of the population– Random sample – each individual in the population has

equal chance of being selected for the sample– The larger the sample, the more reliable our estimates

about the population parameters will be• Because we do not have the entire population,

there is some level of uncertainty about our data• Confidence intervals afford us a way to quantify this

uncertainty

• Imagine you conducted a sample of size n from a population and measured random variable X, say systolic blood pressure

• Calculate the sample mean, X1 from the xi

• Take another sample of size n, calculate X2

• If you repeat for a long time you will have a large collection Xs generated from the samples of size n

• The Xs and standard deviations will differ from sample to sample due to sampling variability – each sample will most likley be different

Sampling distributions

• The collection of all of the possible Xs that can be obtained can be thought of as random variables that themselves follow a distribution

• This distribution is called the sampling distribution

• Imagine having a data set just of the means, the Xs, and plotting the histogram to see the shape of their distribution

• Why bother?– Sampling distributions of sample means have

special properties that allow us to make inference about the mean of a single sample

– We need this theory to be able to calculate confidence intervals or calculate p-values

Central limit theorem• If you have a random variable that comes from

a distribution with mean=µ and standard deviation σ, the following is true for the sampling distribution of the sample means from samples of size n– The mean of the sampling distribution (the

distribution of all of the possible sample means) is µ– The standard deviation of the sampling distribution

is σ/√n– If n is large enough, the shape of the sampling

distribution is approximately normal


Central limit theorem• Central limit theorem (CLT)

– If the underlying population is independently distributed with mean=µ and standard deviation=σ, then if we take a random sample of size n and n is large enough, the distribution of the sample mean is normally distributed.

– The mean of the sampling distribution will be µ and standard deviation will be σ/√n.


Central limit theorem• If we take a sample from any distribution (could be

skewed, or discrete, or whatever) of size n, and take the mean, and we do this over and over, the distribution of the means will be normally distributed with mean=the original distribution mean µ and standard deviation= σ/√n, if n is large enough

• The more symmetric the distribution of the raw data (not the means), the smaller the n needed to become normal-like


• Why does this make sense?– It makes sense that the distribution of means

would cluster around the population mean– It makes sense that the variability in the means is

smaller than in the raw data because the extreme values are already averaged out

– The part about the distribution being normal if n is large enough? Mathematical proof that I can’t do… But we can demonstrate it!

Note• σ is the standard deviation of the original

distribution • σ/√n is called the standard error, or more

precisely, the standard error of the mean, and it is the standard deviation of the distribution of the sample mean.


Central limit theorem example

• http://www.surveymonkey.com/s/F5VVHQZ• Download to excel• Change variable names• Import into stata• Histogram using dayofbirth1• Create average using self and first relative day of

birth (n=2) , dayofbirth_avg2• Make a histogram of dayofbirth_avg2• Repeat for using first 4, then 8 days of birth

Random draws from the chi-square distribution (µ=2)

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0 2 4 6 8 10 12 14 16x

Random draws

Means of 5 random draws from the chi-square distribution (µ=2)

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0 1 2 3 4 5 6 7mean of sample of size 5

Means of 5 random draws


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.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5mean of sample of size 10



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.5 1 1.5 2 2.5 3 3.5 4 4.5mean number of successes



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.5 1 1.5 2 2.5 3 3.5mean of sample of size 30


Distributions of the means of chi-square random draws

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0 2 4 6 8 10 12 14 16x

Random draws

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ctio

n0 2 4 6 8 10 12

mean of sample of size 2


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0 1 2 3 4 5 6 7mean of sample of size 5


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.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5mean of sample of size 10


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.5 1 1.5 2 2.5 3 3.5 4 4.5mean of sample of size 20


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.5 1 1.5 2 2.5 3 3.5mean of sample of size 30


Distributions of the means of binomial random draws0

.1.2

.3.4

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ctio

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0 1 2 3 4 5 6x successes

Random binomial draws

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.2.3

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ctio

n0 .5 1 1.5 2 2.5 3 3.5 4 4.5 5

mean number of successes

Means of 2 random binomial draws

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0 .5 1 1.5 2 2.5 3 3.5mean number of successes


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0 .5 1 1.5 2 2.5mean number of successes


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.2 .4 .6 .8 1 1.2 1.4 1.6 1.8 2mean number of successes


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.4 .6 .8 1 1.2 1.4 1.6 1.8mean number of successes


Using the CLT• Suppose you sampled from a HIV-infected population

with mean µ CD4 count = 250 cells/mm3 and standard deviation σ = 200 cells/mm3.

