This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 7.22. Are the data consistent with the assumed process mean? ...ENGNEERNG STATISTICS HANDBOOK. é TOOLS £ thS lSEARCH iBACK NEXTI 7.2.2. Are the data consistent with the assumed
process mean? The testing
of H0 for a single
population
mean Typical null
hypotheses Test statistic
where the
standard
deviation is
not known Given a random sample of measurements. Y1. YN. there are three types of questions regarding the true mean of the
population that can be addressed with the sample data. They
are: 1. Does the true mean agree with a known standard or assumed mean? 2. Is the true mean of the population less than a given standard?
3. Is the true mean of the population at least as large as a given standard? The corresponding null hypotheses that test the true mean. r“,
against the standard or assumed mean, ,Il'g are: 1. ﬁozrn=rro
2. Howie!)
3 Hornzm The basic statistics for the test are the sample mean and the standard deviation. The form of the test statistic depends on whether the poulation standard deviation, or, is known or is
estimated from the data at hand. The more typical case is
where the standard deviation must be estimated from the data. and the test statistic is fro
six/P7 I where the sample mean is
_ 1 N
r = — 2}: N.
1:1 http://www.itl.nist.gov/div898/handbook/prc/section2/prc22.htm Page 1 of 3 1/28/2012 Comparison
with critical
values Test statistic
where the
standard deviation is
known Caution An
illustrative example of
the ttest The test is 7.22. Are the data consistent with the assumed process mean? and the sample standard deviation is with N  1 degrees of freedom. For a test at signiﬁcance level a, where a is chosen to be
small, typically .01, .05 or .10, the hypothesis associated with each case enumerated above is rejected if: 1* i‘i'li2 Ianew—1
2. Iaiaﬁﬂ 3. t s —ia;ﬁ,_1 where Ema; 3H is the upper M2 critical value from the 2‘ distribution with N—l degrees of freedom and similarly for
cases (2) and (3). Critical values can be found in the tinblc in
Chapter 1. If the standard deviation is known, the form of the test statistic
is Y—ao z=—— Jivﬁ For case (1). the test statistic is compared with Zmz, which is the upper Cir"2. critical value liom the standard normal
distribution, and similarly for cases (2) and (3). If the standard deviation is assumed known for the purpose of this test, this assumption should be checked by a test of
h I othesis for the standard deviation. The following numbers are particle (contamination) counts for
a sample of 10 semiconductor silicon wafers: 50 48 44 56 61 52 53 55 67 51 The mean = 53.7 counts and the standard deviation = 6.567
counts. Over a long run the process average for wafer particle counts http://www.itl.nist.gov/div898/handbook/prc/section2/prc22.htm Page 2 of 3 1/28/2012 7.2.2. Are the data consistent with the assumed process mean? two—sided Critical
values Conclusion NIST
SEMATECH has been 50 counts per wafer. and on the basis of the sample. we want to test whether a change has occurred. The null hypothesis that the process mean is 50 counts is tested against
the alternative hypothesis that the process mean is not equal to 50 counts. The purpose of the two—sided alternative is to rule
out a possible process change in either direction. For a signiﬁcance level of a = .05. the chances of erroneously
rejecting the null hypothesis when it is true are 5% or less. (For a review of hypothesis testing basics. see Chapter 1). Even though there is a history on this process. it has not been
stable enough to justify the assumption that the standard deviation is known. Therefore. the appropriate test statistic is
the rstatistic. Substituting the sample mean. sample standard deviation. and sample size into the formula tor the test
statistic gives a value of t= 1.782 with degrees of freedom = N  I = 9. This value is tested
against the upper critical value t0.025;,9 : 2262 from the ttuble where the critical value is found under the
column labeled 0.025 for the probability of exceeding the critical value and in the row for 9 degrees of freedom. The critical value Cir’2 is used instead of it because of the two
sided alternative (twotailed test) which requires equal
probabilities in each tail of the distribution that add to at. WEE? [Toms a mos" :55“!ch {"hicT "‘e”"'x7‘""i"] http://www.itl.nist.gov/div89S/handbook/prc/sectionZ/prc22.htm Page 3 of3 1/28/2012 1.3.5.2. Conﬁdence Limits for the Mean Page 1 of 5 I I ENGINEEING STATISTICS HANDBOOK
Wm m rim: warm 1.3.5.2. Conﬁdence Limits for the Mean Purpose: Conﬁdence limits for the mean (Sncdecor and Cochran. 1989) are an
Interval interval estimate for the mean. Interval estimates are often desirable
Estimate for because the estimate of the mean varies from sample to sample. Instead
Mean of a single estimate for the mean. a conﬁdence interval generates a lower and upper limit for the mean. The interval estimate gives an
indication of how much uncertainty there is in our estimate of the true mean. The narrower the interval, the more precise is our estimate. Conﬁdence limits are expressed in terms of a conﬁdence coefﬁcient.
