3/11/00 252chisq

E. CHI-SQUARED AND RELATED TESTS.<= /b>

   &nbs= p;            These tests are generalizations of the one-sample and two-sample tests of proportions.    A te= st of Goodness of Fit is necessary when a single sample is to be divided into more than two categories. A Test of Homogeneity is needed when one wants to comp= are more than one sample. A test = of Independence is us= ed to see if two variables or categorizations are related, but is formally identi= cal to a test of homogeneity.

1. Tests of Homogeneity and Independence

Two possible null hypotheses apply here. The observed = data is indicated by _{the expecte=
d data
by _.}

The numbers are obviously identical in these two cases= . In each case the expected values

are done the same way. There are _{rows, _{columns and _{cells. Each=
cell
gets _{. For example, for
the upper left corner the expected value is}}}}

The formula for the chi-squared statistic is _{or _{. The first of these two formulas is shown below. For an
explanation of the equivalence of these two formulas, the reason why the
degrees of freedom are as given below, and to relate the chi-squared test t=
o a _{test of proportions, see 252chisqnot=
e.}}}

       =           =        =

10.0000    10   0.0000  &= nbsp; 0.0000     0.000= 00

8.0000    5&= nbsp; 3.0000     9.00= 00     1.12500

12.0000    15   -3.0000  &= nbsp; 9.0000     0.750= 00

13.3333    15   -1.6667  &= nbsp; 2.7778     0.208= 33

10.6667    10   0.6667  &= nbsp; 0.4445     0.041= 67

16.0000    15   1.0000  &= nbsp; 1.0000     0.062= 50

13.3333    15   -1.6667  &= nbsp; 2.7779     0.208= 34

10.6667    15   -4.3333  &= nbsp; 18.7775     1.76038

16.0000    10   6.0000  &= nbsp; 36.0000     2.25000

13.3333    10   3.3333  &= nbsp; 11.1109     0.83332

10.6667    10   0.6667  &= nbsp; 0.4445     0.041= 67

16.0000=     20   -4.0000     16.0000     1.00000

150.0000    150   0.0000                 8.28121

The degrees of freedom for this application are _{<=
!--[if gte vml 1]>

.}

The most common test is a one-tailed test on the groun= ds that the larger the discrepancy that occurs between _{and _{, the larger will be _{. If our significance level is 5%,
compare _{to _{Since our value of this sum is less than the table
chi-squared, do not reject the=
null
hypothesis.}}}}}

Note: Rule of thumb for _.

All values of _{should be a=
bove 5
and we generally combine cells to make this so. However a number _{is acceptab=
le in _{if i) Our
computed _{turns out t=
o be
less than _{or, (ii) The particular value of _{makes a very
small contribution to _{, relative t=
o the
value of the total.}}}}}}}

Note: Marascuilo Procedure.

The Marascuilo procedure says that, for 2 by c tests, = if (i) equality is rejected and

(ii) , where _{and _{represent 2 groups,
the chi - squared has _{degrees of
freedom and the standard devi=
ation
is _{, you can sa=
y that
you have a significant difference between _{and _{. This is
equivalent to using a confidence interval of}}}}}}

Example: = Pelosi and Sandifer give the data below for satisfaction with local phone service classified by type of company providing the service. 1) Is the proportion of people who rate the phone service as excellent independent of the type of company? 2) If it is not, test= for a difference in the proportion who rate their service as excellent against the best-rated provider. Remember that this is a test of equality of proportion= s. Use a confidence level of 95%.

Service 1 is a long distance company, Service 2 is a l= ocal phone company, Service 3 is a power company, Service 4 is CATV (cable) and Service 5 is Cellular. _{is thus the
proportion of long distance company customers that rate their service as
excellent.}

_{_{Not all ps eq=
ual.}}

Solution:= Set up the O table. To get the number = that rate service as excellent for long distance, note that _{. But this must be a whole number, so round it to 246. The
number that do not rate it as excellent is _{. This gives us our first column.}}

