Chapter 5: Sampling DistributionsChapter 5: Sampling Distributions Tuesday (3/10/15): 5.1 {Sampling...

Chapter 5: Sampling Distributions

Tuesday (3/10/15): 5.1

– Sampling Distributions

– Sample means and the Law of Large Numbers

– The Central Limit Theorem

Thursday (3/12/15): 5.2, Sampling Distributions for Counts andProportionsHomework: WebAssign due Friday, Excel assignment posted today, dueTuesday after Spring Break.

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Parameters and Statistics

Recall:

– A parameter is a number that describes an actual characteristic of apopulation. Assuming we cannot sample the entire population,parameters are never known exactly.

– A statistic is a number that describes a characteristic of a sample.

We use statistics to estimate parameters.

Notation for samples vs. populations:

Symbol Use

µ Population mean

x̄ Sample mean

σ Population standard deviation

s, sx Sample standard deviation

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Recall:

Symbol Use

µ Population mean

x̄ Sample mean

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Recall:

Symbol Use

µ Population mean

x̄ Sample mean

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Why sampling distributions?

Fundamental idea:

We want to use sample statistics to estimate population parameters.Different random samples will yield different statistics, so we must

know the distribution of these sample statistics!

– Sample statistics will be treated like random variables, and wealready know how to find the distribution (PDF) of a random variable(chapter 4).– Recall: given a random variable X, its probability distribution is atable (for discrete RV’s) or function (for continuous RV’s) thatprovides the range of possible outputs of X and their probabilities.

X 2 4 8

pX(x) 1/4 1/2 1/4

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Fundamental idea:

– Sample statistics will be treated like random variables, and wealready know how to find the distribution (PDF) of a random variable(chapter 4).

– Recall: given a random variable X, its probability distribution is atable (for discrete RV’s) or function (for continuous RV’s) thatprovides the range of possible outputs of X and their probabilities.

X 2 4 8

pX(x) 1/4 1/2 1/4

March 12, 2015 3 / 28

Fundamental idea:

– Sample statistics will be treated like random variables, and wealready know how to find the distribution (PDF) of a random variable(chapter 4).– Recall: given a random variable X, its probability distribution is atable (for discrete RV’s) or function (for continuous RV’s) thatprovides the range of possible outputs of X and their probabilities.

X 2 4 8

pX(x) 1/4 1/2 1/4

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Sampling Example

Example: suppose there are 50,000 students at UA. We give everyone aslip of paper with a number written on it. 10,000 students get a 1,10,000 students get a 2, etc, up to 5.

– If we select a student at random and look at their number, this is arandom variable X.– What is the distribution of X?

X 1 2 3 4 5

pX(x) 1/5 1/5 1/5 1/5 1/5

– What is the population mean of X?

5∑i=1

pixi =1

5(1 + 2 + 3 + 4 + 5) = 2.5

– What is the population standard deviation?

5∑i=1

pi(xi − µX)2 =1

((1− 2.5)2 + . . .+ (5− 2.5)2

)= 2.25

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Sampling Example

Example: suppose there are 50,000 students at UA. We give everyone aslip of paper with a number written on it. 10,000 students get a 1,10,000 students get a 2, etc, up to 5.– If we select a student at random and look at their number, this is arandom variable X.

– What is the distribution of X?

X 1 2 3 4 5

pX(x) 1/5 1/5 1/5 1/5 1/5

5∑i=1

pixi =1

5(1 + 2 + 3 + 4 + 5) = 2.5

5∑i=1

pi(xi − µX)2 =1

((1− 2.5)2 + . . .+ (5− 2.5)2

)= 2.25

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Sampling Example

Example: suppose there are 50,000 students at UA. We give everyone aslip of paper with a number written on it. 10,000 students get a 1,10,000 students get a 2, etc, up to 5.– If we select a student at random and look at their number, this is arandom variable X.– What is the distribution of X?

