Sampling & Hypothesis Testing: Population, Sample, Estimators, & Confidence Intervals - Pr, Study notes of Introduction to Econometrics

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Sampling and Hypothesis Testing

Allin Cottrell

Population and sample Population

: an entire set of objects or units of observation of one

sort or another. Sample

: subset of a population.

Parameter

versus

statistic

.

size

mean

variance

proportion

Population:

N

μ

σ

2

π

Sample:

n

¯

x

s

2

p

Properties of estimators: sample mean

¯

x

=

n

n

∑

i

=

1

x

i

To make inferences regarding the population mean,

μ

, we need to

know something about the probability distribution of this samplestatistic, ¯

x

.

The distribution of a sample statistic is known as a

sampling

distribution

. Two of its characteristics are of particular interest, the

mean or expected value and the variance or standard deviation. E(

¯

x)

: Thought experiment: Sample repeatedly from the given

population, each time recording the sample mean, and take theaverage of those sample means.

If the sampling procedure is

unbiased

, deviations of ¯

x

from

μ

in the

upward and downward directions should be equally likely; onaverage, they should cancel out.

E(

¯

x)

=

μ

=

E(X)

The sample mean is then an

unbiased estimator

of the population

mean.

Efficiency One estimator is more

efficient

than another if its values are more

tightly clustered around its expected value.E.g. alternative estimators for the population mean: ¯

x

versus the

average of the largest and smallest values in the sample.The degree of dispersion of an estimator is generally measured by thestandard deviation of its probability distribution (samplingdistribution). This goes under the name

standard error

.

Standard error of sample mean

σ

¯ x

=

σ

√

n

•

The more widely dispersed are the population values around theirmean (larger

σ

), the greater the scope for sampling error (i.e.

drawing by chance an unrepresentative sample whose mean differssubstantially from

μ

).

•

A larger sample size (greater

n

) narrows the dispersion of ¯

x

.

Other statistics Population

proportion

,

π

.

The corresponding sample statistic is the proportion of the samplehaving the characteristic in question,

p

.

The sample proportion is an unbiased estimator of the populationproportion

E(p)

=

π

Its standard error is given by

σ

p

=

√

π (

−

π )

n

Population

variance

,

σ

2

.

σ

2

=

N

N

∑ i

=

1

(x

i

−

μ)

2

Estimator, sample variance:

s

2

=

n

−

n

∑ i

=

1

(x

i

−

¯

x)

2

Shape of sampling distributions Besides knowing expected value and standard error, we also need toknow the

shape

of a sampling distribution in order to put it to use.

Sample mean: Central Limit Theorem implies a Gaussian distribution,for “large enough” samples.Reminder:

0

0

.

05

0

.

1

0

.

15

0

.

2

0

1

2

3

4

5

6

7

P ( ¯

x)

¯ x

Five dice

Not all sampling distributions are Gaussian, e.g. sample variance asestimator of population variance. In this case the ratio

(n

−

)s

2

/σ

2

follows a skewed distribution known as

χ

2

, (

chi-square

) with

n

−

degrees of freedom. If the sample size is large the

χ

2

distribution converges towards the

normal.

Confidence intervals If we know the mean, standard error and shape of the distribution ofa given sample statistic, we can then make definite probabilitystatements about the statistic.Example:

μ

=

100 and

σ

=

12 for a certain population, and we draw a

sample with

n

=

36 from that population.

The standard error of ¯

x

is

σ /

√

n

=

/

=

2, and a sample size of 36

is large enough to justify the assumption of a Gaussian samplingdistribution. We know that the range

μ

±

σ

encloses the central 95

percent of a normal distribution, so we can state

P (

<

¯

x <

)

≈

.

There’s a 95 percent probability that the sample mean lies within 4units (= 2 standard errors) of the population mean, 100.

Population mean unknown If

μ

is unknown we can still say

P (μ

−

<

¯

x < μ

+

)

≈

.

