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Sampling and VariabilityThe Central Limit Theorem

Assumptions

Sampling Distributions

Ken Kelley’s Class Notes

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Lesson Breakdown by Topic

1 Sampling and VariabilitySampling/VariabilityDemonstrationStandard Deviation of the MeanSampling Variability Example

2 The Central Limit TheoremVisualization

3 AssumptionsNotation GlossaryWhat You LearnedA Worked Example

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What You Will Learn from this Lesson

You will learn:

The idea of sampling from a population to obtain pointestimates.

The idea of the variability of point estimates.

How the standard deviation of a mean decreases as the samplesize increases (specifically at a rate of 1/

√n).

How to find probability values for a mean (rather than a singlescore, as was done when discussing the normal distribution).

How the Central Limit Theorem allows the normal distributionto be used when interest concerns a mean (even if thedistribution of scores is not normal).

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Motivation

If the mean effectiveness of a blood pressure medication isestimated to be a 15 point reduction in blood pressure after aweek (i.e., X̄ = −15), does this mean that µ = −15?

What if you discovered that X̄ = −15 when n = 20?

What if three other studies, each with n = 20, reported thefollowing: X̄1 = −1, X̄2 = 3, and X̄3 = −5?

What if you discovered that X̄ = −4.5 when n = 250, 000?

The estimated mean is sample specific, but it represents afixed population value. . . understanding the variability ofestimates is important for making decisions based on a sample.

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Sampling/VariabilityDemonstrationStandard Deviation of the MeanSampling Variability Example

Sampling

Simple random sample: when all members of a populationhave the same probability of being selected.

Unless otherwise stated, we will assume simple randomsampling.

This is not the way all samples are collected – so be aware ofdata collection methods.

There are other sampling methods, which we will discuss at alater time.

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Estimation

Point estimate: the estimated value of a populationparameter of interest based on a sample from the populationof interest.

Point estimates are (essentially) always wrong.

A sample estimate will differ (usually) from the populationparameter it estimates.

Correspondingly, if new or a different sample had beencollected, a different point estimate would likely result.

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Sampling Variability

Multiple samples can be taken and a point estimate calculatedin each instance.

The sample-to-sample variability of the point estimates issampling variability.

Sampling error is the random difference between an estimateand the parameter it estimates.

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Sampling Variability: A Demonstration

The effects of sampling variability on a normal distribution: ADemonstration (Link).

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nd.edu/~kkelley/Teaching/Demonstrations/Class_Demonstration_for_Normal_Distribution.R


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Thought Question

Does Variability Matter in theMarket?

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Thought Question

Does Variability Matter in theMarket?Yes!

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Does Variability Matter in the Market?

Volatility: the standard deviation of the continuouslycompounded returns of a financial instrument (over aspecified period of time).

Volatility is the standard deviation!

Implied volatility, an annualized standard deviation.

The VIX (SP 500 market volatility index).

The VXN (Nasdaq 100 volatility index).

The VXD (DJIA volatility index).

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Financial Prospectus

From Fidelity (Link).

https://www.fidelity.com/retirement-planning/learn-about-iras/contributing-to-ira


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Does Variability Matter in the Market

Example of the effect of volatility (i.e., the standarddeviation) on returns using @Risk Demonstration (Link).

Example performance measures from Fidelity (note thestandard deviation) (Link).

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http://www3.nd.edu/~kkelley/Teaching/Demonstrations/InvestmentsWithAtRisk.xls

https://fundresearch.fidelity.com/mutual-funds/performance-and-risk/316066406


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The Standard Deviation of the Mean

Often, interest is not literally in individual scores, but ratherthe mean of a set of scores (e.g., months, groups, teams).

The variability of the mean is different from the variability ofscores.

The mean is based on more information than individual scoresand is more stable.

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The population standard deviation of the mean is denoted σx̄ ,which equals σx̄ = σ√

n.

Because√n is in the denominator, as the sample size

increases, the variability of the mean decreases.

This fact is extremely important, as it shows how increasingsample size leads to more precise (i.e., less variable) estimates.

Increasing sample size decreases “noise” and magnifies“signal.”

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n.





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Sampling Distributions

Sampling Distribution: the distribution of a statistic ofinterest when when that statistic is calculated from randomsamples of data under the same set of conditions.

This can be for any statistic (e.g., x̄ , p̄, s, s2, r , etc.).

Each statistic has a sampling distribution with its ownproperties.

The sampling distribution always depends on sample size.

The larger the sample size, the less variable the statistic.

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Sampling Variability: 12 Ounce Cans

Beverage manufacturers must ensure that the stated amountof liquid is close to the actual amount.

