Transcript Confidence Intervals and Maximum Errors

Confidence Intervals and
Maximum Errors
By Sidney S. Lewis
For
Baltimore Section, ASQ
February 15, 2005
S S Lewis 2/15/05
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CONFIDENCE INTERVAL
A confidence interval expresses our belief, or
confidence, that the interval we construct
from the data will contain the mean, µ, (for
example) of the population from which the
data were drawn.
Confidence Intervals can be computed on any
population parameter: µ, s, p’, c’, and even
on complex parameters, such as Cp and Cpk.
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CONFIDENCE INTERVAL EXAMPLE
• Example of a confidence Interval (C.I.) on the
population mean µ:
• Statement: The interval 38.0 - 42.0 contains µ
with 90% confidence.
• Alternatively, there is a 5% chance that the
C.I. falls entirely below µ (µ above 42.0), and
likewise, a 5% chance that the C.I. is entirely
above µ (µ below 38.0).
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Example
Pollsters report that 55% of a sample of
1005 members of the voting population
support Proposition A.
The Margin of Error is 3.1%
There is an implied risk of being wrong,
usually 5%
Calcs.: ME  1.96 .5 *.5 / n  3.15%
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POPULATION vs. SAMPLES
PROCESS
POPULATION
PROCESS
SAMPLE
SAMPLE
810
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Distribution of X-Bar
0.4
s=2
n=4
s(xbar) = 1
a = 5%
a/2
m
a/2
0
37
810.05
38
39
40
41
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42
43
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MAXIMUM ERROR
Logic:
Maximum Error
s is known to be 2.0
A sample of 4 yields X-bar = 42.0.
Question? What values of m are probable,
with 95% Confidence (a = 5%).
Xbar
Solution: sXbar = s/on = 1.0
37
ME = Za/2 sXbar = ± 2.0
m2?
m 
Za/2 = 1.96, or about 2
38
39
40
41
42
43
44
45
46
47
m
or from 40.0 to 44.0
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MAXIMUM ERROR of the MEAN
The Maximum Error (ME) is the largest expected
deviation of the sample mean from the
population mean, with the stated confidence
level, 1- 

s/
n if σ is unknown.
ME = z/2 σ/ n if σ is known,
= t/2
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C. I. on the MEAN
Calculation of the C.I. on the mean µ typically uses
either the population standard deviation, s, if known,
or if not, the sample standard deviation, s.
If X-bar is the sample mean, then a 1- a confidence
interval on µ is:
C.I. =X-bar ± Maximum Error (ME)
= X-bar ± za/2 s /n if s is known, or
= X-bar ± ta/2s/n
812.2
if s is unknown.
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Diameters of 3/4" HR Bars
Diameters of 3/4" HR bars
940
X-bar
R
0.7466
0.7457
0.7524
0.7495
0.7489
0.7486
0.0067
0.7496
0.7549
0.7542
0.7566
0.7493
0.7529
0.0073
0.7563
0.7436
0.7475
0.7525
0.7492
0.7498
0.0127
0.7491
0.7508
0.7512
0.7482
0.7520
0.7502
0.0038
0.7498
0.7552
0.7508
0.7477
0.7453
0.7497
0.0099
0.7508
0.7480
0.7498
0.7526
0.7532
0.7509
0.0052
0.7498
0.7491
0.7507
0.7514
0.7527
0.7507
0.0037
0.7526
0.7521
0.7496
0.7507
0.7533
0.7516
0.0037
0.7520
0.7470
0.7550
0.7517
0.7404
0.7492
0.0146
0.7463
0.7554
0.7483
0.7507
0.7474
0.7496
0.0091
0.7537
0.7520
0.7501
0.7522
0.7524
0.7521
0.0036
0.7476
0.7535
0.7542
0.7548
0.7515
0.7523
0.0072
0.7516
0.7442
0.7499
0.7509
0.7472
0.7488
0.0074
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X-Bar Chart of 3/4" HR Bar Diameters
Avg. Diameter
0.756
0.754
0.752
0.75
0.748
0.746
0.744
0
10
20
30
40
Sample No.