With probability .95 the sample mean will be drawn from within 4units of the unknown population mean.We go ahead and draw the sample, and calculate a sample mean of(say) 97. If there’s a probability of .95 that our ¯

x

came from within 4

units of

μ

, we can turn that around: we’re entitled to be 95 percent

confident that

μ

lies between 93 and 101.

We draw up a 95 percent

confidence interval

for the population mean

as ¯

x

±

σ

¯ x

.

Population variance unknown With

σ

unknown, we have to

estimate

the standard error of ¯

x

.

s

¯ x

≡

ˆ

σ

¯ x

=

s

√

n

We can now reformulate our 95 percent confidence interval for

μ

:

¯

x

±

s

¯ x

.

Strictly speaking, the substitution of

s

for

σ

alters the shape of the

sampling distribution. Instead of being Gaussian it now follows the

t

distribution, which looks very much like the Gaussian except that it’sa bit “fatter in the tails”.

The t distribution Unlike the Gaussian, the

t

distribution is not fully characterized by its

mean and standard deviation: there is an additional factor, namelythe

degrees of freedom

(df).

•

For estimating a population mean the df term is the sample sizeminus 1.

•

At low degrees of freedom the

t

distribution is noticeably more

“dispersed” than the Gaussian, meaning that a 95 percentconfidence interval would have to be wider (greater uncertainty).

•

As the degrees of freedom increase, the

t

distribution converges

towards the Gaussian.

•

Values enclosing the central 95 percent of the distribution:

Normal:

μ

±

.

σ

t(

)

:

μ

±

.

σ

Further examples The following information regarding the Gaussian distributionenables you to construct a 99 percent confidence interval.

P (μ

−

.

σ < x < μ

+

.

σ )

≈

0

.

Thus the 99 percent interval is ¯

x

±

.

σ

¯ x

.

If we want greater confidence that our interval straddles the unknownparameter value (99 percent versus 95 percent) then our intervalmust be wider (

±

.

58 standard errors versus

±

2 standard errors).

Estimating a proportion An opinion polling agency questions a sample of 1200 people toassess the degree of support for candidate X.

•

Sample info:

p

=

0

.

•

Our single best guess at the population proportion,

π

, is then 0.56,

but we can quantify our uncertainty.

•

The standard error of

p

is

√

π (

−

π )/n

. The value of

π

is

unknown but we can substitute

p

or, to be conservative, we can

put

π

=

0

.

5 which maximizes the value of

π (

−

π )

.

•

On the latter procedure, the estimated standard error is√

0

.

/

=

0

.

•

The large sample justifies the Gaussian assumption for thesampling distribution; the 95 percent confidence interval is 0

.

±

×

0

.

=

0

.

±

0

.

Generalizing the idea Let

θ

denote a “generic parameter”.

1. Find an estimator (preferably unbiased) for

θ

.

2. Generate

ˆ

θ

(point estimate).

3. Set confidence level, 1

−

α

.

4. Form interval estimate (assuming symmetrical distribution):

ˆ

θ

±

maximum error for

(

−

α)

confidence

“Maximum error” equals so many standard errors of such and such asize. The number of standard errors depends on the chosenconfidence level (possibly also the degrees of freedom). The size ofthe standard error,

σ

ˆ θ

, depends on the nature of the parameter being

estimated and the sample size.

z-scores Suppose the sampling distribution of

ˆ

θ

is Gaussian. The following

notation is useful:

z

=

x

−

μ

σ

The “standard normal score” or “

z

-score” expresses the value of a

variable in terms of its distance from the mean, measured in standarddeviations.Example:

μ

=

1000 and

σ

=

50. The value

x

=

850 has a

z

-score of

−

.

0: it lies 3 standard deviations below the mean.

Where the distribution of

ˆ

θ

is Gaussian we can write the 1

−

α

confidence interval for

θ

as

ˆ

θ

±

σ

ˆ θ

z

α/

2

z

.

975

= −

1

.

96

z

.

025

=

1

.

96

This is about as far as we can go in general terms. The specificformula for

σ

ˆ θ

depends on the parameter.

