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Statistical Inference and Regression Analysis: Stat-GB.3302.30, Stat-UB.0015.01. Professor William Greene Stern School of Business IOMS Department Department of Economics. Part 5 – Hypothesis Testing. Objectives of Statistical Analysis. Estimation How long do hard drives last?

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statistical inference and regression analysis stat gb 3302 30 stat ub 0015 01

Statistical Inference and Regression Analysis: Stat-GB.3302.30, Stat-UB.0015.01

Professor William Greene

Stern School of Business

IOMS Department

Department of Economics

objectives of statistical analysis
Objectives of Statistical Analysis
  • Estimation
    • How long do hard drives last?
    • What is the median income among the 99%ers?
  • Inference – hypothesis testing
    • Did minorities pay higher mortgage rates during the housing boom?
    • Is there a link between environmental factors and breast cancer on eastern long island?
general frameworks
General Frameworks
  • Parametric Tests: features of specific distributions such as the mean of a Bernoulli or normal distribution.
  • Specification Tests (Semiparametric)
    • Do the data arrive from a Poisson process
    • Are the data normally distributed
  • Nonparametric Tests: Are two discrete processes independent?
hypotheses
Hypotheses
  • Hypotheses - labels
    • State 0 of Nature – Null Hypothesis
    • State 1 – Alternative Hypothesis
  • Exclusive: Prob(H0 ∩ H1) = 0
  • Exhaustive: Prob(H0) + Prob(H1) = 1
  • Symmetric: Neither is intrinsically “preferred” – the objective of the study is only to support one or the other. (Rare?)
does the new drug work
Does the New Drug Work?
  • Hypotheses: H0= .50, H1 = .75
  • Priors: P0= .40, P1= .60
  • Clinical Trial: N = 50, 31 patients “respond’” p = .62
  • Likelihoods:
    • L0 (31|  =.50) = Binomial(50,31,.50) = .0270059
    • L1 (31|  =.75) = Binomial(50,31,.75) = .0148156
  • Posterior odds in favor of H0 = (.4/.6)(.0270059/.0148156) = 1.2152 > 1
  • Priors favored H1 1.5 to 1, but the posterior odds favor H0, 1.2152 to 1. The evidence discredits H1even though the ‘data’ seem more consistent with prior P1.
decision strategy
Decision Strategy
  • Prefer the hypothesis with the higher posterior odds
  • A gap in the theory: How does the investigator do the cost benefit test?
    • Starting a new business venture or entering a new market: Priors and market research
    • FDA approving a new drug or medical device. Priors and clinical trials
  • Statistical Decision Theory adds the costs and benefits of decisions and errors.
an alternative strategy
An Alternative Strategy
  • Recognize the asymmetry of null and alternative hypotheses.
  • Eliminate the prior odds (which are rarely formed or available).
slide11

http://query.nytimes.com/gst/fullpage.html?res=9C00E4DF113BF935A3575BC0A9649C8B63http://query.nytimes.com/gst/fullpage.html?res=9C00E4DF113BF935A3575BC0A9649C8B63

classical hypothesis testing
Classical Hypothesis Testing
  • The scientific method applied to statistical hypothesis testing
  • Hypothesis: The world works according to my hypothesis
  • Testing or supporting the hypothesis
    • Data gathering
    • Rejection of the hypothesis if the data are inconsistent with it
    • Retention and exposure to further investigation if the data are consistent with the hypothesis
  • Failure to reject is not equivalent to acceptance.
asymmetric hypotheses
Asymmetric Hypotheses
  • Null Hypothesis: The proposed state of nature
  • Alternative hypothesis: The state of nature that is believed to prevail if the null is rejected.
hypothesis testing strategy
Hypothesis Testing Strategy
  • Formulate the null hypothesis
  • Gather the evidence
  • Question: If my null hypothesis were true, how likely is it that I would have observed this evidence?
    • Very unlikely: Reject the hypothesis
    • Not unlikely: Do not reject. (Retain the hypothesis for continued scrutiny.)
some terms of art
Some Terms of Art
  • Type I error: Incorrectly rejecting a true null
  • Type II error: Failure to reject a false null
  • Power of a test: Probability a test will correctly reject a false null
  • Alpha level: Probability that a test will incorrectly reject a true null. This is sometimes called the size of the test.
  • Significance Level: Probability that a test will retain a true null = 1 – alpha.
  • Rejection Region: Evidence that will lead to rejection of the null
  • Test statistic: Specific sample evidence used to test the hypothesis
  • Distribution of the test statistic under the null hypothesis: Probability model used to compute probability of rejecting the null. (Crucial to the testing strategy – how does the analyst assess the evidence?)
possible errors in testing
Possible Errors in Testing

