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Guidelines One-Sample Tests The t -test Power Analysis Tests for Population Means Chapter 8 Hypothesis Testing t Test 1 / 22
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Guidelines One-Sample Tests The t-test Power Analysis Tests for Population Means

Chapter 8Hypothesis Testing

t Test

1 / 22

Guidelines One-Sample Tests The t-test Power Analysis Tests for Population Means

Outline

Guidelines

One-Sample Tests

Power Analysis

Tests for Population MeansMosquito Life Span

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Guidelines One-Sample Tests The t-test Power Analysis Tests for Population Means

IntroductionThe z-score is

z =x̄ − µσ/√n.

taken under the assumption that the population standard deviation is known.If we are forced to replace the unknown σ2 with its unbiased estimator s2, then thestatistic is known as t:

t =x̄ − µs/√n.

We have previously noted that for independent normal random variables thedistribution of the t statistic can be determined exactly and we used the t distributionto construct a confidence interval for the population mean µ.

We now turn to using the t-statistic as a test statistic for hypothesis tests of thepopulation mean. As with several other procedures we have seen, the two-sided t testis a likelihood ratio test.

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Guidelines One-Sample Tests The t-test Power Analysis Tests for Population Means

Guidelines for Using the t Procedures

• Except in the case of small samples, the assumption that the data are a simplerandom sample from the population of interest is more important that thepopulation distribution is normal.

• For sample sizes less than 15, use t procedures if the data are close to normal.

• For sample sizes at least 15 use t procedures except in the presence of outliers orstrong skewness.

• The t procedures can be used even for clearly skewed distributions when thesample size is large, typically over 40 observations.

These criteria are designed to ensure that x̄ is a sample from a nearly normaldistribution. When these guidelines fail to be satisfied, then we can turn to alternativesthat are not based on the central limit theorem, but rather use the rankings of the data.

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Guidelines One-Sample Tests The t-test Power Analysis Tests for Population Means

One-Sample TestsThe two-sided hypothesis

H0 : µ = µ0 versus H1 : µ 6= µ0,

based on independent normal observations X1. . . . ,Xn with unknown mean µ andunknown variance σ2 is a likelihood ratio test. The parameter space and nullhypothesis space, are, respectively,

Θ = {(µ, σ2);µ ∈ R, σ2 > 0} and Θ0 = {(µ, σ2);µ = µ0, σ2 > 0}.

The critical region is a level set for the t-statistic, T (x) from the data x.

C = {|T (x)| > tn−1,α/2}.

where tn−1,α/2 is the upper α/2 tail probability of the t distribution with n− 1 degreesof freedom.

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Guidelines One-Sample Tests The t-test Power Analysis Tests for Population Means

The t-testTo show that the critical region is a likelihood ratio test, consider

Λ(x) =L(µ0, σ̂

20|x)

L(µ̂, σ̂2|x),

we begin with independent normal observations X1. . . . ,Xn with unknown mean µ andunknown variance σ2. The likelihood function

L(µ, σ2|x) =1

(2πσ2)n/2exp− 1

2σ2

n∑i=1

(xi − µ)2.

We have seen that the maximum over the parameter set Θ occurs for

µ̂ = x̄ and σ̂2 =1

n

n∑i=1

(xi − x̄)2

and over Θ0 for

µ̂0 = µ and σ̂20 =1

n

n∑i=1

(xi − µ0)2.6 / 22

Guidelines One-Sample Tests The t-test Power Analysis Tests for Population Means

The t-test

Substituting back into the likelihood function,

L(µ0, σ̂20|x) =

1

(2πσ̂20)n/2exp− 1

2σ̂20

n∑i=1

(xi − µ0)2 =1

(2πσ̂20)n/2exp−2

n,

L(µ̂, σ̂2|x)) =1

(2πσ̂2)n/2exp− 1

2σ̂2

n∑i=1

(xi − x̄)2 =1

(2πσ̂2)n/2, exp−2

n,

and the likelihood ratio is

Λ(x) =L(µ0, σ̂

20|x)

L(µ̂, σ̂2|x)=

(σ̂20σ̂2

)−n/2=

(∑ni=1(xi − µ0)2∑ni=1(xi − x̄)2

)−n/2

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Guidelines One-Sample Tests The t-test Power Analysis Tests for Population Means

The t-test

The critical region λ(x) ≤ λ0 is equivalent to the fraction in parenthesis above beingsufficiently large. In other words for some value cα,

cα ≤∑n

i=1(xi − µ0)2∑ni=1(xi − x̄)2

=

∑ni=1 ((xi − x̄) + (x̄ − µ0))2∑n

i=1(xi − x̄)2

=

∑ni=1(xi − x̄)2 + 2

∑ni=1(xi − x̄)(x̄ − µ0) +

∑ni=1(x̄ − µ0)2∑n

i=1(xi − x̄)2= 1 +

n(x̄ − µ0)2∑ni=1(xi − x̄)2

Continuing we find that

(cα − 1)(n − 1) ≤ n(x̄ − µ0)2∑ni=1(xi − x̄)2/(n − 1)

=(x̄ − µ0)2

s2/n.

