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Chi-squared (2

) (1.10.5) and F-tests (9.5.2) for thevariance of a normal distribution

2 tests for goodness of fit and indepdendence

(3.5.43.5.5)

Prof. Tesler

Math 283

Nov. 13, 2013

Prof. Tesler 2 andFtests Math 283 / Nov. 13, 2013 1 / 40

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Tests of means vs. tests of variances

Datax1, . . . ,xn, sample mean x, sample var.sX2

Datay1, . . . ,ym, sample mean y, sample var.sY2

Tests for mean

One-sample tests:

H0 :=0 vs.H1 :0

test statistic:

z= x0/

n

ort= x0s/

n

(df= n1)

Tests for variance

One-sample test:

H0 :2 =

0

2 vs.H1 :2

0

2

test statistic: chi-squared

2 = (n 1)s2/0

2 (df= n1)

Two-sample tests:

H0 :X

= Y

vs. H1 :X

Y

test statistic:

z= xy

X2

n +

Y

2

m

ort= xy

sp1

n+ 1

m

(df =n+m 2)

Two-sample test:

H0 :X

2 =Y

2 vs.H1 :X

2

Y

2

test statistic:

F=sY

2/sX

2

(withm 1andn 1 d.f.)


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Application: The fine print in the Zandt-tests

One-sample z-test, H0: =0 vs.H1: 0

This assumes that you know the value of2

, say2

=02

.A2 test could be used to verify that the data is consistent withH0 :

2 =02 instead ofH1 :

20

2.

Two-sample z-test, H0: X= Yvs. H1: XY

This assumes that you know the values of X2

andY2

.Separate2 tests forX

2 andY

2 could be performed to verifyconsistency with the assumed values.

Two-sample t-test, H0: X= Yvs. H1: XY

This assumesX2

=Y2

(but doesnt assume that this commonvalue is known to you).AnF-test could be used to verify that the data is consistent withH0 :X

2 =Y

2 instead ofH1 :X2

Y2.

If the variances are unequal, Welchs t-test can be used instead ofthe regular two-samplet-test (Ewens & Grant pp. 127128).


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The2 (Chi-squared) distribution

Used for confidence intervals and hypothesis tests on the

unknown parameter2 of the normal distribution, based on the

test statistics2

(sample variance).

It has the same degrees of freedom as for the tdistribution.

Point these out on the graphs:The chi-squared distribution withkdegrees of freedom has

Range [0,)Mean =k

Mode 2 =k 2(fork 2, the pdf is maximum for2 =k 2)

2 = 1(fork= 1)

Median

k(1 2

9k)3

Between kandk 23 .Asymptotically decreases k 2

3 ask.

Variance 2 = 2k

PDF x(k/2)1ex/2

2k/2(k/2) : distrib. with shaper= k

2, rate= 1

2

Unlike zandt, the pdf for2

is NOT symmetric.Prof. Tesler 2 andFtests Math 283 / Nov. 13, 2013 4 / 40

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The graphs for 1and 2degrees of freedom are decreasing:

0 1 2 3 40

1

2

3

4

5

!2

1

pd

f

0 1 2 3 4 5 60

0.1

0.2

0.3

0.4

0.5

!2

2

pdf

mean @ = 1 mean @ = 2

mode @ 2

= 0 mode @

2= 0

median @ 2 = chi2inv(.5,1) = qchisq(.5,1) = 0.4549 median @ 2 =chi2inv(.5,2) =qchisq(.5,2) = 1.3863

The rest are hump shaped and skewed to the right:

0 2 4 6 80

0.05

0.1

0.15

0.2

0.25

!2

3

pdf

0 2 4 6 8 10 12 14 160

0.02

0.04

0.06

0.08

0.1

0.12

!2

8

pdf

mean @ = 3 mean @ = 8

mode @ 2

=

1 mode @ 2

=

6median @ 2 =chi2inv(.5,3) =qchisq(.5,3) = 2.3660 median @ 2 =chi2inv(.5,8) =qchisq(.5,8) = 7.3441


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2 (Chi-squared) distribution Cutoffs

0 2 4 6 8 10 12

0.