₌

Long Dist

Local Ph

Power

CATV

Cellular

Total

    = ₌

Excellent

264

444

131

215

198

1252

.2296=

Not

1394

1318

485

431

572

4200

.7704=

Sum

1658

1762

616

646

770

5452

1.0000

Proportion

Excellent

.1592

.2520

.2127

.3328

.2571

₌

.0000807

.0001070

.0002718

.0003437

.0002481

Note that in addition to computing the overall proport= ion of excellent and not excellent service (.2296 and .7704) , the ‘proportion excellent’ has been computed for each type of service as well as the variance _{used in the=
Marascuilo
confidence interval formula. If we apply the proportions in each row to the
column sums we get the following expected values.}

₌

Long Dist

Local Ph

Power<= /p>

CATV<= /p>

Cellul= ar

Total

   ₌

Excellent

380.6= 8

404.5= 6

141.43=

148.3= 2

176.7= 9

1252 .=

.2296=

Not

1277.32

1357.44

474.57=

497.6= 8

593.2= 1

4200 <= /p>

.7704=

sum

1658

1762

616

646

770

5452

1.0000

The chi-squared test follows.

   Row     _{=
_₌}

   1    380.68     264   116.677   13613.5   35.7612

   2   1277.32    1394 -116.677   13613.5   10.6578

   3    404.56     444   -39.445    1555.9    3.8459

   4   1357.44    1318    39.445    1555.9    1.1462

   5    141.43     131    10.434     108.9    0.7697

   6    474.57     485   -10.434     108.9    0.2294

   7    148.32     215   -66.678    4446.0   29.9755

   8    497.68     431    66.678    4446.0    8.9335

   9    176.79     198   -21.208     449.8    2.5441

10   593.21    572    21.208     449.8   0.7582

       5452= .00    5452     0.000        =      94.622

The degrees of freedom are _{and _{, so we reje=
ct the
null hypothesis and say that there is a difference between the proportions =
that
rate their service as excellent. Since the highest satisfaction was with CA=
TV
we compare the proportions with the proportion calculated for CATV using the
confidence interval formula above.}}

Long distance

Local Phone

Po= wer        =

Cellular

Notice that the absolute size of the error term is always smaller than the absolute size of the difference in proportions, so = that we can say that all of these differences are significant. Though I have not checked it, I doubt, if we compare all other proportions with the proportion saying cellular service is excellent we would get such strong results.<= /o:p>

2. Tests of Goodness of Fit

a. Uniform Distribution

Let us pool the data above, that is, treat it all as i= f it were one sample, and ask if it is uniformly distributed.

          =

50   50   0&= nbsp;     0       0

50   40   10   100       2

50<= /u>=    60   -10 &nbs= p; 100       2

150   1= 50   0&= nbsp;           &nbs= p; 4

Since there are 3 numbers here, there are 2 degrees of freedom. Since 4 is less than _{, we cannot =
reject
the null hypothesis. An easier way to do this is to compute _{. Remember}}



50   50   50=

50   40   32

50<= /u>=    60   72=

150   1= 50   154

    =         -150

= &nb= sp;    &n= bsp;      4

For a comb= ined Chi-square test of both uniformity and homogeneity see 252chisqx1

b. Poisson Distribution

Example:

I believe that there is almost a da= ily accident on my corner. To make this into a testable hypothesis, let us sat = that I believe that the distribution is Poisson with a parameter of 0.8 and that= I observe the numbers of accidents shown below over 200 days. For example the= re are 100 days with no accidents, 60 days with 1 etc.

           &nbs= p;

_{=
    &=
nbsp;           &nbs=
p; _{&=
nbsp;           &nbs=
p;       _{=
    &=
nbsp;           &nbs=
p; _{&=
nbsp;           &nbs=
p;            &=
nbsp;           &nbs=
p;            &=
nbsp;           &nbs=
p;            &=
nbsp;           &nbs=
p;            To get _{, I look up
frequencies}}}}}

0        &= nbsp;           10= 0        &= nbsp;           .4493            =       89.86      &n= bsp;            = ;            &n= bsp;            = ;            &n= bsp;            = ;            &n= bsp;            = ;           on the Poisson table and    <= span style=3D'mso-spacerun:yes'>