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Guidelines One-Sample Tests The t-test Power Analysis Tests for Population Means

The t-test

(cα − 1)(n − 1) ≤ T (x)2

where

T (x) =x̄ − µ0s/√n

and s is the square root of the unbiased estimator of the variance.

s2 =1

n − 1

n∑i=1

(xi − x̄)2.

Taking square roots, we have the critical region

C ={x;√

(cα − 1)(n − 1) ≤ |T (x)|}

Thus, we take√

(cα − 1)(n − 1) = tn−1,α/2.9 / 22

Guidelines One-Sample Tests The t-test Power Analysis Tests for Population Means

One-Sample Tests

• Radon is formed as part of the normal radioactive decaychain of uranium.

• It is one of the densest substances that remains a gas undernormal conditions.

• Radon gas from natural sources can accumulate in buildings,especially in confined areas such as attics, and basements.

• Epidemiological evidence shows a clear link betweenbreathing high concentrations of radon and incidence oflung cancer.

• According to the United States Environmental ProtectionAgency, radon is the second most frequent cause of lungcancer, after cigarette smoking, causing 21,000 lung cancerdeaths per year in the United States.

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Guidelines One-Sample Tests The t-test Power Analysis Tests for Population Means

One-Sample Tests

To check the reliability of radon detector, a university placed 12 detectors in achamber having 105 picocuries of radon. (1 picocurie is 3.7× 10−2 decays per second.This is roughly the activity of 1 picogram of radium 226.)

The two-sided hypothesis

H0 : µ = 105 versus H1 : µ 6= 105,

where µ is the mean value of the radon detectors. In other words, we are checking tosee if the detector is biased either upward or downward.

91.9 97.8 111.4 122.3 105.4 95.0 103.8 99.6 96.6 119.3 104.8 101.7

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Guidelines One-Sample Tests The t-test Power Analysis Tests for Population Means

One-Sample Tests

Using R, we find for an α = 0.05 level significance test:

[1] 104.1333

[1] 9.397421

> qt(0.975,11)

[1] 2.200985

The t-statistic is

t =105− 104.1333

9.39742/√

12= −0.3195.

Thus, for a 5% significance test, |t| < 2.200985, the critical value and we fail to rejectH0.

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Guidelines One-Sample Tests The t-test Power Analysis Tests for Population Means

One-Sample Tests

R handles this procedure easily.

One Sample t-test

t = -0.3195, df = 11, p-value = 0.7554

alternative hypothesis: true mean is not equal to 105

Power analyses are based on the noncentral t-distribution. Recall that for Z1,Z2, . . .Zn,independent standard normal random variables with standard deviation sZ ,

T̃ =

√nZ̄ + a

sZ

has a t distribution with n − 1 degrees of freedom and non-centrality parameter a.

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Guidelines One-Sample Tests The t-test Power Analysis Tests for Population Means

Power AnalysisIf we reproduce the calculation we made in determining power for a one-sampletwo-sided z-test, then, in the denominator, we add and subtract the mean µ to obtain

X̄ − µs/√n

=(X̄ − µ) + (µ− µ0)

s/√n

If the Xi have common standard deviation σ, we standardize the variables, writing

Zi =Xi − µσ

.

Thus, sZ = s/σ is the standard deviation of the Zi and upon dividing each term by σ,

(X̄ − µ) + (µ− µ0)

s/√n

=

√n((X̄ − µ)/σ + (µ− µ0)/σ)

s/σ

=

√nZ̄ −

√n(µ0 − µ)/σ

sZ

which has t-distribution with non-centrality parameter a =√n(µ− µ0)/σ.

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Guidelines One-Sample Tests The t-test Power Analysis Tests for Population Means

Power Analysis

To plot the power function, π, we first enter the data.

103.8,99.6,96.6,119.3,104.8,101.7)

> mu0<-105

> mu<-seq(95,115,length=101)

and estimate the standard deviation from the sample.