Hypothesis is Hypothesis is True False

I Do Not Reject the Hypothesis

I Reject the Hypothesis

a legal analogy the null hypothesis is innocent
A Legal Analogy: The Null Hypothesis is INNOCENT

Null Hypothesis Alternative Hypothesis Not Guilty Guilty

Finding: Verdict Not Guilty

Finding: VerdictGuilty

The errors are not symmetric. Most thinkers consider Type I errors to be more serious than Type II in this setting.

jerzy neyman karl pearson methodology
(Jerzy) Neyman – (Karl) Pearson Methodology
  • “Statistical” testing
  • Methodology
    • Formulate the “null” hypothesis
    • Decide (in advance) what kinds of “evidence” (data) will lead to rejection of the null hypothesis. I.e., define the rejection region
    • Gather the data
    • Mechanically carry out the test.
formulating the null hypothesis
Formulating the Null Hypothesis
  • Stating the hypothesis: A belief about the “state of nature”
    • A parameter takes a particular value
    • There is a relationship between variables
    • And so on…
  • The null vs. the alternative
    • By induction: If we wish to find evidence of something, first assume it is not true.
    • Look for evidence that leads to rejection of the assumed hypothesis.
    • Evidence that rejects the null hypothesis is significant
example credit scoring rule
Example: Credit Scoring Rule
  • Investigation: I believe that Fair Isaacs relies on home ownership in deciding whether to “accept” an application.
    • Null hypothesis: There is no relationship
    • Alternative hypothesis: They do use homeownership data.
  • What decision rule should I use?
some evidence
Some Evidence

= Homeowners

5469

5030

1845

1100

hypothesis test
Hypothesis Test
  • Acceptance rate for homeowners = 5030/(5030+1100) = .82055
  • Acceptance rate for renters is .74774
  • H0: Acceptance rate for renters is not less than for owners.
  • H0: p(renters) > .82055
  • H1: p(renters) < .82055
the rejection region
The Rejection Region

What is the “rejection region?”