Manufactures know that the process of filling a can is notexact (e.g., temperature, barometric pressure, viscosity).

The manufacturing specifications for a bottler is such that themachines attempt to fill each can with 12.20 ounces of liquid.

Suppose that data shows that the standard deviation is 0.25of ounces of liquid across the cans.

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Recall, the Standard Normal Distribution

A normal distribution with a mean of µ = 0 and a standarddeviation of σ = 1 is the standard normal distribution.

Such a distribution is standardized because z-scores are formed.

Recall, a z-score for individual i is defined as

zi =xi − µσ

Any normal distribution can be converted into a standardnormal distribution by transforming scores into z-scores.

The distribution shape (e.g., if it is skewed) does not changeby converting to z-scores (but the mean and standarddeviation does).

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Sampling Variability: An Example

What is the probability that a can has less than 12 ounces?

z =x − µσ

=12− 12.2

.25=−.2.25

= −.8

Thus, P(Z ≤ −.8) = .21

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What about the mean of a 6 pack being less than 12 ounces?

z =x̄ − µσ/√n

=12− 12.2

.25/√

6=−.20

.1021= −1.96;P(Z ≤ −1.96) = .025.

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Using Excel Instead of the z-Table

The NORM.S.DIST formula in Excel can be used instead ofthe z-table.

NORM.S.DIST requires one to specify the z value and TRUE(for cumulative).

That is: NORM.S.DIST(z , TRUE)

For the example of a can being less than 12 ounces, in whichz = −.80:NORM.DIST(X , 0, 1, TRUE).

The formula returns: .2118554 (which for our purposes can berounded to four decimal places: P = .2119.

Recall formulas in Excel require an “=” sign in order to returnthe results. Thus, “=NORM.DIST(X , 0, 1, TRUE)” would beentered in an Excel cell.

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Using Excel More Generally: For Any Normal Distribution

The NORM.DIST formula in Excel can be used for anynormal distribution (notice there is not “S”, which denotedstandardized).

NORM.DIST requires one to specify the X value, µ, σ, andTRUE (for cumulative).

That is: NORM.DIST(X , µ, σ, TRUE)

For a standard normal distribution (i.e., z-Distribution):NORM.DIST(X , 0, 1, TRUE)

For the example of a six-pack being less than 12 ounces:NORM.DIST(12, 12.20, .25/sqrt(6), TRUE).

The formula returns: .0250218 (which for our purposes can berounded to four decimal places: P = .0250.

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Rather than using the z-table, the NORM.DIST formula inExcel can be used:

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Sampling Variability Visually (Equal X -Axis)

Distribution of the Weight of Individual Cans

11.2 11.6 12.0 12.4 12.8 13.2

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Weight of One Cans

Distribution of Sample Mean forthe Weight of Cans in a 6−Pack

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Sample Mean Weight of the Cans in a 6 Pack


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Distribution of Sample Mean forthe Weight of Cans in a Case

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Sample Mean Weight of the Cans in a Case

Sampling Variability Visually (Equal X and Y Axes)

Distribution of the Weight of Individual Cans

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Weight of One Can


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Distribution of Sample Mean forthe Weight of Cans in a Case

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Sample Mean Weight of the Cans in a Case


AssumptionsVisualization

The Central Limit Theorem


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Random samples of size n drawn from some population willhave a sampling distribution for x̄ that can be approximatedby a normal distribution as the sample size becomes large.

Note that nothing is stated about the shape of thedistribution in the population!

The distribution of x̄ will have mean µ and standard deviationσx̄ = σ√

n.

This is one of the most important theorems in all of statistics!

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Random samples of size n drawn from some population willhave a sampling distribution for x̄ that can be approximatedby a normal distribution as the sample size becomes large.

Note that nothing is stated about the shape of thedistribution in the population!

The distribution of x̄ will have mean µ and standard deviationσx̄ = σ√

n.

This is one of the most important theorems in all of statistics!

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Histogram of Uniform Distribution

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Histogram of Uniform Distribution

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Sampling Distribution of X̄ from a Uniform Distribution

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Sample Means from Uniform Distribution when n = 2

Value Observed for Sample Mean

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Histogram of Log Normal Distribution

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Histogram of Log Normal Distribution

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Sampling Distribution of X̄ from Log Normal Distribution

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Sample Means from Log Normal Distribution When n = 2

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Sample Mean

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Sample Mean

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Histogram of Mixed Normal Distribution

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Histogram of Mixed Normal Distribution

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Sampling Distribution of X̄ from a Mixed Normal

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Sample Means from Mixtured Normal Distributions When n = 2

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Sample Mean

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Sample Mean

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Said another way, with regards to the sampling distribution ofthe sample mean, the shape of the distribution from whichscores are sampled is irrelevant, because as sample size growslarge the sampling distribution of the sample mean will benormally distributed!