1        &= nbsp;           60=             &nb= sp;       .3595            =       71.90      &n= bsp;            = ;            &n= bsp;            = ;            &n= bsp;            = ;            &n= bsp;            = ;           multip= ly by _{, using the<=
span
style=3D'mso-tab-count:1'>        &=
nbsp;           &nbs=
p;}

2        &= nbsp;           30=             &nb= sp;       .1438            =       28.76      &n= bsp;            = ;            &n= bsp;            = ;            &n= bsp;            = ;            &n= bsp;            = ;           formula

3        &= nbsp;               6            =         .0383            =          7.66                =                        =             &nb= sp;            =             &nb= sp;            =             &nb= sp;   Unfortunately, I cannot

4        &= nbsp;               0            =         .0077            =          1.54(<5)            =       work with _{as it appea=
rs
here.      &n=
bsp;            =
;}

5        &= nbsp;               4            =         .0012            =          0.24(<5)            =       I must have each _{at least         &=
nbsp;           &nbs=
p;}

6        &= nbsp;               0            =         .0002            =       0.04(<5)            =       5. To fix the problem, I add

7+        &= nbsp;             0            =         .0000            =          0.00(<5)        &= nbsp;         the smallest cells together to   = ;            &n= bsp;            = ;

        &= nbsp;           &nbs= p; 200             &n= bsp;     1.0000        &= nbsp;         200.00      &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;         increase _{to 5 or mor=
e.}

    &= nbsp;           &nbs= p;     &= nbsp;           &nbs= p;           &= nbsp;           &nbs= p;        &= nbsp;                    =

0        &= nbsp;           10= 0        &= nbsp;           89= .86        &= nbsp;           11= 1.28        &= nbsp;                Sinc= e I did not estimate the mean

1        &= nbsp;           60=             &nb= sp;       71.90        &= nbsp;           50.07            =             of 0.8 from the data, I have 3

2        &= nbsp;           30=             &nb= sp;       28.76        &= nbsp;           31.29            =             degrees of freedom.

3+        &= nbsp;         10            = ;        9.48           &nb= sp;        10.56             =

        &= nbsp;           &nbs= p; 200        &= nbsp;           20= 0        &= nbsp;           &nbs= p; 203.20        &= nbsp;              _{so I do not=}

        &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;      200.00       &nb= sp;                rejec= t the null hypothesis.

        &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;          3.20            =

But what if my hypotheses are simply _{? Then I would have to estimate the =
mean
from the data. Looking at the=
_{and _{columns I
calculate _{. I would still use the Poisson distribution with a param=
eter
of 0.8 unless I had a computer
handy to compute it with a parameter of 0.79,
but my degrees of freedom are now 3 - 1 =3D 2,<=
/st1:PersonName>
because I used the data to estimate a parameter.}}}}

c. Normal Distribution

A comm= on way to set up a _{test of nor=
mality
is to group data starting at the mean and ending each group one-half of a standard deviation from the mean.<=
span
style=3D'mso-spacerun:yes'> One can proceed outward from the m=
ean
until four or five groups have been sectioned off in each direction. For example,
if our null hypothesis is that _{, we can sta=
rt at
100 and let the width of each group be one half of _{or 5. The groups would be 100-105, 105-110,
etc. going up, and 95-100, 90-95, etc. going down. Then for the high=
est
number in each interval, compu=
te _{.   For
example, for the interval 90-95 compute . Then use =
the
normal distribution to compute _.}}}}

. For example _{Then<=
st1:PersonName
w:st=3D"on">, to
find the frequency of the interval, subtract this _{from the for the pre=
vious
interval. An example of calculating _{this way is=
shown
below.}}}

        &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;           _{=
interval        &=
nbsp;    _{&=
nbsp;           &nbs=
p;           _{&=
nbsp;           &nbs=
p;      _{&=
nbsp;           &nbs=
p;}}}}

        &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;           _{-80   &nbs=
p;            &=
nbsp;           &nbs=
p;            &=
nbsp;           -2=
.0        &=
nbsp;           &nbs=
p;     .0228            =
         .0228        &=
nbsp;           &nbs=
p; 22.8}