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R Chart of 3/4" HR Bars
0.02
Range
0.015
0.01
0.005
0
0
10
20
30
40
Sample No.
840.1
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DATA STATISTICS
840.01
X-bar
R
s
0.7486
0.0067
0.0026
0.7529
0.0073
0.0033
0.7498
0.0127
0.0048
0.7502
0.0038
0.0016
0.7497
0.0099
0.0037
0.7509
0.0052
0.0021
0.7507
0.0037
0.0014
0.7516
0.0037
0.0015
0.7492
0.0146
0.0057
0.7496
0.0091
0.0036
0.7521
0.0036
0.0013
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3/4" HR Bars – MEANS and
CONFIDENCE INTERVALS
s.00300
Sample
840.41
X-bar
s
z(.90) = 1.645
LCI(z)
UCI(z)
t(4,.90) = 2.132
LCI(t)
UCI(t)
1
0.7486
0.00263
0.7464
0.7508
0.7461
0.7511
2
0.7529
0.00328
0.7507
0.7551
0.7498
0.7561
3
0.7498
0.00484
0.7476
0.7520
0.7452
0.7544
4
0.7502
0.00157
0.7480
0.7525
0.7488
0.7517
5
0.7497
0.00371
0.7475
0.7519
0.7462
0.7533
6
0.7509
0.00209
0.7487
0.7531
0.7489
0.7529
7
0.7507
0.00142
0.7485
0.7529
0.7494
0.7521
8
0.7516
0.00148
0.7494
0.7539
0.7502
0.7531
9
0.7492
0.00568
0.7470
0.7514
0.7438
0.7546
10
0.7496
0.00360
0.7474
0.7518
0.7462
0.7530
11
0.7521
0.00129
0.7499
0.7543
0.7509
0.7533
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90% Confidence Limits using s
Confidence Intervals
0.756
0.754
0.752
0.750
0.748
0.746
0.744
0
5
10
15
20
25
30
35
40
Sample No.
840.5
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90% Confidence Limits using s
Confidence Intervals
0.758
0.756
0.754
0.752
0.750
0.748
0.746
0.744
0.742
0
5
10
15
20
25
30
35
40
Sample No.
840.6
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FACTORS AFFECTING THE WIDTH
OF A CONFIDENCE INTERVAL
C.I.  X  z a /2 s / n
C.I.  X  ta /2 s / n
Factors: s or s, n, a
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FACTORS AFFECTING THE WIDTH OF A
CONFIDENCE INTERVAL
s or s: ...................
Width increases as s
or s increases;
Sample size, n .....
Width decreases as n
increases;
C. I. is proportional to 1/on
Confidence level 1 - a, or risk a:
Width increases as confidence
increases, or as risk a decreases.
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CONFIDENCE INTERVALS ON s
Small samples (n<30):
Upper C.I.  s 
2
a / 2
Lower C.I.  s 
2
1 a / 2
/ df
/ df
Large samples (n>30):
C. I. 
812.7
s
1  za / 2 /
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CONFIDENCE INTERVALS ON p’
Large samples (np>5):
C. I.  p  z a / 2
ME  z a / 2
812.7
p1  p / n ,
p1  p / n
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CONFIDENCE INTERVAL ON p’, SMALL n
90% C.I. on p’ for n=50, c=1
Upper C.I. on p'
a/2 c
P’
27.9% 1
5.0%
3.38% 1
10.0%
5.32% 1
9.0%
4.25% 1
9.5%
4.87% 1
9.2%
5.00% 1
9.14%
Excel Function:
861
n
50
50
50
50
50
50
Lower C.I. on p'
a/2 c
P’
82.7% 1
1.5%
91.1% 1
1.0%
97.4% 1
0.5%
96.4% 1
0.6%
95.2% 1
0.7%
95.0% 1
0.72%
n
50
50
50
50
50
50
=BINOMDIST(c,n,p’,1)
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STATISTICAL CALCULATION EXAMPLE A
TECHNIQUE:
1-SAMPLE TEST OF THE MEAN, SIGMA
UNKNOWN: t-TEST
SUBJECT:
Machinability: Increased by a New Practice?