  • Data (evidence) that are inconsistent with my hypothesis
  • Evidence is divided into two types:
    • Data that are inconsistent with my hypothesis (the rejection region)
    • Everything else
my testing procedure
My Testing Procedure
  • I will reject H0 if p(renters) < .815 (chosen arbitrarily)
  • Rejection region is sample values of p(renters) < 0.815
distribution of the test statistic under the null hypothesis
Distribution of the Test Statistic Under the Null Hypothesis
  • Test statistic p(renters) = 1/N i Accept(=1 or 0)
  • Use the central limit theorem:
  • Assumed mean = .82055
  • Implied standard deviation= sqr(.82055*.17945/7413)=.00459
  • Using CLT, normally distributed. (N is very large).
  • Use z = (p(renters) - .82055) / .00459
alpha level and rejection region
Alpha Level and Rejection Region
  • Prob(Reject H0|H0 true) = Prob(p < .815 | H0 is true)= Prob[(p - .82055)/.00459)= Prob[z < -1.209]= .11333
  • Probability of a Type I error
  • Alpha level for this test
the test
The Test
  • The observed proportion is 5469/(5469+1845) = 5469/7314 = .74774
  • The null hypothesis is rejected at the 11.333% significance level (by the design of the test)
application breast cancer on long island
Application: Breast Cancer On Long Island
  • Null Hypothesis: There is no link between the high cancer rate on LI and the use of pesticides and toxic chemicals in dry cleaning, farming, etc.
  • Neyman-Pearson Procedure
    • Examine the physical and statistical evidence
    • If there is convincing covariation, reject the null hypothesis
    • What is the rejection region?
  • The NCI study:
    • Working null hypothesis: There is a link: We will find the evidence.
    • How do you reject this hypothesis?
formulating the testing procedure
Formulating the Testing Procedure
  • Usually: What kind of data will lead me to reject the hypothesis?
  • Thinking scientifically: If you want to “prove” a hypothesis is true (or you want to support one) begin by assuming your hypothesis is not true, and look for evidence that contradicts the assumption.
hypothesis about a mean
Hypothesis About a Mean
  • I believe that the average income of individuals in a population is $30,000.
    • H0 : μ = $30,000 (The null)
    • H1: μ ≠ $30,000 (The alternative)
  • I will draw the sample and examine the data.
  • The rejection region is data for which the sample mean is far from $30,000.
  • How far is far????? That is the test.
application
Application
  • The mean of a population takes a specific value:
  • Null hypothesis: H0: μ = $30,000H1: μ ≠ $30,000
  • Test: Sample mean close to hypothesized population mean?
  • Rejection region: Sample means that are far from $30,000
deciding on the rejection region
Deciding on the Rejection Region
  • If the sample mean is far from $30,000, reject the hypothesis.
  • Choose, the region, for example,

The probability that the mean falls in the rejection region even though the hypothesis is true (should not be rejected) is the probability of a type 1 error. Even if the true mean really is $30,000, the sample mean could fall in the rejection region.

Rejection

Rejection

29,500 30,000 30,500

reduce the probability of a type i error by making the non rejection region wider
Reduce the Probability of a Type I Error by Making the (non)Rejection Region Wider

Reduce the probability of a type I error by moving the boundaries of the rejection region farther out.

Probability outside this interval is large.

28,500 29,500 30,000 30,500 31,500

You can make a type I error impossible by making the rejection region very far from the null. Then you would never make a type I error because you would never reject H0.

Probability outside this interval is much smaller.

setting the level
Setting the α Level
  • “α” is the probability of a type I error
  • Choose the width of the interval by choosing the desired probability of a type I error, based on the t or normal distribution. (How confident do I want to be?)
  • Multiply the z or t value by the standard error of the mean.
testing procedure
Testing Procedure
  • The rejection region will be the range of values greater than μ0 + zσ/√N orless than μ0 - zσ/√N
  • Use z = 1.96 for 1 - α = 95%
  • Use z = 2.576 for 1 - α = 99%
  • Use the t table if small sample, variance is estimated and sampling from a normal distribution.
deciding on the rejection region1
Deciding on the Rejection Region
  • If the sample mean is far from $30,000, reject the hypothesis.
  • Choose, the region, say,

Rejection

Rejection

I am 95% certain that I will not commit a type I error (reject the hypothesis in error). (I cannot be 100% certain.)

the test procedure
The Test Procedure
  • Choosing z = 1.96 makes the probability of a Type I error 0.05.
  • Choosing z = 2.576 would reduce the probability of a Type I error to 0.01.
  • Reducing the probability of a Type I error reduces the power of the test because it reduces the probability that the null hypothesis will be rejected.
p value
P Value
  • Probability of observing the sample evidence assuming the null hypothesis is true.
  • Null hypothesis is rejected if P value < 
p value prob p renter 74774 prob z 74774 82055 00459 15 86 59946942854362260 10 56 impossible
P value < Prob[p(renter) < .74774] = Prob[z < (.74774 - .82055)/.00459] = (-15.86) = .59946942854362260 * 10-56Impossible

=.11333

confidence intervals
Confidence Intervals
  • For a two sided test about a parameter, a confidence interval is the complement of the rejection region. (Proof in text, p. 338)
confidence interval
Confidence Interval
  • If the sample mean is far from $30,000, reject the hypothesis.
  • Choose, the region, say,