• If we select repeated samples of size 50, what proportion of the samples will have a mean value of less than 100 cells/mm3 ?

• Using the CLT, we know that the mean of all the samples, X, will follow a normal distribution with mean µ=250 and standard error σ/ √n=200/ √50

• Then we know that (X-250)/(200/ √50) ~ N(0,1)• If the mean cutoff value is 100, we want P(Z<100) then

z=(100-250)/(200/ √50)= -150 / (200/ 7.07) = -150/28.3 = -5.3

P(Z<-5.3) = ?

Using the CLT• What level of CD4 count is the lower 10th

percentile of the mean values?• P(Z<=z)=.10 for what value of z?• Table A.3 give the value P(Z>=z)• P(Z<=z) = - P(Z>=z) for the same value of z• The value of z for which P(Z>=z) = 0.10 is ____• The lower 10th percentile cutoff for z is ____• Now we need to transform back to get X• Using

nZ X

/

_

Using the CLT• What level of CD4 count is the lower 2.5th

percentile of the mean values?• P(Z<=z)=.025 for what value of z?

– Remember the tails of the standard normal distribution or look up in Table A.3

• The value of z for which P(Z<=z) = 0.025 is ____

• The value of z for which P(Z>z) = 0.025 is ____• Now we need to transform back to get X• Using

nZ X

/

_

Using the CLT• What level of CD4 count is the upper 2.5th

percentile of the mean values?• P(Z<=z)=.025 for what value of z?• The value of z for which P(Z>=z) = 0.025 is

____• The lower 2.5th percentile cutoff for z is ____• Now we need to transform back to get X• Using

nZ X

/

_

• Now we have the lower and upper 2.5% percentiles of the distribution of the sample means.

• The interior area contains 95% of the sample means.

• 95% of the means from sample size 50 lie within the 95% confidence bounds (194.6, 305.4)

• If we selected a sample of size 50 and the sample mean was outside these percentiles, we might suspect it came from an underlying population with a different population mean and standard deviation, or that a rare (5% probability) event had occurred.

• The confidence interval for the mean depends on the sample size, n. If the sample size was 300, what would be the interval?

• -1.96 <= (X – 250 )/(200/ √ 300) <= 1.96

• The lower and upper limits would be:

227.4 <= X <= 272.6

Which are narrower than the limits for n=50

(194.6, 305.4)

• As n increases, the width of the interval decreases

Confidence intervals for means

• X , the sample mean, is a point estimate of , the population mean

• Different samples will yield different Xs, so we cannot be certain how our estimate differs from

• Interval estimation provides a range of reasonable values that contain the population parameter (in this case ) with a certain degree of confidence

• This interval is called a confidence interval



• We put together what we learned about the normal distribution and the central limit theorem in order to construct confidence intervals

• By the CLT, X follows a normal distribution if n is sufficiently large X ~ N(,/√n)

• So, follows a standard normal distribution Z ~ N(0,1)


nZ X

/

_


• We know from examining the standard normal distribution that P(-1.96 ≤ Z ≤ 1.96) = 0.95


-5 -4 -3 -2 -1 0 1 2 3 4 5x

Standard normal distribution

95%2.5%2.5%


• P(-1.96 ≤ Z ≤ 1.96) = 0.95Substituting the formula for Z into the above we get

95.0)96.1/

96.1(

_

n

P X

Multiplying by σ/√n , adding X , and multiplying by -1, we get

95.0)/96.1/96.1(__

nnP XX



Thus the lower 95% confidence limit for µ is

And the upper 95% confidence limit for µ is

So we are 95% confident that the interval we calculate using the above includes

nX /96.1_

nX /96.1_


An important subtlety:X is a random variable

is a population parameter that is fixed in perpetuity; it has the same value irrespective of the sample

is either in the interval you calculate or it is not

What is random is the interval because it is based on the sample (X - 1.96/√n , X + 1.96/√n )



Interpreting confidence intervals for means

• The probability that the interval contains the true population mean is 95%

• If we were to select 100 random samples from the population and calculate confidence intervals for each, approximately 95 of them would include the true population mean µ (and 5 would not)



• 90% confidence interval– Replace 1.96 in the formula with 1.64

• 99% confidence interval– Replace 1.96 in the interval with 2.58

Generic formula:

Where 100%*(1-) is the % of the confidence interval

E.g. for a 95% confidence interval, =0.05, and we use z0.025 =1.96


)/,/( 2/

_

2/

_

nznz XX

Confidence intervals for means• How to get a tighter interval?

