Diving into the Deep:Advanced Concepts

Module 9

Violation of Distributional Assumptions

• What distribution do the following charts assume?– i-Chart– Xbar-Range– Xbar-Sigma– p Chart– np Chart– u Chart– c Chart

Violation of Distributional Assumptions

• What distribution do the following charts assume?– i-Chart—near normal– Xbar-Range—near normal– Xbar-Sigma—near normal– p Chart--binomial– np Chart--binomial– u Chart--Poisson– c Chart--Poisson

Making your distribution normal:Trial and Error Steps

• Plot original data in histogram and on a probability plot of some kind.

• If skewed, try lower power (i.e., take the square root of each value).

• Make probability plot on transformed data.– If looks better, take even lower powers.– If looks worse, take higher powers.

• Continue until probability plot looks reasonably straight.

Ladder ofPowers

Cubic

Square

Identity

SQRT

Log

1/SQRT

1/X

1/Square

1/Cubic

This is yourraw data

Lower Powers

Higher Powers

Try thesetransformationsuntil you geta distributionthat looks likeit comes froma normaldistribution.

Average monthly number of preventable hospitalizations due to chronic disease in one IPC site, 2004-2005

02

46

8F

requ

ency

0 10 20 30 40Hospitalizations

of M

onth

s

Histograms of different power transformations.

01.0

e-05

2.0

e-05

3.0

e-05

4.0

e-05

5.0

e-05

0 20000 40000 60000 80000

cubic

05.0

e-04.0

01.0

015.

002

0 500 1000 1500 2000

square

0.0

2.0

4.0

6

0 10 20 30 40

identity0

.1.2

.3

2 3 4 5 6 7

sqrt

0.2

.4.6

1 2 3 4

log

02

46

-.6 -.5 -.4 -.3 -.2 -.1

1/sqrt

02

46

810

-.3 -.2 -.1 0

inverse

010

2030

40

-.1 -.05 0

1/square

050

100

150

-.04 -.03 -.02 -.01 0

1/cubic

De

nsity

HospitalizationsHistograms by transformation

Quantiles of preventable hospitalizations against quantiles of the normal distribution:

Dots are right on the line for normally distributed data

513

.540

010

2030

40H

osp

italiz

atio

ns

16.7917 34.8622-1.27882

0 10 20 30 40Inverse Normal

Grid lines are 5, 10, 25, 50, 75, 90, and 95 percentiles

Quantile Normal Probability Plots for Different Transformations

125

2470

.56400

0

-200

0002000

040

000

6000

080

000

11962.79 47721.51-23795.93

-20000 0 200004000060000

cubic

25182.

516

00

-5000

500

100015

002000

397.625 1246.94-451.6897

-500 0 500 1000 1500

square

513.5

40

0102

0304

0

16.79167 34.86215-1.278818

0 10 20 30 40

identity

2.2

3606

83.

673

604

6.3

2455

5

23

45

67

3.904662 5.9933441.815981

2 3 4 5 6

sqrt

1.6

0943

82.

602

003

3.6

8887

9

12

34

2.622609 3.7059711.539247

1.5 2 2.5 3 3.5 4

log

-.4

4721

36-.

272

3057

-.1

5811

39

-.6-.

5-.4-.

3-.2-.

1

-.2841066 -.122298-.4459152

-.5 -.4 -.3 -.2 -.1

1/sqrt-.

2-.

074

1758

-.0

25

-.3-.

2-.10

.1

-.0899905 .0218869-.2018679

-.2 -.15 -.1 -.05 0 .05

inverse

-.0

4-.

005

5096

-.0

0062

5

-.1-.

050

.05

-.0125318 .0251665-.0502301

-.06 -.04 -.02 0 .02

1/square

-.0

08-.

000

4098

-.0

0001

56

-.0

4-.0

3-.0

2-.0

10.01

-.0025449 .0098794-.0149693

-.015 -.01 -.005 0 .005 .01

1/cubic

HospitalizationsQuantile-Normal plots by transformationGrid lines are 5,10,25,50,75,90, and 95 percentiles

Which transformation is the closest to normally distributed?

Quantile Normal Probability Plots for Different Transformations

125

2470

.56400

0

-200

0002000

040

000

6000

080

000

11962.79 47721.51-23795.93

-20000 0 200004000060000

cubic

25182.

516

00

-5000

500

100015

002000

397.625 1246.94-451.6897

-500 0 500 1000 1500

square

513.5

40

0102

0304

0

16.79167 34.86215-1.278818

0 10 20 30 40

identity

2.2

3606

83.

673

604

6.3

2455

5

23

45

67

3.904662 5.9933441.815981

2 3 4 5 6

sqrt

1.6

0943

82.

602

003

3.6

8887

9

12

34

2.622609 3.7059711.539247

1.5 2 2.5 3 3.5 4

log

-.4

4721

36-.

272

3057

-.1

5811

39

-.6-.

5-.4-.

3-.2-.

1

-.2841066 -.122298-.4459152

-.5 -.4 -.3 -.2 -.1

1/sqrt-.

2-.

074

1758

-.0

25

-.3-.

2-.10

.1

-.0899905 .0218869-.2018679

-.2 -.15 -.1 -.05 0 .05

inverse

-.0

4-.

005

5096

-.0

0062

5

-.1-.

050

.05

-.0125318 .0251665-.0502301

-.06 -.04 -.02 0 .02

1/square

-.0

08-.