00

0.

05

0.

10

0.

15

!!2

5

pdf

!!2"",df

left!sided critical region

0 5 10 15

0.

00

0.

05

0.

10

0.

15

!!2

5

pdf

!!2

0.025,5 == 0.8312116

!!20.975,5 == 12.83250

2!sided acceptance region: df=5, "" == 0.05

Define2,dfas the number where the cdf (area leftof it) is: P(2

df 2

,df) =

Different notation thanz andt,df (areaon right) since pdf isnt symmetric.

Matlab R

20.025,5 = chi2inv(.025,5) = qchisq(.025,5) = 0.83122

0.975,5 = chi2inv(.975,5) = qchisq(.975,5) = 12.8325

chi2cdf(0.8312,5) = pchisq(0.8312,5) = 0.025

chi2cdf(12.8325,5)= pchisq(12.8325,5)= 0.975

chi2pdf(0.8312,5) = dchisq(0.8312,5) = 0.0665

chi2pdf(12.8325,5)= dchisq(12.8325,5)= 0

.0100


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Two-sided cutoff

0 5 10 15

0.

00

0.

05

0.

10

0.

15

!!25

pdf

!!2

0.025,5 == 0.8312116

!!20.975,5 == 12.83250

2!sided acceptance region: df=5, "" == 0.05

The mean, median, and mode are different, so it may not be

obvious what values of 2 are more consistent with

the nullH0: 2

= 10000vs. the alternative2

10000.Closer to the median of 2 is "more consistent" with H0.

For 2-sided hypothesis tests or confidence intervals with = 5%,

we still put 95%of the area in the middle and 2.5%at each end,

but the lower and upper cutoffs are determined separately instead

of each other.Prof. Tesler 2 andFtests Math 283 / Nov. 13, 2013 7 / 40

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Two-sided hypothesis test for varianceTest H0 :

2 = 10000 vs. H1: 2 10000 at sig. level = .05

(In general, replace 10000 by02; here,0 = 100)

Decision procedure

1 Get a samplex1, . . . ,xn.

650, 510, 470, 570, 410, 370 with n= 6

2 Calculate m= x1++xn

n

and s2 = 1

n1

ni=1(xi m)

2.

m= 496.67, s2 = 10666.67, s= 103.28

3 Calculate the test-statistic2 = (n1)s2

0 2 =n

i=1(xim)

2

0 2

2 = (n1)s2

0 2 =

(61)(10666.67)

10000 = 5.33

4 AcceptH0 if2

is between2

/2,n1 and2

1/2,n1.RejectH0 otherwise.

2.025,5 =.8312, 2.975,5 = 12.8325

Since2 = 5.33is between these, we accept H0.

(Or, there is insufficient evidence to reject2 = 10000.)


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P-value

We can do the same test with a P-value:

P(25 5.33) = 0.6231is the area left of 5.33 on the2 curve with

5 degrees of freedom.

Matlab: chi2cdf(5.33,5) R: pchisq(5.33,5)

Values at least as extreme as this (compared to the median) are

those at the 62.31th percentile or higher, OR at the 37.69th

percentile or lower:

P= 2 min(0.6231, 1 0.6231) = 2(0.3769) = 0.7539

P> (0.75> 0.05) so acceptH0.


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Two-sided 95%confidence interval for the variance

Continue with data 650, 510, 470, 570, 410, 370

which has n= 6, m= 496.67, s2 = 10666.67, s= 103.28.

Get bounds on2 in terms ofs2 for the two-sided test:

0.95 = P(2.025,5 < 2 < 2.975,5)

= P(0.8312< 2 < 12.8325)

= P

0.8312< (61)S2

2 < 12.8325

= P(61)S2

0.8312 > 2 >

(61)S2

12.8325

A two-sided 95%confidence interval for the variance 2 is(

6

1)S

2

12.8325 , (6

1)S

2

0.8312

= (4156.11, 64164.26)

A two-sided 95%confidence interval for is(61)S2

12.8325,

(61)S2

0.8312

= (64.47, 253.31)


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Properties of Chi-squared distribution

1 Definition of Chi-squared distribution:

LetZ1, . . . ,Zkbe independent standard normal variables.Let2k=Z1

2 + +Zk2.The pdf of the random variable2k is the chi-squared distribution

withkdegrees of freedom.