X 1 2 3 4 5

pX(x) 1/5 1/5 1/5 1/5 1/5

5∑i=1

pixi =1

5(1 + 2 + 3 + 4 + 5) = 2.5

5∑i=1

pi(xi − µX)2 =1

((1− 2.5)2 + . . .+ (5− 2.5)2

)= 2.25

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Sampling Example

X 1 2 3 4 5

pX(x) 1/5 1/5 1/5 1/5 1/5

5∑i=1

pixi =1

5(1 + 2 + 3 + 4 + 5) = 2.5

5∑i=1

pi(xi − µX)2 =1

((1− 2.5)2 + . . .+ (5− 2.5)2

)= 2.25

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Sampling Example

X 1 2 3 4 5

pX(x) 1/5 1/5 1/5 1/5 1/5

5∑i=1

pixi =1

5(1 + 2 + 3 + 4 + 5) = 2.5

5∑i=1

pi(xi − µX)2 =1

((1− 2.5)2 + . . .+ (5− 2.5)2

)= 2.25

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Sample Means and the Law of Large Numbers

– Now, suppose we select 100 students at random and record theirnumber. You can simulate this in excel using RANDBETWEEN(1,5).

1, 5, 2, 2, 5, 4, 3, 1, 3, 5, . . .

– We think of each of the above numbers as a random variable Xi. Inother words, we have 100 random variables!– We can compute the sample mean x̄ = 1

∑ni=1Xi, with n = 100.

Law of large numbers: as the number of samples (n) gets large,x̄→ µX .

– We are more interested in the variability of x̄ across allpossible samples of size 100.– In particular, x̄ is a random variable, since it is a linear combinationof random variables! What is its distribution?

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1, 5, 2, 2, 5, 4, 3, 1, 3, 5, . . .

– We think of each of the above numbers as a random variable Xi. Inother words, we have 100 random variables!

– We can compute the sample mean x̄ = 1n

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1, 5, 2, 2, 5, 4, 3, 1, 3, 5, . . .

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1, 5, 2, 2, 5, 4, 3, 1, 3, 5, . . .

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1, 5, 2, 2, 5, 4, 3, 1, 3, 5, . . .

– We are more interested in the variability of x̄ across allpossible samples of size 100.

– In particular, x̄ is a random variable, since it is a linear combinationof random variables! What is its distribution?

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1, 5, 2, 2, 5, 4, 3, 1, 3, 5, . . .

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Bias, Variability, Unbiased Estimators

Recall a concept from chapter 3:

If we compute a statistic (such as x̄ or sX) over all possible samples ofa certain size, this statistic will vary.

– Bias is when the mean of the statistic, over all samples, is differentfrom the population parameter. A statistic is an unbiasedestimator if its mean over all samples is actually equal to thepopulation parameter.

– Variability is how spread out the statistic is over all samples

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Mean and standard deviation of the sample mean

Yes, that’s right - we are going to compute the mean of the mean andthe standard deviation of the mean! Here is what we mean:

– For a given sample size n, where n is small relative to thepopulation size, each individual sample is a random variable Xi

which is independent from the others and identicallydistributed. In other words, the distribution of X10 is the same asthe distribution of X1000.

– In particular, the population mean and standard deviation of each Xi

is the same:

µXi = µX = µ, σXi = σX = σ

Another way of saying this: choosing a random sample of 100without replacement from a very large population (e.g. 50,000) willresult in the same behavior as if we sampled with replacement.

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is the same:

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is the same:

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– Knowing that µXi = µ for all i, we can compute the mean of x̄ (recallthat x̄ is a random variable!)

µx̄ = E[x̄] = E

n∑i=1

nn · µ = µ

So, this says that x̄ is an unbiased estimator of µ, since the mean ofthe sampling distribution is the same as the mean of the population!

– Note: the assumption that n is small compared to the population iscrucial. If n is not small compared to population, Law of LargeNumbers is more useful.– We can also compute the sample variance/standard deviation:

σ2x̄ =

n∑i=1

σ2 =σ2

so, σx̄ = σ/√n

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µx̄ = E[x̄] = E

n∑i=1

nn · µ = µ

So, this says that x̄ is an unbiased estimator of µ, since the mean ofthe sampling distribution is the same as the mean of the population!– Note: the assumption that n is small compared to the population iscrucial. If n is not small compared to population, Law of LargeNumbers is more useful.

– We can also compute the sample variance/standard deviation:

σ2x̄ =

n∑i=1

σ2 =σ2

so, σx̄ = σ/√n

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µx̄ = E[x̄] = E

n∑i=1

nn · µ = µ

So, this says that x̄ is an unbiased estimator of µ, since the mean ofthe sampling distribution is the same as the mean of the population!– Note: the assumption that n is small compared to the population iscrucial. If n is not small compared to population, Law of LargeNumbers is more useful.– We can also compute the sample variance/standard deviation:

σ2x̄ =

n∑i=1

σ2 =σ2

so, σx̄ = σ/√n

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Interpretation

– Recall the setting: we are taking simple random samples of size nfrom a population of size much bigger than n.