GOAL: Determine whether the average machinability of steel made
using a new practice in the Melt Shop can increase the
machinability, with 95% CONFIDENCE (5% a).
HISTORIC DATA: The recent past average machinability is 85.0 = m0 ;
s = 12.2 .
DATA:
Machinability data of steel made with the new practice are
91 99 83 87 98 94 86 92 85 81
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STATISTICAL CALCULATION EXAMPLE A
Calcs.: X-bar = 89.6; s = 6.196; n = 10;
Maximum Error, ME = ta/2df) * SX-bar
= 1.833 * 1.957 = 3.6 units
90% C.I. = X-bar ± ME
= 89.6 ± 3.6 = 86.0 to 93.2 .
That means that the machinability should increase
by at least 1.0 units, and may increase by 8
units.
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STATISTICAL CALCULATION EXAMPLE B
TECHNIQUE: 1-SAMPLE TEST OF PROPORTIONS – Z-TEST
SUBJECT:
Cap leakers – reduction trial
GOAL: Determine whether the rates of leaking caps are lower if a new
cap design is used, with 95% CONFIDENCE (a = 5%).
HISTORIC DATA: Cap leaker rate = 1.2% = p’.
DATA: A trial using 2000 caps of a new design found 18 leaking caps.
p = 0.90%.
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STATISTICAL CALCULATION EXAMPLE B
p  p d0
FORMULAS:
z 0
; and s p 
sp
where p is in percent.
p0(100  p0 )
,
n
CALCS: p = 18/2000 = 0.90%; D = p0 – p = 1.2 - 0.90 = 0.30%
sp 
12
. (100  12
. )/ 2000  0.243%
Maximum Error, ME = Z.05 * sp = 1.645 * 0.243 = 0.400%
90% C.I. (2-tail) on the difference = (D - d0) ± ME = (0.90 – 1.20) ±
0.400 = +0.10% to -0.70%;
90% C.I. on p’: p ± ME = 0.90 ± 0.40 = 0.5% to 1.3%
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STATISTICAL CALCULATION EXAMPLE B
CONCLUSION:
The long term leaker rate of the new caps may be 0.7%
lower than the old caps, but it may also be 0.1% higher,
which if true, says to avoid the new caps. Therefore the
data are insufficient to show, with 95% confidence, that
the new caps are definitely better, which confirms the
test of hypothesis.
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STATISTICAL CALCULATION EXAMPLE C
TECHNIQUE: 1-SAMPLE TEST OF A SAMPLE STANDARD
DEVIATION–CHI-SQUARED TEST
SUBJECT: XYZ Digital Blood Pressure Monitor measures of systolic
blood pressure
GOAL: To determine if the monitor has become more variable than
when new.
HISTORIC DATA: Early evaluation of this monitor found the standard
deviation to be 2.5 units.
DATA : Using the monitor, the systolic blood pressure of a patient was
measured 7 times over a ten minute period. The patient sat quietly
throughout the testing. The results were: 144, 147, 147, 149, 140,
140, 144, from which s = 3.51.
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STATISTICAL CALCULATION EXAMPLE C
CONFIDENCE INTERVAL: a 2-tail, 90% confidence
interval will be calculated
The critical values of 2 are:
.205 (6)  12.59; .295 (6)  1635
.
.
Upper CI  s   .205 / df  3.51 12.59 / 6  5.08
2
Lower CI  s   .95
/ df  3.51 1.635 / 6  1.83
With 90% confidence, the true standard deviation lies between
1.83 and 5.05 units, which includes the earliest determined
standard deviation of 2.5.
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