Because the distribution of the sample means is normal,provided sample size is not too small, inferences about themean can legitimately be made using a normal distribution!

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Assumptions

Notation GlossaryWhat You LearnedA Worked Example

The Normality Assumption

The standard normal distribution (i.e., the z-table) assumesthat the normally distributed.If the normality assumption is violated, the probabilitiesobtained will not be correct.If interest concerns a mean, which is often the case, theCentral Limit Theorem tells us that the distribution of samplemeans will be normally distributed (if sample size is not toosmall).

Thus, the z-distribution can be used when interest concernsmeans, regardless of the shape of the distribution of the scores.However, the z-distribution should not be used for probabilitiesof individual scores if the distribution from which the scoresare sampled is not normal.

We have also assumed that σ is known.

We relax this assumptions in the next topic (by using at-distribution instead of z). 49 / 57


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Sampling Error

The random discrepancy between a statistic (from a sample)and the parameter (the population value).

Sampling error is the reason for inferential statistics.

If there were no sampling error, the estimate from a particularsample would equal the population value.

Because sample error (essentially) always exists, we need toquantify the uncertainty of an estimate.

This is why confidence intervals and hypothesis tests are soimportant (much more to come on these procedures later).

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Finite Sample Calculations

For small populations in which the population size (N) isknown, there is a correction factor that can be used whencalculating the standard error of the mean.

Rather than using σx̄ = σ√n

as would be typical,using

σx̄ =√

N−nN−1

(σ√n

)provides a more appropriate value of the

standard error of the sample mean in situations in which thepopulation is small.

In general, this might be used only if the sample size is 5% ormore of the population size.

Usually we assume that the data come from an infinitepopulation or process (or we acknowledge we do not know thepopulation size), and thus this correction is not often needed.

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Notation Glossary

σx̄ - population standard deviation of the sample mean(σx̄ = σ√

n)

n - sample size

N - population size

x̄ - the mean of a sample

µ - the mean of a population

z - z-score for a score of interest (z = x−µσ )

z - z-score when the sample mean is of interest (z = x̄−µσ/√n

)

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What You Learned from this Lesson

You learned:

The idea of sampling from a population to obtain pointestimates.

The idea of the variability of point estimates.

Understanding how the standard deviation of a mean decreasesas the sample size increases (i.e., σX̄ = σ/

√n).

How to find probability values for a mean.

How the Central Limit Theorem allows for normal distributionto be used when interest concerns a mean (even if thedistribution of scores is not normal).

How the Central Limit Theorem allows for the assumption ofnormality not to be met, but still allows the use of the normaldistribution when finding probabilities for a mean.

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A Worked Example

From a statement made by a automotive dealership in a courtproceeding, it was stated that “40% of the time they sell plusor minus 8 vehicles per month around the population mean.”That is, p(µ± 8) = .40.

Assume that the statement is based on the relevant populationvalues and the vehicle sales follow a normal distribution.

What is σ, which you would find, as a competitor to thisdealership, very valuable for comparison purposes?

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A Worked Example, Continued

First, we need to determine the relevant z-values based onwhat is known.

If there is a 40% chance of selling plus/minus 8 vehiclesaround mean, we know can find the corresponding z-scores;60% of the time the dealership falls outside of the limits (30%less and 30% more).

There thus being 30% of the time on either size:p(Z ≤ z) = .30 is for z = .524; meaning for this problem therelevant quantiles (i.e., z-values) are ±.524 (which can befound using qnorm(.30) and qnorm(.70)).

The relevant z-values are thus ±.524.

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Recalling that z = X−µσ , and from the problem we have

±.524 = ±8σ , we are solve for σ (because we know everything

else):σ = 8/.524 = 15.267.

Thus, we have found that the population standard deviation isaround 15.267 cars per month.

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What is the probability of the dealership having a “summermean” (i.e, of the three months of summer) that is within 10vehicles of the overall mean?Here we use X̄ instead of X , and thus σX̄ instead of σ:

z = X̄−µσ/√n

.

From earlier we already know that σ = 15.267 and we knowthe other values:

z =±10

15.267/√

3=±10

8.81.

The z-values of interest are ±1.135.Thus, p(± 10 Cars around µ) = p(Z ≥ −1.135 & Z ≤1.135) = .7436.

The above probability can be obtained from1− 2× pnorm(−1.135) or, equivalently yet more formally,1-[pnorm(-1.135)+1-pnorm(1.135)]. 57 / 57

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