Introduction to Statistical Process Control Module 4

Introduction to Statistical Process Control

Module 4

History of Statistical Process Control

• Quality Control in Industry– Shewhart and Bell Telephones

• Deming & Japan after WWII

• Use in Health Care & Public Health

The Run Chart

The Count

Sunday 12

Monday 2

Tuesday 4

Wednesday 3

Thursday 5

Friday 4

Saturday 2

Day Cups of Coffee

The Mean

The mean of 4, 7, 8 , and 2 is equal to:

4+7+8+24

The Median-Odd Numbers

• = the middle value in an ordered series of numbers.

• To take the median of 1, 7, 3, 10, 19, 4, 8

• Order these numbers: 1,3,4,7,8,10,19

• The median is zth number up the series where z=(k+1)/2 and k=number of numbers.

• What is the median in this case?

The Median-Even k

• Order the numbers, i.e., 1,7,10,14,15, 17.

• Find the middle values, i.e., 10 and 14.

• Take the average between these two values.

• What is the median?

The ProportionYou have these 10 values representing 10 people:

0,0,0,1,0,1,0,0,1,0.

Zero means person did not get sick.One means person did get sick.

What is the mean of these 10 values?

(0+0+0+1+0+1+0+0+1+0)/10 = .333

Proportion= n/N, where n=number of people who got sick andN=total number of people. n=numerator, N=denominator.

What is a population?

• A group of people? • A group of people over time?• Hospital visits?• Motor vehicle crashes?• Ambulance Calls?• Vehicle-Miles?• X-Rays Read?• Other?

Populations take onDistributions

In simple statistical process control, we dealwith 4 distributions.

From central tendency to variation.

The Normal Distribution

How do we describe variation about the red line in the normal curve?

• In other words, how fat is that distribution?• How about the average difference between each

observation and the mean?– Oops, can’t add those differences, some are positive

and some are negative.

• How about adding up the absolute values of those differences? – Bad statistical properties.

• How about the average squared difference?– Now we are talking!

Population Variance

∑ i=1

N1N

(xi - µ)2Population Variance =

The average squared deviation!

Population Standard Deviation

∑ i=1

N1N

(xi - µ)2

Population Standard Deviation =

The square root of the average squared deviation!

√

Standard Deviation has Nice Properties!

.025025

It’s Time to Dance

How do I estimate the standard deviation of the means of repeated samples?

• Estimate the standard deviation of the population with your sample using the sample standard deviation.

• Estimate the standard deviation of the mean of repeated samples by calculating the standard error.

Sample Standard Deviation

∑ i=1

N1

N-1 (xi - x)2S = √How is this different from the Population StandardDeviation?

Standard Error

SE =

How is this different from the Sample StandardDeviation?

s

√ n

Z-Score for Distribution of Sample Means

Z = x -

SE

You can convert any group of numbers to z-scores.

μ

X = mean observed in your sampleμ = is the population mean you believe in.

Z = number of standard errors x is away from μ,

If we kept dancing for hundreds of timesHere is the distribution of our sample means

(standardized)

Wait a minute!

• When you do a survey, you only have one sample, not hundreds of repeated samples.

• How confident can you be that the mean of your one sample represents the mean of the population?

• If you think reality is a normal distribution with mean y and standard deviation s, how likely is your observed mean of x?

Welcome to

• Confidence Intervals

• P-Values

• Let’s focus on p-values for now.

Here is our distribution of sample means—WE BELIEVE

.025.025

Here is the mean we observed(1.96 ≈ 2)

What is the probability of observing a mean at least as far away as zero (on either side) as 1.96 standard errors? 2.72?

Area under curveis probability andit adds up to one.

Remember the p-value question?

• If you think reality is a normal distribution with mean y and standard deviation s, how likely is your observed mean of x?

Let’s ask it again.• We have systolic blood pressure measurements on a sample of 50

patients for each of 25 months. For each of those months, we a mean blood pressure and a sample standard deviation.

• “You think reality for each month should be a normal distribution with a mean blood pressure that equals the average of the 25 mean blood pressures.

• You also think that for each month, this normal distribution should have a standard error based on the average sample standard deviation across the 25 months.

• How likely is your observed mean in month 4 of 220 if the average mean across the 25 months was 120?

• How many standard errors is 220 away from 120? What is the probability of being at least that many standard errors away from 120?

Welcome to the Shewart Control Chart

1 2 3 4 5 6 7 . . . 25

120

Indian Health Service, DHHS

Anatomy of the control chart:

From Amin, 2001