– We can estimate the population mean by computing x̄. This is anunbiased estimate because if we took x̄ for every possible sample ofsize n, the distribution of x̄ would be centered around µ.

– The larger we take n (but not too big...), the smaller σx̄ gets. Inother words, the distribution of x̄ becomes less variable if wetake larger samples.

– Intuition: averaging (computing x̄ over a sample smooths out thevariability, making sample means less variable than individualobservations.

– The question still remains:

What does the distribution of x̄ look like?

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Interpretation

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Interpretation

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Interpretation

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Interpretation

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The distribution of x̄

– Using excel, we generate uniform random whole numbers from{1, 2, 3, 4, 5} using RANDBETWEEN(1,5). This is like drawingrandom people from our UA student example - recall that each of50,000 people had a number between 1 and 5, with the same numberof people holding each number.

– To simulate random sampling of size n, we can simply use a range ofcells of size n, i.e. A1:A5 or A11:A15 are simple random samples ofsize 5.

– We compute x̄ for different random samples of size n, then make ahistogram of x̄

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We know that for samples of size n, x̄ has mean µ (same as thepopulation mean) and standard deviation σ/

√n. The distributions

look like:

These look suspiciously normal!

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look like:

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look like:

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look like:

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look like:

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look like:

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The Central Limit Theorem

Suppose that X1, X2, . . . , Xn are independent, identically distributedrandom variables. For instance, drawing random samples from a largepopulation, with n� population size, each individual shares the samedistribution as X, the population parameter. Then, the distribution of

x̄ =1

n(X1 +Xn + . . .+Xn)

is approximately normal (unless the Xi are normal) with mean µXand standard deviation σX/

√n. If the Xi are normal, then x̄ is exactly

N(µ, σ/√n).

– In other words,

x̄ ∼ N(µ, σ/√n) (Approximately)

– Note: It does not matter what the probability distribution of thevariables Xi is! This works with normally distributed RVs, uniformlydistributed RVs, and RVs with crazy skewed distributions.

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x̄ =1

n(X1 +Xn + . . .+Xn)

N(µ, σ/√n).

– In other words,

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x̄ =1

n(X1 +Xn + . . .+Xn)

N(µ, σ/√n).

– In other words,

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Example

Suppose that we draw random real numbers uniformly from 0 to 10:

0.1 Mean = 5, standard deviation = 2.88

a) What is the probability that a given number drawn is between 1and 2?

b) Suppose we select 25 numbers randomly. Approximately what is theprobability that the mean of these numbers is between 1 and 2?

Answer :

a) Find the area under the PDF between 1 and 2.

b) Using the CLT, we know that x̄ ∼ N(5, 2.88/√

25), so we use

normalcdf(1,2,5,2.88/5)

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Example

Suppose that we draw random real numbers uniformly from 0 to 10:

0.1 Mean = 5, standard deviation = 2.88

a) What is the probability that a given number drawn is between 1and 2?

b) Suppose we select 25 numbers randomly. Approximately what is theprobability that the mean of these numbers is between 1 and 2?

Answer :

a) Find the area under the PDF between 1 and 2.

b) Using the CLT, we know that x̄ ∼ N(5, 2.88/√

25), so we use

normalcdf(1,2,5,2.88/5)

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Example

Suppose ACT scores across all incoming freshmen in the US arenormally distributed with a mean of 20.8 and a standard deviation of4.8.

a) Are the given means and standard deviations for the population orfor a sample?

b) What is the probability that a randomly selected student had ascore higher than 30?

c) What is the probability that 5 randomly selected students havetheir average score higher than 30?

Answers:

a) These are for the population

b) Since this random variable is normally distributed, we can usenormalcdf(30,1000,20.8,4.8) = 0.028 (2 sig figs). As usual, the1000 is just a ‘large’ number to indicate that we want all numbersto the right of 30.

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Example

Suppose ACT scores across all incoming freshmen in the US arenormally distributed with a mean of 20.8 and a standard deviation of4.8.

a) Are the given means and standard deviations for the population orfor a sample?

b) What is the probability that a randomly selected student had ascore higher than 30?

c) What is the probability that 5 randomly selected students havetheir average score higher than 30?

Answers:

a) These are for the population

b) Since this random variable is normally distributed, we can usenormalcdf(30,1000,20.8,4.8) = 0.028 (2 sig figs). As usual, the1000 is just a ‘large’ number to indicate that we want all numbersto the right of 30.