Although the choice of conﬁdence coefﬁcient is somewhat arbitrary. in
practice 90%. 95%. and 99% intervals are often used. with 95% being
the most commonly used. As a technical note. a 95% conﬁdence interval does not mean that there is a 95% probability that the interval contains the true mean. The
interval Wag} givengsagmpleeﬁither containsthe true mean or ._o— __ ___,_ I._,—_
_—_ it does not. Instead. the level of conﬁdence is associatedwith the
method of calculating the interval. The COnﬁdenCe coefﬁcient is simply
the proportion of samples of a given size that may be expected to contain the true mean. That is. for a 95% conﬁdence interval, if many samples are collected and the conﬁdence interval computed. in the long
run about 95% of these intervals would contain the true mean. Deﬁnition: Conﬁdence limits are deﬁned as:
Conﬁdence _
Interval y i tmﬂﬁ_l)3/¢N where Y is the sample mean. 5 is the sample standard deviation. N is the
sample size. or is the desired signiﬁcance level. and threw—.1} is the u ) Jcr critical value of the 1 distribution with N— 1 degrees of freedom.
Note that the conﬁdence coefﬁcient is l  r1. From the formula. it is clear that the width of the interval is controlled
by two factors: 1. As N increases. the interval gets narrower from the ‘JN term. http://www.itl.nist.gov/div898/handbook/eda/section3/eda3 52.htm 1/28/2012 1.3.5.2. Conﬁdence Limits for the Mean Page 2 of 5 That is. one way to obtain more precise estimates for the mean is
to increase the sample size. 2.. The larger the sample standard deviation, the larger the
conﬁdence interval. This simply means that noisy data. i.e., data with a large standard deviation. are going to generate wider
intervals than data with a smaller standard deviation. Deﬁnition: To test whether the population mean has a Speciﬁc value. #0. against the
Hypothesis twosided alternative that it does not have a value Ho. the conﬁdence
Test interval is converted to hypothesis—test form. The test is a onesample { test, and it is deﬁned as: H0: :3 2 Ho Ha: a 5'5 #0 Test Statistic: T = [i7 — gm)/(s/VIE) where I”, N, and a are deﬁned as above. Signiﬁcance u: The most commonly used value for u: is 0.05.