2 Pooling property: IfUandVare independent2 random

variables withqandrdegrees of freedom respectively, thenU+V

is a2 random variable withq+rdegrees of freedom.

3

Sample variance: PickX1, . . . ,X

n from a normal distributionN(, 2). It turns out thatn

i=1

(XiX)2

2 =

SS

2 =

(n 1)S2

2

has a2 distribution withdf =n 1, so we test on 2 = (n1)s2

02 .


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Fdistribution

LetUandVbe independent

2

random variables withqandrdegreesof freedom, respectively. Then the random variable

F=Fq,r= U/q

V/r

is called the Fdistribution withqandrdegrees of freedom.

Range [0,)Mean r

r2 ifr> 2

Mode r(q2)

q(r+2) ifq> 2

Median 1 ifq=r

> 1ifq>r

< 1ifq 4

PDF messy


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Fdistribution

0 1 2 30

0.5

1

1.5

2

2.5

0 1 2 30

0.2

0.4

0.6

0.8

1

0 1 2 30

0.2

0.4

0.6

0.8

1

0 1 2 30

0.5

1

1.5

2

F1,1

F1,2

F1,5F

1,10

F1,30

F1,100

F1,!

F3,1

F3,2

F3,5F

3,10

F3,30

F3,100

F3,!

F10,1

F10,2

F10,5

F10,10

F10,30

F10,100

F10,!

F30,1

F30,2

F30,5

F30,10

F30,30

F30,100

F30,!


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FdistributionDefineF,q,ras the number where P(F F,q,r) =(left-hand area)

! " # $ % & ' ( )!

!*"

!*#

!*$

!*%

!*&

!*'

+!*!#&,(,&

-!*"). +

!*.(&,(,&-'*)&$

#!/0121 4552674852 92:0;8

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Two-sided hypothesis test for F

Test at significance level = 5%:

H0 :X2 =Y2 vs. H1 :X2 Y2

Data for X: 650, 510, 470, 570, 410, 370

x= 496.67, sX

2 = 10666.67, sX

= 103.28, df = 6 1 = 5

Data for Y: 510, 420, 520, 360, 470, 530, 550, 490

y= 481.25, sY

2 = 4012.5, sY

= 63.3443, df = 8 1= 7

Test statistic: F=F7,5 =sY2/sX2 = 4012.510666.67 = 0.3762.

Since our test statistic 0.3762lies between the cutoffs

F.025,7,5 = 0.1892and F.975,7,5 = 6.8531, we acceptH0 / reject H1.


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P-values

The CDF is P(F7,5 0.3762) = 0.1174

Matlab: fcdf(.3762,7,5)

R: pf(.3762,7,5)

To make it two sided,

P= 2 min(0.1174, 1 0.1174) = 2(0.1174) = 0.2348

SinceP> (0.2348> 0.05), we accept the null hypothesis.


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Fstatistic to compare variances (two-sample data)Theoretical setup

First sample: x1, . . . ,xn

V=(n 1)s

X2

X

2 =

ni=1

(xi x)2

X

2 withn 1 d.f.

Second sample: y1, . . . ,ym

U=

(m 1)sY

2

Y

2 =

m

j=1

(yiy)2

Y

2 withm 1d.f.

Assuming the null hypothesisX

2 =Y

2, the variances cancel:

F=

U/(m 1)

V/(n 1) =

sY

2

sX2 =m

i=1(yiy)2

/(m 1)

nj=1(xix)

2

/(n 1)

withm 1andn 1 degrees of freedom.

For a null hypothesisX

2 =CY

2 whereC> 0is constant, use

F=CsY2

/sX2

instead.Prof. Tesler 2 andFtests Math 283 / Nov. 13, 2013 17 / 40

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k-sample experiments

ANOVA(Analysis of Variance) is a procedure to compare the

meansofk-sample data (analagous to two-sample data, but for k

independent sets of data).