Rejection

Confidence

Rejection

I am 95% certain that the confidence interval contains the true mean of the distribution of incomes. (I cannot be 100% certain.)

one sided tests
One Sided Tests
  • H0 = 0, H10 Rejection region is sample mean far from 0 in either direction
  • H0 = 0, H1>0. Sample means less than 0 cannot be in the rejection region.
  • Entire rejection region is above 0.
  • Reformulate: H0<0, H1>0.
carrying out the lr test
Carrying Out the LR Test
  • In most cases, exact distribution of the statistic is unknown
  • Use -2log  Chi squared [1]
  • For a test about 1 parameter, threshold value is 3.84 (5%) or 6.45 (1%)
specification tests
Specification Tests
  • Generally a test about a distribution where the alternative is “some other distribution.”
  • Test is generally based on a feature of the distribution that is true under the null but not true under the alternative.
poisson specification tests
Poisson Specification Tests
  • 3820 observations on doctor visits
  • Poisson distribution?
deviance test
Deviance Test
  • Poisson Distribution p(x) = exp(-)x/x!
  • H0: Everyone has the same Poisson Distribution
  • H1: Everyone has their own Poisson distribution
  • Under H0, observations will tend to be near the mean. Under H1, there will be much more variation.
  • Likelihood ratio statistic (Text, p. 348)
dispersion test
Dispersion Test
  • Poisson Distribution p(x) = exp(-)x/x!
  • H0: The distribution is Poisson
  • H1: The distribution is something else
  • Under H0, the mean will be (almost) the same as the variance
  • Approximate Likelihood ratio statistic (Text, p. 348) = N * Variance / Mean
  • For the doctor visit data, this is 22,348.6 vs. chi squared with 1 degree of freedom. H0 is rejected.
specification test normality
Specification Test - Normality
  • Normal Distribution is symmetric and has kurtosis = 3.
  • Compare observed 3rd and 4th moments to what would be expected from a normal distribution.
kurtosis t 5 vs normal
Kurtosis: t[5] vs. Normal

Kurtosis of normal(0,1) = 3

Kurtosis of t[k] = 3 + 6/(k-4); for t[5] = 3+6/(5-4) = 9.

testing for a distribution
Testing for a Distribution
  • H0: The distribution is assumed
  • H1: The assumed distribution is incorrect
  • Strategy: Do the features of the sample resemble what we would observe if H0 were correct
    • Continuous: CDF of data resemble CDF of the assumed distribution
    • Discrete: Sample cell probabilities resemble predictions from the assumed distribution
chi squared test for a discrete distribution
Chi Squared Test for a Discrete Distribution
  • Outcomes = A1, A2,…, AM
  • Predicted probabilities based on a theoretical distribution = E1(), E2(),…,EM().
  • Sample cell frequencies = O1,…,OM
v2 rocket hits
V2 Rocket Hits

Adapted from Richard Isaac, The Pleasures of Probability, Springer Verlag, 1995, pp. 99-101.

576 0.25Km2 areas of South London in a grid (24 by 24)

535 rockets were fired randomly into the grid = N

P(a rocket hits a particular grid area) = 1/576 = 0.001736 = θ

Expected number of rocket hits in a particular area = 535/576 = 0.92882

How many rockets will hit any particular area? 0,1,2,… could be anything up to 535.

The 0.9288 is the λ for a Poisson distribution:

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poisson process
Poisson Process
  • θ = 1/169
  • N = 144
  • λ = 144 * 1/169 = 0.852
  • Probabilities:
    • P(X=0) = .4266
    • P(X=1) = .3634
    • P(X=2) = .1548
    • P(X=3) = .0437
    • P(X=4) = .0094
    • P(X>4) = .0021
interpreting the process
λ = 0.852

Probabilities:

P(X=0) = .4266

P(X=1) = .3634

P(X=2) = .1548

P(X=3) = .0437

P(X=4) = .0094

P(X>4) = .0021

There are 169 squares

There are 144 “trials”

Expect .4266*169 = 72.1 to have 0 hits/square

Expect .3634*169 = 61.4 to have 1 hit/square

Etc.