Level: Critical Region: Reject the null hypothesis that the mean is a speciﬁed
value. #0. if T <1 —t(a,'2,.v_1) or
T I? tfrxli2.N—l} Sample Dataplot generated the following output for a conﬁdence interval from Output for the ZARR'13.DAT data set: Conﬁdence Interval CONFIDENCE LIMITS FOR MEAN
(2SIDED) NUMBER OF OBSERVATIONS
MEAN STANDARD DEVIATION
STANDARD_DEV 0. 8881E—01J/ 0.163194OEO2 ll  CONFIDENCE T T x SD(MEAN) LOWER UPPER
VALUE l%) VALUE LIMIT LIMIT
50.000 0.676 i0.110279E02 9.26036 9.26256
75.000 1.154 0.188294E02 9.25958 9.26334
90.000 1.653 0.269718E02 9.25876 9.26416
95.000 1.972 0.321862E02 9.25824 9.26468
99.000 2.601 0.424534E02 9.25721 9.26571 99.900 3.341: 0.545297E02 9.25601 9.26691 http://www.itl.nistgov/div898/handbook/eda/section3/eda3 52 .htm 1/28/2012 1.3.5.2. Conﬁdence Limits for the Mean Page 3 of 5 99.990 3.973 0.648365E—02 9.25498 9.26794
99.999 4.536 0.740309E02 9.25406 9.26886 Interpretation The ﬁrst few lines print the sample statistics used in calculating the
of the Sample conﬁdence interval. The table shows the conﬁdence interval for several Output different signiﬁcance levels. The ﬁrst column lists the conﬁdence level (which is 1  fl“ expressed as a percent), the second column lists the t
value (i.e.. Harmer—1}), the third column lists the tvalue times the standard error (the standard error is s/VN). the fourth column lists the lower conﬁdence limit. and the ﬁfth column lists the upper conﬁdence limit. For example, for a 95% conﬁdence interval, we go to the row
identiﬁed by 95.000 in the ﬁrst column and extract an interval of
(9.25824, 9.26468) from the last two columns. Output from other statistical software may look somewhat different
from the above output. Sample Dataplot generated the following output for a onesample t—test from the
Output for t ZA RR] SDAT data set:
Test
T TEST
(1—SANRLE)
MUO = 5.000000 NULL HYPOTHESIS UNDER TESTMEAN MU = 5.000000 SAMPLE:
NUMBER OF OBSERVATIONS = 195
MEAN = 9.261460 0.2278881E01
0.1631940E02 STANDARD DEVIATION
STANDARD DEVIATION OF MEAN TEST:
MEAN—M00 = 4.261460
T TEST STATISTIC VALUE = 2611.284
DEGREES OF FREEDOM = 194.0000
T TEST STATISTIC CDF VALUE = 1.000000
ALTERNATIVE ALTERNATIVE
ALTERNATIVE HYPOTHESIS HYPOTHESIS
HYPOTHESIS ACCEPTANCE INTERVAL CONCLUSION
MU <> 5.000000 (0.0.025) (0.975.1) ACCEPT
MU < 5.000000 (0.0.05) REJECT
MU > 5.000000 (0.95.1) ACCEPT
Interpretation
gsamﬂe We are testing the hypothesis that the population mean is 5. The output
“PM is divided into three sections. 1. The ﬁrst section prints the sample statistics used in the
computation of the ttest. http://www.itl.nist.gov/div898/handbook/eda/section3/eda352 .htm 1/28/2012 1.3.5.2. Conﬁdence Limits for the Mean Questions Related Techniques Case Study Sofnvare The second section prints the t~test statistic value, the degrees of
freedom, and the cumulative distribution function cdl value of
the ttest statistic. The ttest statistic cdf value is an alternative
way of expressing the critical value. This cdf value is compared
to the acceptance intervals printed in section three. For an upper
onetailed test. the alternative hypothesis acceptance interval is (l
 (1,1), the alternative hypothesis acceptance interval for a lower
onetailed test is (0.0:), and the alternative hypothesis acceptance
interval for a twotailed test is (l  (Jr/2,1) or (0.05/2). Note that accepting the alternative hypothesis is equivalent to rejecting the
null hypothesis. The third section prints the conclusions for a 95% test since this is
the most common case. Results are given in terms of the alternative hypothesis for the twotailed test and for the onetailed
test in both directions. The alternative hypothesis acceptance interval column is stated in terms of the cdf value printed in
section two. The last column speciﬁes whether the alternative hypothesis is accepted or rejected. For a different signiﬁcance
level, the appropriate conclusion can be drawn from the ttest
statistic cdf value printed in section two. For example. for a signiﬁcance level of 0.10. the corresponding alternative hypothesis acceptance intervals are (0.005) and (095.1). (0.
0.10), and (0.90.1). Output from other statistical software may look somewhat different
from the above output. Conﬁdence limits for the mean can be used to answer the following
questions: 1. What is a reasonable estimate for the mean? 2. How much variability is there in the estimate of the mean?