The logic of hypothesis testing Analogy between the set-up of a hypothesis test and a court of law.Defendant on trial in the statistical court is the

null hypothesis

, some

definite claim regarding a parameter of interest.Just as the defendant is presumed innocent until proved guilty, thenull hypothesis (

H

0

) is assumed true (at least for the sake of

argument) until the evidence goes against it.

H

0

is in fact:

Decision:

True

False

Reject

Type I error

Correct decision

P

=

α

Fail to reject

Correct decision

Type II error

P

=

β

−

β

is the

power

of a test; trade-off between

α

and

β

.

Choosing the significance level How do we get to

choose

α

(probability of Type I error)?

The calculations that compose a hypothesis test are condensed in akey number, namely a conditional probability:

the probability of

observing the given sample data, on the assumption that the nullhypothesis is true

.

This is called the

p-value

. If it is small, we can place one of two

interpretations on the situation:

(a) The null hypothesis is true and the sample we drew is an

improbable, unrepresentative one.

(b) The null hypothesis is false.

The smaller the p-value, the less comfortable we are with alternative(a). (Digression) To reach a conclusion we must specify the limit ofour comfort zone, a p-value below which we’ll reject

H

0

.

Say we use a cutoff of .01: we’ll reject the null hypothesis if thep-value for the test is

≤

.

If the null hypothesis is in fact true, what is the probability of ourrejecting it? It’s the probability of getting a p-value less than or equalto .01, which is (by definition) .01.In selecting our cutoff we selected

α

, the probability of Type I error.

Example of hypothesis test A maker of RAM chips claims an average access time of 60nanoseconds (ns) for the chips. Quality control has the job ofchecking that the production process is maintaining acceptableaccess speed: they test a sample of chips each day.Today’s sample information is that with 100 chips tested, the meanaccess time is 63 ns with a standard deviation of 2 ns. Is this anacceptable result?Sould we go with the symmetrical hypotheses

H

0

:

μ

=

versus

H

1

:

μ

≠

60?

Well, we don’t mind if the chips are faster than advertised.So instead we adopt the asymmetrical hypotheses:

H

0

:

μ

≤

versus

H

1

:

μ >

Let

α

=

0

.

The p-value is

P (

¯

x

≥

|

μ

≤

)

where

n

=

100 and

s

=

•

If the null hypothesis is true,

E(

¯

x)

is no greater than 60.

•

The estimated standard error of ¯

x

is

s/

√

n

=

/

=

.

•

With

n

=

100 we can take the sampling distribution to be normal.

•

With a Gaussian sampling distribution the

test statistic

is the

z

-score.

z

=

¯

x

−

μ

H

0

s

¯ x

=

−

.

=

Variations on the example Suppose the test were as described above, except that the sample wasof size 10 instead of 100.Given the small sample and the fact that the population standarddeviation,

σ

, is unknown, we could not justify the assumption of a

Gaussian sampling distribution for ¯

x

. Rather, we’d have to use the

t

distribution with df = 9.The estimated standard error,

s

¯ x

=

/

√

=

0

.

632, and the test

statistic is

t(

)

=

¯

x

−

μ

H

0

s

¯ x

=

−

.

=

.

The p-value for this statistic is 0.000529—a lot larger than for

z

=

15,

but still much smaller than the chosen significance level of 5 percent,so we still reject the null hypothesis.

In general the test statistic can be written as

test

=

ˆ

θ

−

θ

H

0

s

ˆ θ

That is, sample statistic minus the value stated in the nullhypothesis—which by assumption equals

E(

ˆ

θ)

—divided by the

(estimated) standard error of

ˆ

θ

.

The distribution to which “test” must be referred, in order to obtainthe p-value, depends on the situation.

Another variation We chose an asymmetrical test setup above. What difference would itmake if we went with the symmetrical version,

H

0

:

μ

=

versus

H

1

:

μ

≠

60?

We have to think:

what sort of values of the test statistic should count

against the null hypothesis? In the asymmetrical case only values of ¯

x

greater than 60 counted

against

H

0

. A sample mean of (say) 57 would be consistent with

μ

≤

60; it is not even

prima facie

evidence against the null.