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Example, cont’d

Solution to part c): by the central limit theorem, we know that

x̄ ∼ N(µ,

σ√5

)Since our original random variable is normal, this is in fact exact - x̄will have exactly a normal distribution with mean µ = 20.8 andstandard deviation σ/

√5 (5 is the sample size). Thus the probability

that 5 randomly selected students has an average score higher than 30

is found with normalcdf(30,1000,20.8,4.8/√

5) = 9.11 · 10−6

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Review of LLN and CLT

Assumptions:

1) I have n independent identically distributed random variablesX1, . . . , Xn, each with mean µ and standard deviation σ.

In most examples, the situation is this: each Xi represents avariable for an individual drawn from a very large population, sothat the collection {X1, . . . , Xn} models a sample of size n. Forexample, each Xi might be a randomly selected person’s age or

height or blood O2 level.

2) The random variables must have finite mean and standarddeviation (not an issue in this class, but it does come up!)

Define the sample mean as the random variable

X̄ =1

n(X1 +X2 + . . .+Xn) =

n∑i=1

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Assumptions:

X̄ =1

n(X1 +X2 + . . .+Xn) =

n∑i=1

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Assumptions:

X̄ =1

n(X1 +X2 + . . .+Xn) =

n∑i=1

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Conclusions:

1) Law of large numbers: X̄ → µ. Loosely speaking we mean that aswe collect larger and larger samples, the sample mean ‘goes to’ thepopulation mean.

2) Central limit theorem: the variation of X̄ around µ follows anormal distribution. More precisely,

X̄ ≈ N(µ,

σ√n

)– This means that if we collected many samples of a fixed size n, wewould expect the sample means of these samples to be clusteredaround the actual population mean. The variability decreases withlarger samples, as long as we still satisfy the IID assumption.

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5.2: Repeated Trials and the Binomial Distribution

– A Bernoulli trial is an experiment where the outcome is one of twopossibilities - i.e. ‘yes’ or ‘no’, ‘heads’ or ‘tails’, etc.

– When we repeat Bernoulli trials and record the number of positiveresults, we are in a Binomial setting.

More precisely, we need:

– A binary experiment (outcome is one of two options)

– Each ‘trial’ is independent

– We fix a number of trials at the start

– Each trial has the same probability p of success and probability 1− pof failure.

Example: flip a fair coin 5 times and count the number of heads. Whatis p? Answer: p = 1/2 since the coin is fair.

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Example: flip a fair coin 5 times and count the number of heads. Whatis p?

Answer: p = 1/2 since the coin is fair.

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Example

Suppose we roll a (fair) 6-sided die. A ‘success’ (S) will be if the dieshows a 6. Any other number is considered a fail (F) We will roll thedie 3 times and count the number of successes.

– What is the probability of a success on a given trial? Answer:p = 1/6.

– What is the probability of the following: SFF? Answer:Pr(SFF ) = (1/6) · (5/6) · (5/6) = 25

216 ≈ 0.116

– What is the probability of getting exactly one success in 3 rolls?Answer: there are 3 different ways to get one success:SFF, FSF, FFS. Each has the same probability, 25/216. So, theprobability of getting exactly one success is

216≈ 0.347

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Example

– What is the probability of a success on a given trial?

Answer:p = 1/6.

216 ≈ 0.116

216≈ 0.347

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Example

216 ≈ 0.116

216≈ 0.347

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Example

– What is the probability of the following: SFF?

Answer:Pr(SFF ) = (1/6) · (5/6) · (5/6) = 25

216 ≈ 0.116

216≈ 0.347

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Example

216 ≈ 0.116

216≈ 0.347

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Example

216 ≈ 0.116

– What is the probability of getting exactly one success in 3 rolls?

Answer: there are 3 different ways to get one success:SFF, FSF, FFS. Each has the same probability, 25/216. So, theprobability of getting exactly one success is

216≈ 0.347

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Example

216 ≈ 0.116

216≈ 0.347

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Example

Suppose flip an unfair coin 100 times. The probability of heads is 0.4.What is the probability that we get exactly 20 heads in 100 flips?