3. Does a given target value fall within the conﬁdence limits? Two—Sample T—Test Conﬁdence intervals for other location estimators such as the median or midmean tend to be mathematically difﬁcult or intractable. For these
cases. conﬁdence intervals can be obtained using the bootstrap... Heat 'llrjiw meter data. Conﬁdence limits for the mean and onesample ttests are available in
just about all general purpose statistical software programs. including Datap I o t. http://www.itl.nistgov/divS98/handbook/eda/section3/eda352.htm Page 4 of 5 1/28/2012 1.3.5.3. TwoSample <i>t</i>—Test for Equal Means Page 1 of 4 __E_NGINERING STATISTICS HANDBOOK
W I'sen—ncﬁ' m Tie—am I
I3 EDA Technicues
1.3.5. uantitutive Techni ues 1.3.5.3.. TwoSample t—Test for Equal Means Purpose: The two—sample ttest (Sucdccor and Cochran. 1989) is used to Test if two determine if two population means are equal. A common application of
population this is to test if a new process or treatment is superior to a current
means are process or treatment. aqua! There are several variations on this test. 1. The data may either be paired or not paired. By paired. we mean
that there is a onetoone correspondence between the values in the two samples. That is. ile. X2. ....XH and Y1. Y . . Y” are
the two samples. then X . corresponds to Y 1.. For paired samples. the difference X1.  Y]. is usually calculated. For unpaired samples. the sample sizes for the two samples may or may not be equal.
The formulas for paired data are somewhat simpler than the formulas for unpaired data. 2. The variances of the two samples may be assumed to be equal or
unequal. Equal variances yields somewhat simpler formulas.
although with computers this is no longer a signiﬁcant issue. 3. In some applications. you may want to adOpt a new process or treatment only if it exceeds the current treatment by some
threshold. In this case. we can state the null hypothesis in the form that the difference between the two populations means is equal to
some constant (#1 — fig 2 (in) where the constant is the desired threshold.
Deﬁnition The two sample t test for unpaired data is defined as: #11 : fltz
H3: #1 Té #2 Test
Statistic: T _ Y1 — Y3 — if Sir/N1 + where N1 and N2 are the sample sizes. 17. and E are the
sample means. and and are the sample variances. http://mvw.itl.nist.gov/div898/handbook/eda/section3/eda3 53 .htm 1/28/2012 Page 2 of 4 t0:
17 —}7
T: 1 2
Spqulle + where Signiﬁcance or.
Level:
Crltrcal RBJBCt the null hypothesis that the two means are equal 1f
Region
T (C —t(‘1lf2:1)
or
T 3‘" tram»)
where Hagan) is the critical value of the I distribution with
.12 degrees of freedom where
U __ (Si/N1 + Sﬁ/Nzlz
(sf/Ntlzrlel — 1) + (SS/Nzlg/UE — 1)
If equal variances are assumed, then
tJ::.Aﬁl+—JV§—EZ
Sample Dataplot generated the following output for the t test from the
Output A UT(_)8.SB.DAT data set:
T TEST
(2SAMPLE) NULL HYPOTHESIS UNDER TEST—POPULATION MEANS MUl = MU2 SAMPLE 1: NUMBER OF OBSERVATIONS = 249 MEAN = 20 . 14458 STANDARD DEVIATION = 6. 414700 STANDARD DEVIATION OF MEAN = 0.4065151
SAMPLE 2: NUMBER OF OBSERVATIONS = 79 MEAN = 30 . «ﬂ 8 1 0 l http://www.itl.nist.gov/diVS98/handbook/eda/section3/eda3 5 3 .htm 1/28/2012 1.3.5.3. Two—Sample <i>t</i>Test for Equal Means Page 3 of 4 STANDARD DEVIATION = 6 107710
STANDARD DEVIATION OF MEAN = 0.6871710
IF ASSUME SIGMA1 = SIGMA2:
POOLED STANDARD DEVIATION = 6.342600
DIFFERENCE (DEL) IN MEANS = —10.33643
STANDARD DEVIATION OF DEL = 0.8190135
T TEST STATISTIC VALUE = 12 62059