> tstar<-qt(0.975,11)

> pi<-1-(pt(tstar,11,a)-pt(-tstar,11,a))

>plot(mu,pi,type="l",col="brown",lwd=2)

95 100 105 110 115

0.2

0.4

0.6

0.8

mu

pi

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Guidelines One-Sample Tests The t-test Power Analysis Tests for Population Means

Power Analysis

The power.t.test command considers five issues

• the sample size n,

• the difference between the null and a fixed value of the alternative delta,

• the standard deviation s,

• the significance level α, and

• the power 1− β.

We can use power.t.test to drop out any one of these five and use the remainingfour to determine the remaining value. For example, if we want to assure an 80%power against an alternative difference of 5 piconewtons,

The output shows that we need to make 30 measurements.

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Guidelines One-Sample Tests The t-test Power Analysis Tests for Population Means

Power Analysis

Exercise. Fill in the tables based on the radon detector data set.

1. Consider a significance level α = 0.05 and the standard deviation s from the data.Find the necessary number of observations.

2. Consider a significance level α = 0.05 and a difference delta of 5 piconewtons.Find the power.

power

delta 0.7 0.8 0.9

510

observations

s 10 20 30

5.009.40

3. Which two values in each table are most extreme? Explain why.

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Guidelines One-Sample Tests The t-test Power Analysis Tests for Population Means

Tests for Population Means

• Use the z-statistic when the standard deviations are known.

• Use the t-statistic when the standard deviations are computed from the data.

null hypothesis

t or z-procedure one-sided two-sided

single sample H0 : µ ≤ µ0 H0 : µ = µ0H0 : µ ≥ µ0

two samples H0 : µ1 ≤ µ2 H0 : µ1 = µ2H0 : µ1 ≥ µ2

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Guidelines One-Sample Tests The t-test Power Analysis Tests for Population Means

Tests for Population Means

The test statistic,

t =estimate− parameter

standard error.

t-procedure parameter estimate standard error degrees of freedom

one sample µ x̄ s√n

n − 1

two sample µ1 − µ2 x̄1 − x̄2

√s21n1

+s22n2

ν in W-S equation

pooled two sample µ1 − µ2 x̄1 − x̄2 sp

√1n1

+ 1n2

n1 + n2 − 2

For one-sample and two-sample z procedures, replace the values s with σ and s1 ands2 with σ1 and σ2, respectively. Use the normal distribution for these tests.

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Guidelines One-Sample Tests The t-test Power Analysis Tests for Population Means

Tests for Population MeansFor a two-sample t-test, the test statistic does not have a t-distribution. TheBehrens-Fisher problem is the problem of hypothesis testing concerning the differencebetween the means of two independent normally distributed populations when thevariances of the two populations are not assumed to be equal.

One approach solves this by approximating a t distribution with the effective degrees offreedom ν calculated using the Welch-Satterthwaite equation:

ν =(s21/n1 + s22/n2)2

(s21/n1)2/(n1 − 1) + (s22/n2)2/(n2 − 1)

We have thatmin{n1, n2} − 1 ≤ ν ≤ n1 + n + 2− 2.

The value ν is determined by approximating a linear combination of squares of normalusing a χ2

ν distribution.20 / 22

Guidelines One-Sample Tests The t-test Power Analysis Tests for Population Means

Mosquito Life SpanAnopheles mosquitoes are the carrier of par-asitic protozoans of the genus Plasmodium.The blood obtained from a bite from a fe-male mosquito is used as a source of proteinfor the production of eggs. In this way in-fracted mosquitoes transmit malaria to hu-mans.

We test to see if overstimulation of theinsulin signaling cascade in the midgutof transgenic mosquitoes reduces the µt ,the mean life span of these transgenicmosquitoes from that of the wildtype µwt .

H0 : µwt ≤ µt versus H1 : µwt > µt .

transgenic wildtype

010

2030

4050

0 10 20 30 40 50

0.0

0.2

0.4

0.6

0.8

1.0

fract

ion

livin

g lo

nger

0 10 20 30 40 50

0.0

0.2

0.4

0.6

0.8

1.0

fract

ion

livin

g lo

nger

wildtypetransgenic

Boxplot and survival function of lifespan in days

for transgenic and wildtype mosquitoes.

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Guidelines One-Sample Tests The t-test Power Analysis Tests for Population Means

Mosquito Life Span

R easily handles the analysis.

> t.test(transgenic,wildtype,alternative = c("less"))

Welch Two Sample t-test

data: transgenic and wildtype

t = -2.4106, df = 169.665, p-value = 0.008497

alternative hypothesis: true difference in means is less than 0

95 percent confidence interval:

-Inf -1.330591

sample estimates:

mean of x mean of y

16.54545 20.78409

NB. the number of degrees of freedom ν = 169.665.

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