Solution: Let X be the number of heads in 100 flips. The probabilityof getting exactly 20 heads will be

P (X = 20) = N0.4200.680

where N is the number of different ways to get 20 heads. We can findN using the ‘choose’ function:(

)(‘100 choose 20’)

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Example

Suppose flip an unfair coin 100 times. The probability of heads is 0.4.What is the probability that we get exactly 20 heads in 100 flips?Solution: Let X be the number of heads in 100 flips. The probabilityof getting exactly 20 heads will be

P (X = 20) = N0.4200.680

where N is the number of different ways to get 20 heads. We can findN using the ‘choose’ function:(

)(‘100 choose 20’)

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The Binomial Distribution

When we have repeated Bernoulli trials that satisfy the conditions fora binomial setting (independence, fixed number of trials n, fixedprobability of success p), the distribution for the number of success isthe Binomial Distribution:

P (X = k) =

)pk(1− p)n−k

Intuitively,

– pk is the probability of getting k successes– (1− p)n−k is the probability of the rest (n− k) being failures– pk(1− p)n−k is the probability of getting exactly k success and n− k

failures–(nk

)is the number of different ways this can happen.

The number(nk

)is the binomial coefficient:(n

k!(n− k)!, n! = n · (n− 1) · · · 2 · 1

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P (X = k) =

)pk(1− p)n−k

Intuitively,

failures–(nk

The number(nk

k!(n− k)!, n! = n · (n− 1) · · · 2 · 1

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P (X = k) =

)pk(1− p)n−k

Intuitively,

failures–(nk

The number(nk

k!(n− k)!, n! = n · (n− 1) · · · 2 · 1

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Computing Binomial Coefficients

– On the TI-84, you can compute(nk

)by inputting n, then selecting

MATH:PRB:nCr, then inputting k. It will look like this:

5 nCr 3

– You can compute binomial coefficients by hand using Pascal’striangle:

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Computing Binomial Coefficients

– On the TI-84, you can compute(nk

)by inputting n, then selecting

MATH:PRB:nCr, then inputting k. It will look like this:

5 nCr 3

– You can compute binomial coefficients by hand using Pascal’striangle:

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Example of Binomial Distribution?

– Example: suppose we are going to test a lot of 1000 climbing ropesfor strength. We select 5 ropes at random and test them. Supposethe probability of a rope passing a test is 0.999 (i.e. 999/1000 ropeswill pass). Define the random variable X to be the number of ropesthat fail the test in our sample.

– Is this an example of a binomial setting?

– Answer: No, not quite - each trial is not independent since thepopulation gets smaller each time.

– What is P (X = 0)?

P (X = 0) =999

1000· 998

999· 997

998· 996

997· 995

996= 0.995

– Try computing this again using a binomial distribution.

P (X = 0) ≈(

)0.0010(0.999)5 ≈ 0.99500999

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P (X = 0) =999

1000· 998

999· 997

998· 996

997· 995

996= 0.995

P (X = 0) ≈(

)0.0010(0.999)5 ≈ 0.99500999

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P (X = 0) =999

1000· 998

999· 997

998· 996

997· 995

996= 0.995

P (X = 0) ≈(

)0.0010(0.999)5 ≈ 0.99500999

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P (X = 0) =999

1000· 998

999· 997

998· 996

997· 995

996= 0.995

P (X = 0) ≈(

)0.0010(0.999)5 ≈ 0.99500999

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P (X = 0) =999

1000· 998

999· 997

998· 996

997· 995

996= 0.995

P (X = 0) ≈(

)0.0010(0.999)5 ≈ 0.99500999

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Sampling Distribution for a ‘Count’

Suppose we sample from a population and count how many individualssatisfy a condition. For example, polling for a yes/no vote on aninitiative. We are interested in the distribution of the number of‘successes’.

If we select a simple random sample of size n from a population of sizemuch larger than n, where the probability of success is known to be p,the number of successes k in the sample is approximately binomial

with parameters p and n:

P (X = k) ≈(n

)pk(1− p)n−k

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Mean and Standard Deviation of a Binomial RV

The mean and standard deviation of a binomial RV are:

µ = np

σ =√np(1− p)

Derivation is tedious, uses the binomial theorem:

(a+ b)n =

n∑k=0

)akbn−k

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Approximating Binomial Distributions with Normal

If the number of trials is large compared to either the probability ofsuccess or failure (say both np and n(1− p) are bigger than 10), butstill small compared to the population size, we have

Binom(n, p) ≈ N(np,√np(1− p))

Example: A survey asks a nationwide random sample of 2500 adultstheir opinion on same-sex marriage. Suppose that 55% of Americans 1

actually support same-sex marriage. Estimate the probability that1400 people or more out of the 2500 support same-sex marriage.– Check for validity of normal approximation:np = 2500 · 0.55 = 1375� 10, n(1− p) = 1125� 10.– Compute P (X ≥ 1400) using z-scores or normalcdf:

µ = np = 1375, σ =√np(1− p) ≈ 24.875

P (X ≥ 1400) ≈ normalcdf(1400, 10000, 1375, 24.875) ≈ 0.157

1see Gallup.comMarch 12, 2015 26 / 28

actually support same-sex marriage. Estimate the probability that1400 people or more out of the 2500 support same-sex marriage.