Therefore the

critical region

of the sampling distribution (the region

containing values that would cause us to reject the null) lies strictly inthe upper tail.But if the null hypothesis were

μ

=

60, then values of ¯

x

both

substantially below and substantially above 60 would count against it.The critical region would be divided into two portions, one in each tailof the sampling distribution.

H

0

:

μ

=

60. Two-tailed test. Both high and low values count against

H

0

.

α/

α/

H

0

:

μ

≤

60. One-tailed test. Only high values count against

H

0

.

α

Practical consequence We must double the p-value

, before comparing it to

α

.

•

The sample mean was 63, and the p-value was defined as theprobability of drawing a sample “like this or worse”, from thestandpoint of

H

0

.

•

In the symmetrical case, “like this or worse” means “with a samplemean this far away from the hypothesized population mean, orfarther, in either direction”.

•

So the p-value is

P (

¯

x

≥

∪

¯

x

≤

)

, which is double the value we

found previously.

More on p-values Let

E

denote the sample evidence and

H

denote the null hypothesis

that is “on trial”. The p-value can then be expressed as

P (E

|

H)

.

This may seem awkward. Wouldn’t it be better to calculate theconditional probability the other way round,

P (H

|

E)

?

Instead of working with the probability of obtaining a sample like theone we in fact obtained, assuming the null hypothesis to be true, whycan’t we think in terms of the probability that the null hypothesis istrue, given the sample evidence?

Recall the multiplication rule for probabilities, which we wrote as

P (A

∩

B)

=

P (A)

×

P (B

|

A)

Swapping the positions of

A

and

B

we can equally well write

P (B

∩

A)

=

P (B)

×

P (A

|

B)

And taking these two equations together we can infer that

P (A)

×

P (B

|

A)

=

P (B)

×

P (A

|

B)

or

P (B

|

A)

=

P (B)

×

P (A

|

B)

P (A)

This is

Bayes’ rule

. It provides a means of converting from a

conditional probability one way round to the inverse conditionalprobability.

Substituting

E

(Evidence) and

H

(null Hypothesis) for

A

and

B

, we get

P (H

|

E)

=

P (H)

×

P (E

|

H)

P (E)

We know how to find the p-value,

P (E

|

H)

. To obtain the probability

we’re now canvassing as an alternative,

P (H

|

E)

, we have to supply in

addition

P (H)

and

P (E)

.

P (H)

is the marginal probability of the null hypothesis and

P (E)

is

the marginal probability of the sample evidence.Where are these going to come from??

Confidence intervals and tests The symbol

α

is used for both the significance level of a hypothesis

test (the probability of Type I error), and in denoting the confidencelevel (

−

α

) for interval estimation.

There is an equivalence between a two-tailed hypothesis test atsignificance level

α

and an interval estimate using confidence level

−

α

.

Suppose

μ

is unknown and a sample of size 64 yields ¯

x

=

50,

s

=

The 95 percent confidence interval for

μ

is then

±

.

(

√

)

=

±

.

=

.

55 to 52

.

Suppose we want to test

H

0

:

μ

=

55 using the 5 percent significance

level. No additional calculation is needed. The value 55 lies outside ofthe 95 percent confidence interval, so we can conclude that

H

0

is

rejected.

In a two-tailed test at the 5 percent significance level, we fail to reject H

0

if and only if ¯

x

falls within the central 95 percent of the sampling

distribution, according to

H

0

.

But since 55 exceeds 50 by more than the “maximum error”, 2.45, wecan see that, conversely, the central 95 percent of a samplingdistribution centered on 55 will not include 50, so a finding of ¯

x

=

must lead to rejection of the null.“Significance level” and “confidence level” are complementary.



Digression

0

0.

0.

0.

0.

0.

0.

0.

0.

0.

0.

0

0.

normal pdf

normal p-value

The further we are from the center of the sampling distribution,according to

H

0

, the smaller the p-value.

Back to the main discussion.