        &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;           80-85            &= nbsp;                &= nbsp;           &nbs= p;           -1.5<= span style=3D'mso-tab-count:1'>        &= nbsp;           &nbs= p;     .0668            =          .0440        &= nbsp;           &nbs= p; 44.0

        &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;           85-90            &= nbsp;                &= nbsp;           &nbs= p;           -1.0<= span style=3D'mso-tab-count:1'>        &= nbsp;           &nbs= p;     .1587            =          .0919        &= nbsp;           &nbs= p; 91.9

        &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;           90-95            &= nbsp;                &= nbsp;           &nbs= p;           -0.5<= span style=3D'mso-tab-count:1'>        &= nbsp;           &nbs= p;     .3085            =          .1498        &= nbsp;           &nbs= p; 149.8

        &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;           95-100            =         &n= bsp;            = ;            &n= bsp;      0.0           &nb= sp;            = .5000            =          .1915        &= nbsp;           &nbs= p; 191.5

        &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;           100-105            = ;      0.5            &n= bsp;            .6915            =          .1915        &= nbsp;           &nbs= p; 191.5

        &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;           105-110            = ;      1.0            &n= bsp;            .8413            =          .1498        &= nbsp;           &nbs= p; 149.8

        &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;           110-115            = ;      1.5            &n= bsp;            .9332            =          .0919        &= nbsp;           &nbs= p; 91.9

        &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;           115-120            = ;      2.0            &n= bsp;            .9772            =          .0440        &= nbsp;           &nbs= p; 44.0

        &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;           120-_{&=
nbsp;          _{&=
nbsp;           &nbs=
p;        1.0000        &=
nbsp;            .0228        &=
nbsp;           &nbs=
p; 22.8}}

For smaller values of= _{we may find=
that
some numbers in _{are less th=
an 5, so that we have to combine some intervals. I=
n the
above example the degrees of freedom for _{are 10 - 1 =
=3D
9 if the mean and variance are
known. If they both had to be computed from data the degrees of freedom wou=
ld
be reduced by 2 to 7.}}}

&nb= sp;

3. Kolmogorov-Smirnov Test

a. Kolmogorov-Smirnov One-Samp= le Test

This is a more powerful test of goodness of fit than the Chi-Squared test. Unfortunately, it can only be used when the distribution in the null hypothesis is totally specified. For example, if we = wanted to do the test for Poisson(0.8) above, we would look up the cumulative distribution _{for
Poisson(0.8) and proceed as below. Note that this would not work if our
hypothesis was that the distribution was Poisson without the mean specified=
.}

        &= nbsp;           &nbs= p;             =     &= nbsp;           &nbs= p;     &= nbsp;           &nbs= p;     &= nbsp;              &= nbsp;               &= nbsp;

        &= nbsp;           &nbs= p;             = 0        &= nbsp;           10= 0        &= nbsp;        .50            &n= bsp; .50            = .4493            =    .0507

        &= nbsp;           &nbs= p;             = 1        &= nbsp;           60=             &nb= sp;    .30            &n= bsp; .80            = .8088            =    .0088

        &= nbsp;           &nbs= p;             = 2        &= nbsp;           30=             &nb= sp;    .15            &n= bsp; .95            = .9526            =    .0026

        &= nbsp;           &nbs= p;             = 3        &= nbsp;               6            =      .03            &n= bsp; .98            = .9909            =    .0109

        &= nbsp;           &nbs= p;             = 4        &= nbsp;               0            =      .00            &n= bsp; .98            = .9986            =    .0106

        &= nbsp;           &nbs= p;             = 5        &= nbsp;               4            =      .02            &n= bsp; 1.00        &= nbsp;   .9998            =    .0002

        &= nbsp;           &nbs= p;             = 6        &= nbsp;               0            =      .00            &n= bsp; 1.00        &= nbsp;   1.0000        &= nbsp;      .0000

        &= nbsp;           &nbs= p;             = 7+        &= nbsp;             0                =   .00            =    1.00        &= nbsp;   1.0000        &= nbsp;      .0000

        &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;           200        &= nbsp;        1.00        &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;            &= nbsp;

The maximum difference is _{, which must=
be
checked against the Kolmogorov-Smirnoff table for _{. According to the table,
for _{, the critic=
al
value is _{. Since _{is less than
.0962, accept the null hypothe=
sis.}}}}}

b. Lilliefors Test.