– Check for validity of normal approximation:np = 2500 · 0.55 = 1375� 10, n(1− p) = 1125� 10.– Compute P (X ≥ 1400) using z-scores or normalcdf:

µ = np = 1375, σ =√np(1− p) ≈ 24.875

P (X ≥ 1400) ≈ normalcdf(1400, 10000, 1375, 24.875) ≈ 0.157

actually support same-sex marriage. Estimate the probability that1400 people or more out of the 2500 support same-sex marriage.– Check for validity of normal approximation:np = 2500 · 0.55 = 1375� 10, n(1− p) = 1125� 10.

– Compute P (X ≥ 1400) using z-scores or normalcdf:

µ = np = 1375, σ =√np(1− p) ≈ 24.875

P (X ≥ 1400) ≈ normalcdf(1400, 10000, 1375, 24.875) ≈ 0.157

actually support same-sex marriage. Estimate the probability that1400 people or more out of the 2500 support same-sex marriage.– Check for validity of normal approximation:np = 2500 · 0.55 = 1375� 10, n(1− p) = 1125� 10.– Compute P (X ≥ 1400) using z-scores or normalcdf:

µ = np = 1375, σ =√np(1− p) ≈ 24.875

P (X ≥ 1400) ≈ normalcdf(1400, 10000, 1375, 24.875) ≈ 0.157

Sample Proportions

Sometimes, we care more about the proportion of a population thatsatisfies some condition rather than the number of people. For example‘what percentage of people drink coffee every day’?

p̂ =number of successes in sample

size of sample=X

– 0 ≤ p̂ ≤ 1, whereas 0 ≤ X ≤ n.

– If X is binomial (or is approximately binomial), then

µp̂ = p

σp̂ =

√p(1− p)

– Careful to distinguish p̂, which is a sample proportion, from p, whichis a probability. p̂ is like an approximation of p.

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Sample Proportions

size of sample=X

– 0 ≤ p̂ ≤ 1, whereas 0 ≤ X ≤ n.

µp̂ = p

σp̂ =

√p(1− p)

March 12, 2015 27 / 28

Sample Proportions

size of sample=X

– 0 ≤ p̂ ≤ 1, whereas 0 ≤ X ≤ n.

µp̂ = p

σp̂ =

√p(1− p)

March 12, 2015 27 / 28

Sample Proportions

size of sample=X

– 0 ≤ p̂ ≤ 1, whereas 0 ≤ X ≤ n.

µp̂ = p

σp̂ =

√p(1− p)

March 12, 2015 27 / 28

Normal Approximation for Proportion

Making the binomial assumption again - i.e. for sample sizes muchsmaller than the population size (book uses population = 20*sample asa rule-of-thumb), we can use a normal distribution to approximate aproportion:

p̂ ≈ N

√p(1− p)

Example: suppose we toss a fair coin 500 times. Let X be the numberof heads.

a) Find the mean and standard deviation of X

b) Find the mean and standard deviation of p̂, the proportion of headsout of 500 tosses.

c) Find the probability that between 45% and 50% of the coins showedheads.

March 12, 2015 28 / 28

Normal Approximation for Proportion

Making the binomial assumption again - i.e. for sample sizes muchsmaller than the population size (book uses population = 20*sample asa rule-of-thumb), we can use a normal distribution to approximate aproportion:

p̂ ≈ N

√p(1− p)

Example: suppose we toss a fair coin 500 times. Let X be the numberof heads.

a) Find the mean and standard deviation of X

b) Find the mean and standard deviation of p̂, the proportion of headsout of 500 tosses.

c) Find the probability that between 45% and 50% of the coins showedheads.

March 12, 2015 28 / 28

Chapter 5: Sampling DistributionsChapter 5: Sampling Distributions Tuesday (3/10/15): 5.1 {Sampling...

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