– Decrease the confidence level

– Increase n


Confidence level Z/2

99% .01 2.58

95% .05 1.96

90% .10 1.64

80% .20 1.28

n 95% confidence limits Length of interval

10 X 1.96/√10 = X 0.620 1.240100 X 1.96/√100 = X 0.3920 0.7841000 X 1.96/√1000 = X 0.062 0.124

• Uniform distribution demonstration

Confidence intervals for means• What to do when σ is not known? (In practice, always)

• By the Central limit theorem, follows a normal distribution, if n is sufficiently large

• Can we substitute s, the sample standard deviation for ?• s is not a reliable estimate of if n is small• If X is normally distributed, and a sample of size n is chosen, then

follows a Student’s t distribution

with n-1 degrees of freedomThis is denoted tn-1


nZ X

/

_

1

)(1

2

n

xxs

n

ii

nst X

/

_

Student’s t distribution

• The mean of the t distribution is 0 and the standard deviation is 1

• The t distribution is symmetric and bell-shaped, but has heavier tails than the standard normal – extreme values are more likely to occur

• For small n, the tails are fatter• For large n, the t distribution approaches (i.e. becomes

indistinguishable from) the standard normal distribution


The t-distribution

Student’s t distribution• There are separate curves for each degree of freedom (df)

– Table A.4 gives t value for selected P(T>t) and selected df

• Better to use Stata:• P(T>t) is calculated using ttail *****

**** note that normal() is P(Z<z)!!!• The code is ttail(df,t)• E.g., P(T>1.95) n=20

display ttail(19,1.95)

.03304428

• To find the value for which P(T>t) use invttail(df,p)• The answer for this the t cutoff is denoted t19,.05

display invttail(19,.05)

1.7291328


Confidence intervals for means• So using the t-distribution, the general formula for a 1-

% confidence interval for a mean is:

• The formula for a 95% confidence interval for a mean is:

Where df =n-1


)/,/( 2/,

_

2/,

_

nstnst dfdf XX

)/,/( 025.0,

_

025.0,

_

nstnst dfdf XX


• Remember that when n is large, the t distribution approaches the normal distribution– E.g. z0.025 = 1.96

– While tn-1,0.025 =


n tn-1,0.025

2 12.706

3 4.303

5 2.776

9 2.262

50 2.010

100 1.984

200 1.972

300 1.968

500 1.965

1000 1.962


• Example• CD4 cell count among HIV positives diagnosed at Mulago

Hospital– N=270– Sample mean = 296.9– Sample SD = 255.4– t269,.025=1.969

– 95% CI = ( 296.6 – 1.969*255.4/√270, 296.6 + 1.969*255.4/√270)= (266.0-327.2)


• Note that some statistical output gives you the SE or the SEM, which stand for standard error or standard error of the mean.

• This is s/ √n which is the standard deviation of the distribution of X

• Remember, if X is a random variable with mean µ and standard deviation , if n is large enough, X is normally distributed with mean µ and standard deviation / √n

. summ ncigs

Variable | Obs Mean Std. Dev. Min Max-------------+-------------------------------------------------------- ncigs | 545 .1963303 1.405081 0 20

. mean ncigs

Mean estimation Number of obs = 545

-------------------------------------------------------------- | Mean Std. Err. [95% Conf. Interval]-------------+------------------------------------------------ ncigs | .1963303 .060187 .0781028 .3145577--------------------------------------------------------------

. ci ncigs

Variable | Obs Mean Std. Err. [95% Conf. Interval]-------------+------------------------------------------------------------- ncigs | 545 .1963303 .060187 .0781028 .3145577

Normal approximation to the binomial distribution

• Remember that binomial distributions are used to describe the number of success in n trials P(X=x)

• The parameters of the binomial distribution are n and p, and the mean=np and standard deviation=square root of (np(1-p))

• As n increases, the binomial distribution more closely resembles the normal distribution

0.2

.4.6

Bin

omia

l pro

babi

lity

0 5 10 15 20n successes

n=10 p=0.05

0.1

.2.3

.4B

inom

ial p

roba

bilit

y


n=20 p=0.050

.05

.1.1

5.2

.25

Bin

omia

l pro

babi

lity


n=50 p=0.05

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.2B

inom

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roba

bilit

y


n=100 p=0.05

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.15

.2.2

5B

inom

ial p

roba

bilit

y


n=10 p=0.35

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.2B

inom

ial p

roba

bilit

y


n=20 p=0.350

.05

.1.1

5B

inom

ial p

roba

bilit

y


n=50 p=0.35

0.0

2.0

4.0

6.0

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inom

ial p

roba

bilit

y

0 20 40 60n successes

n=100 p=0.35

Binomial approximation to normal distribution

• Note that the binomial distribution approaches normality at smaller sample sizes when p is closer to 0.5

