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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