confidence intervals and maximum errors
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Confidence Intervals and Maximum Errors. By Sidney S. Lewis For Baltimore Section, ASQ February 15, 2005. CONFIDENCE INTERVAL - PowerPoint PPT PresentationTRANSCRIPT
S S Lewis 2/15/05 1
Confidence Intervals and Maximum Errors
By Sidney S. Lewis
For
Baltimore Section, ASQ
February 15, 2005
S S Lewis 2/15/05 2
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, , p’, c’, and even on complex parameters, such as Cp and Cpk.
812.1
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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.: %15.3n/5.*5.96.1ME
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POPULATIONPROCESS
SAMPLESAMPLE
PROCESS
POPULATION vs. SAMPLES
810
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Distribution of X-Bar
37 38 39 40 41 42 43
0
0.4
/2 /2
= 2n = 4(xbar) = 1 = 5%
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Logic:
is known to be 2.0
A sample of 4 yields X-bar = 42.0.
Question? What values of are probable,
with 95% Confidence ( = 5%).
Solution: Xbar = /n = 1.0
Z/2 = 1.96, or about 2
ME = Z/2 Xbar = ± 2.0
or from 40.0 to 44.0
Maximum Error
37 38 39 40 41 42 43 44 45 46 47
2?
Xbar
MAXIMUM ERROR
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MAXIMUM ERROR of the MEAN
The Maximum Error (ME) is the largest expecteddeviation of the sample mean from thepopulation mean, with the stated confidencelevel, 1-
ME = z /2 σ/n if σ is known,
= t /2 s/n if σ is unknown.
ME = z /2 σ/n if σ is known,
= t /2 s/n if σ is unknown.
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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, , if known, or if not, the sample standard deviation, s.
If X-bar is the sample mean, then a 1- confidence interval on µ is:
C.I. =X-bar ± Maximum Error (ME)
= X-bar ± z/2 /n if is known, or
= X-bar ± ta/2s/n if is unknown.812.2
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Diameters of 3/4" HR BarsDiameters of 3/4" HR bars 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
940
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X-Bar Chart of 3/4" HR Bar Diameters
0.744
0.746
0.748
0.75
0.752
0.754
0.756
0 10 20 30 40
Sample No.
Avg
. Dia
met
er
840.1
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R Chart of 3/4" HR Bars
0
0.005
0.01
0.015
0.02
0 10 20 30 40
Sample No.
Ran
ge
840.1
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DATA STATISTICS
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
z(.90) = 1.645 t(4,.90) = 2.132
Sample X-bar s LCI(z) UCI(z) 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 0.7533840.41
S S Lewis 2/15/05 15
90% Confidence Limits using
0.744
0.746
0.748
0.750
0.752
0.754
0.756
0 5 10 15 20 25 30 35 40
Sample No.
Con
fiden
ce In
terv
als
840.5
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90% Confidence Limits using s
0.742
0.744
0.746
0.748
0.750
0.752
0.754
0.756
0.758
0 5 10 15 20 25 30 35 40
Sample No.
Con
fiden
ce In
terv
als
840.6
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FACTORS AFFECTING THE WIDTH OF A CONFIDENCE INTERVAL
C I X z n. . // 2
C I X t s n. . // 2
812.4
Factors: or s, n,
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FACTORS AFFECTING THE WIDTH OF A CONFIDENCE INTERVAL
or s: ...................Width increases as or s increases;
Sample size, n .....Width decreases as n increases;
C. I. is proportional to 1/n
Confidence level 1 - , or risk :
Width increases as confidence increases, or as risk decreases.
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CONFIDENCE INTERVALS ON
Small samples (n<30):
Upper C.I. s df / /22
Low er C.I. s df 1 22
/ /
C Is
z n. .
//
1 22
Large samples (n>30):
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CONFIDENCE INTERVALS ON p’
Large samples (np>5):
C I p z p p n
M E z p p n
. . / ,
/
/
/
2
2
1
1
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CONFIDENCE INTERVAL ON p’, SMALL n
Upper C.I. on p' Lower C.I. on p'
c P’ n c P’ n
27.9% 1 5.0% 50 82.7% 1 1.5% 50
3.38% 1 10.0% 50 91.1% 1 1.0% 50
5.32% 1 9.0% 50 97.4% 1 0.5% 50
4.25% 1 9.5% 50 96.4% 1 0.6% 50
4.87% 1 9.2% 50 95.2% 1 0.7% 50
5.00% 1 9.14% 50 95.0% 1 0.72% 50
Excel Function: =BINOMDIST(c,n,p’,1)861
90% C.I. on p’ for n=50, c=1
S S Lewis 2/15/05 22
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% ).
HISTORIC DATA: The recent past average machinability is 85.0 = 0 ; = 12.2 .
DATA: Machinability data of steel made with the new practice are
91 99 83 87 98 94 86 92 85 81890.53
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STATISTICAL CALCULATION EXAMPLE A
Calcs.: X-bar = 89.6; s = 6.196; n = 10;
Maximum Error, ME = tdf) * 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 (= 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
FORMULAS: where p is in percent.
CALCS: p = 18/2000 = 0.90%; = p0 – p = 1.2 - 0.90 = 0.30%
Maximum Error, ME = Z.05 * p = 1.645 * 0.243 = 0.400%
90% C.I. (2-tail) on the difference = ( - 0) ± 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%
zp p
a ndp p
npp
0 0 0 0100
;( )
,
890.83
p 12100 12 2000 0 243%. ( . ) / .
S S Lewis 2/15/05 26
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.
STATISTICAL CALCULATION EXAMPLE B
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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:
. .( ) . ; ( ) . .052
9526 12 59 6 1635
Upper C I s df. / . . / .052 3 51 12 59 6 5 08
Lower C I s / df 3.51 1.635 / 6 1.83.952
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.
890.72