   &nbs= p;            Because the Kolmogorov-Smirnov Test is so limited in application, it proved advantageous to develop a special version of that test to use to test for a normal distribution when the mean and variance are unknown. Once a sample mean and variance are found, this test is identical to the K-S Test except for the use of a speci= al table.

Problem E9: Is the following data normal?

         420, 440, 445, 450, 460, 475, 480, 500, 520, 530

Solution: Assume _{_{The only pr=
actical
method is the Lilliefors method. Question: Why is Chi-squared impractical a=
nd
Kolmogorov-Smirnov impossible?}}

The numbers must be in order before we begin computi= ng cumulative probabilities! Checking the data we find that _{and _{. We compute _{. (This is really a _{.) _{is the cumu=
lative
distribution, gotten from the =
Normal table by adding or subtracting 0.5. _{comes from =
the
fact that there are 10 numbers, so
that each number is one-tenth of the distribution.}}}}}}

For and _{the critical value from the Lilliefors table is 0.258. Si=
nce
the largest deviation here is .1293,
we do not reject _.}

Remember that the Lilliefors method is a specialized version of the KS method used only in situations where you are testing for a Normal distribution and using a sample mean and standard deviation estimated from the data. The KS method can only be used in situations where the null hypothesis including parameters is specified in advance. A Chi-squared test= of goodness of fit is usually considered a large sample test, but can be adjus= ted for estimation of parameters.

_{<=
!--[if gte vml 1]>

=}

_=
=

_=
=

_=
=

_=
=

420

-1.45

0.1000

.0735

.0265

440

-0.89

0.2000

.1867

.0133

445

-0.75

0.3000

.2266

.0734

450

-0.61

0.4000

.2702

.1291

460

-0.33

0.5000

.3707

.1293

475

0.08

0.6000

.5319

.0681

480

0.22

0.7000

.5871

.1129

500

0.78

0.8000

.7823

.0177

520

1.34

0.9000

.9099

.0099

530

1.61

1.0000

.9463

.0537

©2002 Roger Even Bove

₌	Long Dist	Local Ph	Power	CATV	Cellular	Total	= ₌
Excellent	264	444	131	215	198	1252	.2296=
Not	1394	1318	485	431	572	4200	.7704=
Sum	1658	1762	616	646	770	5452	1.0000
Proportion Excellent	.1592	.2520	.2127	.3328	.2571
₌	.0000807	.0001070	.0002718	.0003437	.0002481

₌	Long Dist	Local Ph	Power<= /p>	CATV<= /p>	Cellul= ar	Total	₌
Excellent	380.6= 8	404.5= 6	141.43=	148.3= 2	176.7= 9	1252 .=	.2296=
Not	1277.32	1357.44	474.57=	497.6= 8	593.2= 1	4200 <= /p>	.7704=
sum	1658	1762	616	646	770	5452	1.0000

_{<= !--[if gte vml 1]> =}	_= =	_= =	_= =	_= =
420	-1.45	0.1000	.0735	.0265
440	-0.89	0.2000	.1867	.0133
445	-0.75	0.3000	.2266	.0734
450	-0.61	0.4000	.2702	.1291
460	-0.33	0.5000	.3707	.1293
475	0.08	0.6000	.5319	.0681
480	0.22	0.7000	.5871	.1129
500	0.78	0.8000	.7823	.0177
520	1.34	0.9000	.9099	.0099
530	1.61	1.0000	.9463	.0537

1. Tests of Homogeneity and Independence

Note: Rule of thumb for .

Note: Marascuilo Procedure.

2. Tests of Goodness of Fit

a. Uniform Distribution

b. Poisson Distribution

c. Normal Distribution

3. Kolmogorov-Smirnov Test

a. Kolmogorov-Smirnov One-Samp= le Test

b. Lilliefors Test.

©2002 Roger Even Bove

Note: Rule of thumb for _.