Expect the average number of hits/square to = .852.

Interpreting The Process
difference in means of two populations
Difference in Means of Two Populations
  • Two Independent Normal Populations
    • Common known variance
    • Common unknown variance
    • Different Variances
    • One and two sided tests
  • Paired Samples
    • Means of paired observations
    • Treatments and Controls – Diff-in-Diff SAT
  • Nonparametric – Mann/Whitney
  • Two Bernoulli Populations
household incomes equal variances
Household Incomes, Equal Variances

------------------------------------------------------

t test of equal means INCOME by MARRIED

------------------------------------------------------

MARRIED = 0 Nx = 817 MARRIED = 1 Ny = 3057

t [ 3872] = 3.7238 P value = .0002

------------------------------------------------------

Mean Std.Dev. Std.Error

INCOME ----------------------------------------------

MARRIED = 0 .27982 .12939 .00453

MARRIED = 1 .30145 .15194 .00275

------------------------------------------------------

2 proportions
2 Proportions
  • Two Bernoulli Populations:Xi ~ Bernoulli with Prob(xi=1) = xYi ~ Bernoulli with Prob(yi=1) = y
  • H0: x = y
  • The sample proportions arepx = (1/Nx)ixi and py = (1/Ny)iyi
  • Sample variances are px(1-px) and py(1-py).
  • Use the Central Limit Theorem to form the test statistic.
z test for equality of proportions
z Test for Equality of Proportions

Application: Take up of public health insurance.

------------------------------------------------------

t test of equal means PUBLIC by FEMALE

------------------------------------------------------

FEMALE =0 Nx = 1812 FEMALE =1 Ny = 1565

t [ 3375] = 5.8627 P value = .0000

------------------------------------------------------

Mean Std.Dev. Std.Error

PUBLIC ----------------------------------------------

FEMALE = 0 .84713 .35996 .00846

FEMALE = 1 .91310 .28178 .00712

paired sample t and z test
Paired Sample t and z Test
  • Observations are pairs (Xi,Yi), i = 1,…,N
  • Hypothesis x = y.
  • Both normal distributions. May be correlated.
    • Medical Trials: Smoking vs. Nonsmoking (separate individuals, probably independent)
    • SAT repeat tests, before and after. (Definitely correlated)
  • Test is based on Di = Xi – Yi. Same as earlier with H0:D = 0.
treatment effects
Treatment Effects
  • SAT Do Overs
    • Experiment: X1, X2, …, XN = first SAT score, Y1, Y2, …, YN = second
    • Treatment: T1,…,TN = whether or not the student took a Kaplan (or similar) prep score
    • Hypothesis, y > x.
  • Placebo: In Medical trials, N1 subjects receive a drug (treatment), N2 receive a placebo.
    • Hypothesis: Effect is greater in the treatment group than in the control (placebo) group.
treatment effects in clinical trials
Treatment Effects in Clinical Trials
  • Does Phenogyrabluthefentanoel (Zorgrab) work?
  • Investigate: Carry out a clinical trial.
    • N+0 = “The placebo effect”
    • N+T – N+0 = “The treatment effect”
    • The hypothesis is that the difference in differences has mean zero.

Placebo Drug Treatment

No Effect N00 N0T

Positive Effect N+0 N+T

a test of independence
A Test of Independence
  • In the credit card example, are Own/Rent and Accept/Reject independent?
  • Hypothesis: Prob(Ownership) and Prob(Acceptance) are independent
  • Formal hypothesis, based only on the laws of probability: Prob(Own,Accept) = Prob(Own)Prob(Accept) (and likewise for the other three possibilities.
  • Rejection region: Joint frequencies that do not look like the products of the marginal frequencies.
contingency table analysis
Contingency Table Analysis

The Data: Frequencies Reject Accept TotalRent 1,845 5,469 7,214Own 1,100 5,030 6,630Total 2,945 10,499 13,444

Step 1: Convert to Actual Proportions Reject Accept TotalRent 0.13724 0.40680 0.54404Own 0.08182 0.37414 0.45596Total 0.21906 0.78094 1.00000

independence test
Independence Test

Step 2: Expected proportions assuming independence: If the factors are independent, then the joint proportions should equal the product of the marginal proportions.