DEGREES OF FREEDOM = 326.0000
T TEST STATISTIC CDF VALUE = 0.000000
IF NOT ASSUME SIGMAI = SIGMAZ:
STANDARD DEVIATION SAMPLE 1 = 6.414700
STANDARD DEVIATION SAMPLE 2 = 6.107710
BARTLETT CDF VALUE = 0.402799
DIFFERENCE (DEL) IN MEANS = —10 33643
STANDARD DEVIATION OF DEL = 0.7984100
T TEST STATISTIC VALUE = —12.94627
EQUIVALENT DEG. OF FREEDOM = 136.8750
T TEST STATISTIC CDP VALUE = 0.000000
ALTERNATIVE— ALTERNATIVE—
ALTERNATIVE— HYPOTHESIS HYPOTHESIS
HYPOTHESIS ACCEPTANCE INTERVAL CONCLUSION
M01 <> M02 (0.0.025) (0.975.l) ACCEPT
MUl < MU2 (0.0.05) ACCEPT
MU1 > MU2 (0.95.1) REJECT
Interpretation
ofsampge We are testing the hypothesis that the population mean is equal for the
Output two samples. The output is divided into ﬁve sections. 1. The ﬁrst section prints the sample statistics for sample one used in
the computation of the Host. 2. The second section prints the sample statistics for sample two
used in the computation of the ttest. 3. The third section prints the pooled standard deviation. the
difference in the means, the ttest statistic value, the degrees of
freedom, and the cumulative distribution function Cdﬂ value Of the Itcst statistic under the assumption that the standard
deviations are equal. The Host statistic cdf value is an alternative way of expressing the critical value. This cdf value is compared to
the acceptance intervals printed in section ﬁve. For an upper one— tailed test, the acceptance interval is (0.1  or). the acceptance
interval for a twotailed test is (Cr/2. 1  nun/2), and the acceptance interval for a lower onetailed test is ((1.1). 4. The fourth section prints the pooled standard deviation. the
difference in the means. the t—test statistic value. the degrees of freedom, and the emmilstive distribution function (cdl) value of
the (—test statistic under the assumption that the standard http://www.it1.nist.gov/div898/handbook/eda/section3/eda3 53 .htm 1/28/2012 1.3.5.3. Two—Sample <i>t</i>Test for Equal Means Questions Related
Techniques Case Study Software NIST
SEMATECH deviations are not equal. The ttest statistic cdf value is an
alternative way of expressing the critical value. cdf value is
compared to the acceptance intervals printed in section ﬁve. For
an upper onetailed test. the alternative hypothesis acceptance
interval is (1  [1.1). the alternative hypothesis acceptance interval
for a lower one—tailed test is (0.:1). and the alternative hypothesis acceptance interval for a twotailed test is (1  (Jr/2.1) or (0.0/2).
Note that accepting the alternative hypothesis is equivalent to rejecting the null hypothesis. The ﬁfth section prints the conclusions for a 95% test under the assumption that the standard deviations are not equal since a 95%
test is the most common case. Results are given in terms of the
alternative hypothesis for the twotailed test and for the onetailed
test in both directions. The alternative hypothesis acceptance
interval column is stated in terms of the cdf value printed in
section four. The last column speciﬁes whether the alternative
hypothesis is accepted or rejected. For a different signiﬁcance
level. the appropriate conclusion can be drawn from the Host statistic Cdf value printed in section four. For example. for a
significance level of 0.10. the corresponding alternative hypothesis acceptance intervals are (0.005) and (095.1). (0.
0.10). and (090.1). Output from other statistical software may look somewhat different
from the above output. Twosample ttests can be used to answer the following questions: 1. 2.