• Therefore you could use the normal distribution to look up the probability of observing X or more (or less) successes, using np as the mean and square root[ np(1-p)] as the standard deviation

• Considered valid when np and np(1-p) ≥5

• It is easier to use the normal distribution than to use table A.1. For example, if n=50 and you wanted to know the P(X>=30), using table A.1 which gives you P(X=x), you would need to find

P(X=30) + P(X=31) + .... + P(X=50)

(Although in Stata the binomialtail function does actually give you P(X≥x)

• Continuity correction – because you are approximating a discrete distribution with a continuous one, add 0.5 to X

Sampling distribution of a proportion

• Previous slides were about estimating X, the number of successes

• We often are more interested in the proportion of successes, rather than the number of successes

• The true population proportion p is estimated by

x = the number of successes/eventsn=the number of trials/people/observations

hat) (p

/ˆ nxp


• In our data set, we use 1s to represent the event and 0 to represent no event for each person or observation

• These are Bernoulli random variables – they have the value 0 or 1 with probability p

• The mean of the population of event/no events is p (this is the Bernoulli mean)

• The standard deviation σ of the population of events/no events is √p(1-p) (this is the Bernoulli standard deviation)


• If we take repeated samples of size n, and calculate p hat=x/n for each of the samples, if n is large enough, the p hats will follow a normal distribution (by the central limit theorem)– Remember that the sampling distribution is needed to

be able to make inference about a single sample

• The mean of this sampling distribution is p• The standard deviation is σ/√n = which is

also called the standard errorn

pp )1(

Sampling distribution of proportions• Then ~ N(0,1)

• This holds when np≥5 and np(1-p) ≥5 and also by the CLTThen we can use the normal distribution to calculate probabilities

of observing certain proportions in a sample

• E.g. What proportion of samples of size 50 from a population with p=.10 will have a p hat of .20 or higher?

• What is P(p hat ≥ 0.20)? • Mean=0.10; SD= √(.10*.90) =0.3; SEM = SD/√50 = 0.0424• P(Z ≥ ((.20-.10)/.0424)) = P(Z ≥ 2.36)

. display 1-normal(2.36)

.00913747


npp

ppZ

)1(

ˆ

Confidence intervals for proportions ~ N(0,1)

– So

• Rearranging, we get

• • Lower 95% confidence limit: or

• Upper 95% confidence limit: or


n

ppp

)1(*96.1ˆ

n

ppp

)1(*96.1ˆ

npp

ppZ

)1(

ˆ

95.0)96.1/)1(

ˆ96.1(

npp

ppP

95.0)/)1(96.1ˆ/)1(96.1ˆ( nppppnpppP

Confidence intervals for proportions

• However we don’t know p (if we did we wouldn’t be calculating these intervals). So we substitute p hat into the formula for the SEM.

• Lower 95% confidence limit:

• Upper 95% confidence limit:

• This interval has a 95% chance of containing the true population parameter p


n

ppp

)ˆ1(ˆ*96.1ˆ

n

ppp

)ˆ1(ˆ*96.1ˆ

Confidence intervals for proportions

• HIV prevalence in those testing at Mulago Hospital– N=933– n HIV+ = 269– Prevalence = 269/933 = 0.288– Standard error estimate = sqrt [ .288*(1-.288)/933 ] = 0.015

– 95% CI : (.288 – 1.96*.015, .288 + 1.96*.015 )• = (.259, .317)

– Interpretation: we are 95% confident that the interval 0.259-0.317) includes the true HIV prevalence in the population


Key points

• It is not practical or feasible to study an entire population, so we take a sample

• We need to make inference from our sample to the population

• We use the properties of repeated samples to do so• For any random variable X with mean µ and standard

deviation σ, if the sample n is large enough, the distribution of the sample mean is normally distributed with mean µ and standard deviation σ/ √n

• We use this to calculate intervals with known probability of containing the population mean

For next time

• Read Pagano and Gauvreau– Chapter 8, 9, and 14 (pages 324-329) (Review of

today’s material)

– Chapter 10 and 14 (pages 329-330)

Lecture 4. For assignment 2, recode the smoking variable replace smoke=smoke-1.

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Transcript of Lecture 4. For assignment 2, recode the smoking variable replace smoke=smoke-1.