000

4098

-.0

0001

56

-.0

4-.0

3-.0

2-.0

10.01

-.0025449 .0098794-.0149693

-.015 -.01 -.005 0 .005 .01

1/cubic

HospitalizationsQuantile-Normal plots by transformationGrid lines are 5,10,25,50,75,90, and 95 percentiles

Probability of the observed distribution under the hypothesis of a normal distribution

ladder Hospitalizations

Transformation chi2(2) P(chi2)

cubic 16.36 0.000square 12.38 0.002raw 6.17 0.046square-root 2.01 0.366log 0.29 0.864reciprocal root 8.04 0.018reciprocal 18.25 0.000reciprocal square 31.62 0.000reciprocal cubic 36.56 0.000

The approximateprobability that theoriginal data arenormal.

The approximateprobability that the log of each datapoint is normallydistributed.

Which one would you use?

In STATA: Simply typein this command!

Results

How do you use this transformation?

• Take the log of each value.

• Calculate centerline and control limits for the log.

• Exponentiate the centerline and control limits.

• Plot these values.

i-Chart assuming normal distribution of raw data.

i-Chart of the log of the number of hospitalizations.

i-Chart of No. of hospitalizations with control limits derived from the log of hospitalizations.

Note: UCL for log of hospitalizations = 4.81, UCL for this graph=e4.81 = 122.74

But what about theCentral

LimitTheorem?

How do I know if I can use the Poisson distribution?C-Chart and U-Chart?

Practical Way…..

• If there is a large “n” and very small “p.”

• If the mean ≈ variance– Calculate variance using formula for

estimation of population variance with a sample (sum of squared deviations divided by n-1) and compare to mean.

• Then you probably have a Poisson distributed variable.

How do I know if I can use the Poisson distribution?C-Chart and U-Chart?

• You can get technical if you need to.– It is theoretically possible to compare the

distribution of units of area of opportunity (i.e., bed-days) over the range of possible events (0,1,2,3….50) to the distribution predicted by the mathematical formulae for the Poisson distribution.

• Save that for another course, please.

When can I assume the binomial distribution?

• If you have a situation where each observation can be classified as yes or no, 0 or 1, etc.

• If average p is not close to 0 or 1.

• If n*p*(1-p) >/= 5– THEN you can probably use the binomial, i.e.,

p-chart, np-chart.

What about sample size?

• Need 25 subgroups, more or less

• For p and np charts, n >/= 4/pBar

• For u and c charts, cBar >/= 4

• See Tables by Benneyan.

Extra-binomial & Extra-Poisson Dispersion

• Can occur when subgroup sizes are too large.

• Your observed points may be spread out more widely than predicted by the binomial or Poisson distribution.

Is this data over-dispersed?

How do we solve this?

• Use smaller increments of time.

• Check with subject-matter experts to be sure that there are no special causes.

• Then use the p’ method available in CHARTrunner 3.6.– Adjusts the control limits by combining within-

subgroup variation with between sub-group variance.

Notice the wider control limits with the p’ chart.

This method is available in CHARTrunner 3.6

• p’

• np’

• u’

• c’

Exercise E9

Enjoy

What did we learn from the exercise?

• Time intervals or subgroups can be “roped together”, i.e., not independent.

• This restricts their freedom to take on all possible values at any given time.

• This reduces the real sample size.• Control limits are then too narrow.

• Question: Are monthly measures on the same 50 patients for percent screened for cancer over the last year roped together? Autocorrelated?

What can we do?

• Separate the time between subgroups, e.g., sample once every three months.

• Change what you are measuring.• Use advanced charts to adjust for

autocorrelation (limited).• Switch to time-series analysis and model the

autocorrelation. (Usually need help here.)• Don’t worry unless autocorrelation is really high.• Use a run chart without run chart rules.

Hazards of Pooling Unlike Streams of Data

• Can we combine – Males with females? – Data from the private sector with the public?– Hospital A with Hospital B?

• Confusing the effect of time with the effect of another variable is the hazard.

• Confusion=Confounding

Example30

Day

Mor

talit

y fo

llow

ing

CA

BG

Hospital A

Hospital B.5

1

n=500

n=5250

n=10000

N=5250

Example: Pooled Hospitals A and B30

Day

Mor

talit

y fo

llow

ing

CA

BG

Hospitals A & B

.5

1

n=10500

n=10500

How is this possible?

Pooled Results

• Confused the effect of time with the effect of changing patient load of hospital A vs B.

Solutions

• Don’t combine unless there is no special cause between subgroups.

• Use indirect standardization.– See Hart and Hart, 2002, Appendix 2.

P-Chart comparing reporting sites on 1 executive order.

What is wrong with this comparison?

Needs Risk Adjustment

• When comparing outcomes sensitive to the severity of the patient’s condition, and

• When comparing across sites.• Risk adjustment not as important when

comparing the same patient population over time.

• Risk adjustment not as important for process measures that should be done regardless of severity.

Approaches

• Simple: Adjust for Age and Sex

• Complex: Age, Sex, Comorbidity, Other.– Requires many variables and proprietary or

government software.

Read about risk-adjusted control charts.

Converted into a system withits own software.

Other Risk Adjustment Programs

• CMS

• JCAHCO

• AHRQ

Problems and Solutions

• We don’t collect enough variables centrally to risk adjust.

• Solutions:– Collect more data and apply software.– Stratify/Restrict your comparisons

• Males, aged 65-74, diagnosed with diabetes 5 years ago.

– Compare over time (“percent improvement”)– Investigate case-mix as explanation for

difference.