It involves theFdistribution with a formula for Fthat takes into

account allksamples instead of just two samples.


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2 tests for goodness of fit and

independence (3.5.43.5.5)


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Multinomial test

Consider ak-sided die with faces 1, 2, . . . , k.

We want to simultaneously test that the probabilities p1,p2, . . . ,pkof rolling 1, 2, . . . , kare specified values.

To test if a 6-sided die is fair,

H0: (p1, . . . ,p6) = (1/6, . . . , 1/6)

H1: At least one pi 1/6

Decision rule is based counting # 1s, 2s, etc. onnindependent

rolls of the die.

For the fair coin problem, the exact distribution was binomial, andwe approximated it with a normal distribution.

For this problem, the exact distribution is multinomial.

We will combine the separate counts of 1, 2, . . .into a single test

statistic whose distribution is approximately a 2 distribution.Prof. Tesler 2 andFtests Math 283 / Nov. 13, 2013 20 / 40

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Goodness of fit tests for Mendels experiments

In Mendels pea plant experiments,

yellow seeds (Y) are dominant and green (y) recessive;round seeds (R) are dominant and wrinkled (r) are recessive.

Consider the phenotypes of the offspring in a dihybrid cross

YyRr YyRr:

Expected ObservedType fraction number

yellow & round 9/16 315yellow & wrinkled 3/16 101green & round 3/16 108

green & wrinkled 1/16 32Total: n=556

Hypothesis test:

H0: (p1,p2,p3,p4) = ( 9

16, 3

16, 3

16, 1

16)

H1: At least one pi disagreesProf. Tesler 2 andFtests Math 283 / Nov. 13, 2013 21 / 40

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Does the data fit the expected distribution?

Expected Observed

Type fraction numberyellow & round 9/16 315yellow & wrinkled 3/16 101green & round 3/16 108green & wrinkled 1/16 32

Total: n=556

The observed number of yellow & round plants is O= 315.

(Dont confuse the letter Owith the number 0.)

The expected number isE= (9/16) 556= 312.75.

The goodness of fit test requires that we convert all the expected

proportions into expected numbers.


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Goodness of fit test

Observed number Expected number

Type O E OE (OE)2/E

yellow & round 315 (9/16)556= 312.75 2.25 0.0161871yellow & wrinkled 101 (3/16)556= 104.25 3.25 0.1013189green & round 108 (3/16)556= 104.25 3.75 0.1348921green & wrinkled 32 (1/16)556= 34.75 2.75 0.2176259

Total 556 556 0 0.4700240

k= 4categories givek 1 = 3degrees of freedom.

(TheOandEcolumns both total 556, so the OEcolumn totals

0; thus, any 3 of the(OE)s dictate the fourth.)

The test statistic is the total of the last column, 23

= 0.4700240.

The general formula is2k1 =k

i=1

(OiEi)2

Ei.

Warning: Technically, that formula only has an approximate

chi-squared distribution. When E 5in all categories, the

approximation is pretty good.Prof. Tesler 2 andFtests Math 283 / Nov. 13, 2013 23 / 40

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Goodness of fit testSmaller values of2 indicate better agreement between theOand

Evalues (so supportH0 better). Larger values supportH1 better.

Its a one-sided test.

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0 5 10 15

0.

00

0.

05

0.

10

0.

15

0.

20

0.

25

Supports H0better

Supports H1better

Observed !2

!2

pdf

TheP-value is the probability, under H0, of a test statistic that

supports H1 as well as or better than the observed value:

P= P(23 0.4700240) =.9254259

Matlab:1-chi2cdf(.4700240,3)

R:1-pchisq(.4700240,3)Prof. Tesler 2 andFtests Math 283 / Nov. 13, 2013 24 / 40

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Goodness of fit test

P=.9254259is not too extreme.

It means that if H0 is true and the experiment is repeated a lot,

about 7.5%of the time, a 23

value supportingH0 better (lower

values of23

) will be obtained, and about 92.5%of the time, values

supporting H1 better (higher values of 2

3) will be obtained.


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Ronald Fisher (18901962)

He made important contributions to both statistics and genetics.