[Rent,Reject] 0.54404 x 0.21906 = 0.11918[Rent,Accept] 0.54404 x 0.78094 = 0.42486[Own,Reject] 0.45596 x 0.21906 = 0.09988[Own,Accept] 0.45596 x 0.78094 = 0.35606

when is the chi squared large
When is the Chi Squared Large?
  • Critical values from chi squared table
  • Degrees of freedom = (R-1)(C-1).

Critical chi squaredD.F. .05 .01 1 3.84 6.63 2 5.99 9.21 3 7.81 11.34 4 9.49 13.28 5 11.07 15.09 6 12.59 16.81 7 14.07 18.48 8 15.51 20.09 9 16.92 21.6710 18.31 23.21

analyzing default
Analyzing Default
  • Do renters default more often (at a different rate) than owners?
  • To investigate, we study the cardholders (only)

DEFAULT

OWNRENT 0 1 All

0 4854 615 5469

46.23 5.86 52.09

1 4649 381 5030

44.28 3.63 47.91

All 9503 996 10499

90.51 9.49 100.00

multiple choices travel mode
Multiple Choices: Travel Mode
  • 210 Travelers between Sydney and Melbourne
  • 4 available modes, air, train, bus, car
  • Among the observed variables is income.
  • Does income help to explain mode choice?
  • Hypothesis: Mode choice and income are independent.
travel mode choices and income
Travel Mode Choices and Income

+----------------------------------------------------------+

| Travel MODE Data |

+--------+-------------------------------------------------+

|INCOME | AIR TRAIN BUS CAR || Total |

+--------+-------------------------------------++----------+

|LOW | 10 36 9 8 || 63 |

| | 0.04761 0.17143 0.04286 0.03810 || 0.30000 |

|----------------------------------------------++----------+

|MEDIUM | 19 20 13 24 || 76 |

| | 0.09048 0.09524 0.06190 0.11429 || 0.36190 |

|----------------------------------------------++----------+

|HIGH | 29 7 8 27 || 71 |

| | 0.13810 0.03333 0.03810 0.12857 || 0.33810 |

|==============================================++==========+

|Total | 58 63 30 59 || 210 |

| | 0.27619 0.30000 0.14286 0.28095 || 1.00000 |

+--------+-------------------------------------+-----------+

contingency table
Contingency Table

+----------------------------------------------------------+

| Travel MODE Data |

+--------+-------------------------------------------------+

|INCOME | AIR TRAIN BUS CAR || Total |

+--------+-------------------------------------++----------+

| | 10 36 9 8 || 63 |

|LOW | 0.04761 0.17143 0.04286 0.03810 || 0.30000 || | 0.08286 0.09000 0.04286 0.08429 ||

|----------------------------------------------++----------+

| | 19 20 13 24 || 76 |

|MEDIUM | 0.09048 0.09524 0.06190 0.11429 || 0.36190 || | 0.09995 0.10857 0.05170 0.10168 ||

|----------------------------------------------++----------+

| | 29 7 8 27 || 71 |

|HIGH | 0.13810 0.03333 0.03810 0.12857 || 0.33810 || | 0.09338 0.10143 0.04830 0.09499 ||

|==============================================++==========+

|Total | 58 63 30 59 || 210 |

| | 0.27619 0.30000 0.14286 0.28095 || 1.00000 |

+--------+-------------------------------------+-----------+

Assuming independence, P(Income,Mode) = P(Income) x P(Mode).

computing chi squared
Computing Chi Squared

For our transport mode problem, R = 3, C = 4, so DF = 2x3 = 6. The critical value is 12.59. The hypothesis of independence is rejected.