Connection: he invented statistical methods while working on

genetics problems.

Our way of using the normal, Student t,2, andFdistributions in

the same framework, plus ANOVA, is due to him.

In genetics, he reconciled continuous variations (heights andweights) with Mendelian genetics (discrete traits), and developed

much of population genetics.


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Did Mendel fudge his data?

Forindependent experiments,the values of2 may be pooled by

adding the2

values and adding the degrees of freedom.

Fisher pooled the data from Mendels experiments and got

2 = 41.6056with 84 degrees of freedom.

Assuming Mendels laws are true, how often would we get 284

supporting H0/H1 better than this?

SupportH0 better:

P(284 41.6056) = 0.00002873

SupportH1 better:

P-valueP=P(2

84 41.6056) = 1 0.00002873=.99997127.

So if Mendels laws hold and 1 million researchers independently

conducted the same experiments as Mendel, about 29 of them

would get data with as little or even less variation than Mendel had.


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0 50 100 150

0.

00

0

0.

010

0.

020

0.

030

Supports H0better

Prob. =0.00002873

Supports H1better

Prob. =0.99997127

!84

2=41.6056

!84

2

pd

f


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Based on this and similar tests, Fisher believed that something was

fishy with Mendels data:

The values are too good in the sense that they are too close to

what was expected.

At the same time, they are bad in the sense that there is too little

random variation.

Some people have accused Mendel of faking data.

Others speculate that he only reported his best data.

Other people defend Mendel by speculating on biologicalexplanations for why his results would be better than expected.

All pro and con arguments have later been rebutted by someone

else.


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Tests of independence (contingency tables)

A study in 1899 examined 6800 German men to see if hair color and

eye color are related.

Observed countsO: Hair color

Brown Black Fair Red Total

Eye Brown 438 288 115 16 857

Color Gray/Green 1387 746 946 53 3132

Blue 807 189 1768 47 2811Total 2632 1223 2829 116 6800

Hypothesis test (at= 0.05)

H0: eye color and hair color are independent, vs.

H1: eye color and hair color are correlated

Meaning of independence

For all eye colors xand all hair colors y:

P(eye color=xand hair color=y) =P(eye color=x)

P(hair color=y)


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ComputingEtable


H0: eye color and hair color are independent, vs.H1: eye color and hair color are correlated

The fraction of people with red hair is 116/6800.

The fraction with blue eyes is 2811/6800.

Use these as point estimates: P(hair color=red) 116/6800andP(eye color=blue) 2811/6800.Under the null hypothesis, the fraction with red hair and blue eyes

would be (116 2811)/68002.The expected number of people with red hair and blue eyes is

6800(116 2811)/68002 = (116 2811)/6800= 47.95.(Row total times column total divided by grand total.)

ComputeEthis way for all combinations of hair and eye color.

As long asE 5in every cell (here it is) and the data is normally

distributed (an assumption), the 2

test is valid.Prof. Tesler 2 andFtests Math 283 / Nov. 13, 2013 31 / 40

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ComputingEandOEtables

Expected countsE: Hair color

Brown Black Fair Red

Eye Brown 331.71 154.13 356.54 14.62

Color Gray/Green 1212.27 563.30 1303.00 53.43

Blue 1088.02 505.57 1169.46 47.95

In each position, computeOE.

For red hair and blue eyes, this is OE= 47 47.95= .95:

OE: Hair color

Brown Black Fair Red

Eye Brown 106.29 133.87 241.54 1.38

Color Gray/Green 174.73 182.70 357.00 0.43Blue 281.02 316.57 598.54 0.95

Note all the row and column sums in the OEtable are 0, so if we hid

the last row and column, we could deduce what they are. Thus, this

3 4table has(3 1)(4 1) = 6degrees of freedom.Prof. Tesler 2 andFtests Math 283 / Nov. 13, 2013 32 / 40

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Computing test statistic2

Compute(OE)2/Ein each position.

For red hair and blue eyes, this is (.95)2

/47.95= 0.0189.(You could go directly to this computation after the Ecomputation,

without doingOEfirst.)

(OE)2/E: Hair color

Brown Black Fair RedEye Brown 34.0590 116.2632 163.6301 0.1304

Color Gray/Green 25.1852 59.2571 97.8139 0.0034

Blue 72.5845 198.2220 306.3398 0.0189

Add all twelve of these to get

2 = 34.0590+ +0.0189= 1073.5076

There are 6 degrees of freedom, so 26

= 1073.5076.


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Performing the test of independence

2 would be 0 if the traits were truly independent.

Smaller values supportH0 better (traits independent).

Larger values supportH1 better (traits correlated).Its a one-sided test.

At the 0.05level of significance, we reject H0 if

26 2

0.95,6 = 12.5916

Indeed, 1073.5076> 12.5916so we reject H0 and conclude thathair color and eye color are linked in this data.

This doesnt prove that a particular hair color causes one to have

a particular eye color, or vice-versa; it just says theres a

correlation in this data.Using P-values: P=P(2

6 1073.5076) 1.1 10228

soP = 0.05and we rejectH0.Matlab: cant compute this (givesP= 0).

R: pchisq(1073.5076,6,lower.tail=FALSE)


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Performing the test of independence

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0 200 400 600 800 1000 1200 1400

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Supports H0better

Prob. = 1 !"

Supports H1better

Prob. = "= 1.1e!228

!6

2= 1073.508

!6

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pdf


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Mendels pea plants revisited: Are lociYandRlinked?

We will use the same data as in the goodness-of-fit test but for a

different purpose. Consider the phenotypes of the offspring in adihybrid crossYyRr YyRr:

Observed countsO: Seed Shape

Round (R) Wrinkled (r) Total

Seed Yellow (Y) 315 101 416

Color Green (y) 108 32 140

Total 423 133 556


H0: Seed color and seed shape are independent, vs.

H1: Seed color and seed shape are correlated


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Seed Color Seed ShapeRound (R) Wrinkled (r) Total

O: Yellow (Y) 315 101 416(Observed #) Green (y) 108 32 140

Total 423 133 556

E: Yellow (Y) 316.4892 99.5108 416

(Expected #) Green (y) 106.5108 33.4892 140Total 423 133 556

OE: Yellow (Y) -1.4892 1.4982 0(Deviation) Green (y) 1.4892 -1.4892 0

Total 0 0 0

(OE)2/E: Yellow (Y) 0.0070 0.0223 0.0293(2 contrib.) Green (y) 0.0208 0.0662 0.0870

Total 0.0278 0.0885 0.1163


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Using 2 as the test statistic:

df = (2 1)(2 1) = 1

21

=.0070+.0223+.0208+.0662= 0.1163

cutoff: 20.95,1 = chi2inv(.95,1)= qchisq(.95,1)= 3.8415

0.1163< 3.8415so its not significant

Using P-values:

P=P(21

> 0.1163) = 1-chi2cdf(0.1163,1)= 0.7331

1-pchisq(0.1163,1)

P> 0.05so AcceptH0 (genes not linked)


C i f h

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Comparison of the two tests

At fertilization, if genesRandYare not linked, then in an

RrYyRrYycross, the expected proportions areRY:Ry : rY: ry= 1 : 1 : 1 : 1.

If linked, it would be different.

Some genotypes may not survive to the points at which the

phenotype counts are made; e.g., hypothetically, 40%of

individuals withRrmight not be born, might die before reproducing

(affecting multigenerational experiments), etc.This would change the ratio of

RR :Rr: rr from 1 : 2 : 1 to 1 : 1.2 : 1= 5 : 6 : 5,

and round:wrinkled from 3 : 1to 2.2 : 1= 11 : 5.


C i f h

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Comparison of the two tests

The goodness-of-fit test assumed all genotypes are equally viable.

Whether the genes are linked or not should be a separate matter.

If you know the yellow:green and round:wrinkled viability ratios,you can use the goodness-of-fit test on 4 phenotypes with 3

degrees of freedom by adjusting the proportions.

If you dont know these viability ratios, you can estimate the ratios

from data via contingency tables, at the cost of dropping to 1degree of freedom.

283 chi2 f13-handout

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