3. Is process 1 equivalent to process 2? Is the new process better than the current process?
Is the new process better than the current process by at least some predetermined threshold amount? Conﬁdence Limits for the Mean
Analvsis of Variance Ceramic strength data. Twosample t—tests are available in just about all general purpose
statistical software programs. including Dataplot. ,pﬁnwwm 1”! Wm H http ://www.itl.nist. gov/div898/handbook/eda/section3 /eda3 5 3 .htm Page 4 of4 1/28/2012 1.3.5.8. ChiSquare Test for the Standard Deviation Page 1 of 3 iTOOLS 3‘ AIDS 1$EAICH HACK NEXT! l. .
.3. EDA Techniques
1 3.5. Quantitative Techniques 1.3.5.8. ChiSquare Test for the Standard Deviation Purpose: A chi—square test ( Snedecor and C ochran. 1983) can be used to test if the T est if standard deviation of a population is equal to a speciﬁed value. This test can be
standard either a twosided test or a onesided test. The twosided version tests against
deviation is the alternative that the true standard deviation is either less than or greater than
equal to a the speciﬁed value. The onesided version only tests in one direction. The
speciﬁed choice of a two—sided or one—sided test is determined by the problem. For value example. if we are testing a new process. we may only be concerned if its variability is greater than the variability of the current process. Deﬁnition The chisquare hypothesis test is deﬁned as:
H0: 0 = 00
H3: 5 <5 Us for a lower onetailed test IT 3" Us for an upper onetailed test 0 E’é an for a twotailed test Test T=(N — 1)[:3/Ja)2
Statistic: where N is the sample size and s is the sample standard
deviation. The key element of this formula is the ratio 3/03 which compares the ratio of the sample standard deviation to the target standard deviation. The more this ratio deviates from 1,
the more likely we are to reject the null hypothesis. Signiﬁcance :1.
Level: C 't' al _ _ _ _ _
Rzlglign: Re} ect the null hypotheSIS that the standard devtatlon IS a speciﬁed value, mu. if T L)» xii, N_ 1} for an upper onetailed alternative
T «c: Xﬁ_mN_n for a lower onetailed alternative T <1 x3 _m;21N_1} for a twotailed test
01' http://www.itl.nist.gov/divS98/handbook/eda/section3/eda358.htm 1 02/701 ’3 1.3.5.8. ChiSquare Test for the Standard Deviation Page 2 of 3 T > Xian—1} where xii{N4} 1s the crlucal value of the ChiSt; uare d'lSll‘lbLllmn with N  1 degrees of freedom. In the above formulas for the critical regions, the Handbook
follows the convention that X: is the upper critical value from the chisquare distribution and Xf—n: is the lower critical value
from the chisquare distribution. Note that this is the opposite of some texts and software programs. In particular, Dataplot uses
the opposite convention. The formula for the hypothesis test can easily be converted to form an interval
estimate for the standard deviation: Sample Dataplot generated the following output for a Chisquare test from the
Output GEARDAT data set: CHITSQUARED TEST SIGMAO = 0.1000000
NULL HYPOTHESIS UNDER TESTSTANDARD DEVIATION SIGMA = .1000000 SAMPLE:
NUMBER OF OBSERVATIONS MEAN
STANDARD DEVIATION S 100
0.9976400 0.6278908E02 ll 11 TEST:
S/SIGMAO = 0.6278908E—01
CEISQUARED STATISTIC = 0 3903044
DEGREES OF FREEDOM = 99.00000
CHI—SQUARED COE VALUE = 0.000000
ALTERNATIVE* ALTERNATIvE~
ALTERNATIVE HYPOTHESIS HYPOTHESIS
HYPOTHESIS ACCEPTANCE INTERVAL CONCLUSION
SIGMA <> .1000000 (0.0.025), (0.975.I) ACCERT
SIGMA < .1000000 (0.0.05) ACCEPT
SIGMA > .lOOOOOO (0.95.1) REJECT
Interpretation
ofSampZe We are testing the hypothesis that the population standard deviation is 0.1. The
Output output is divided into three sections. 1. The ﬁrst section prints the sample statistics used in the computation of
the chisquare test. http ://www.itl.nist.gov/div898/handbook/eda/section3/eda3 58 .htm 1/30/2012 1.3.5.8. ChiSquare Test for the Standard Deviation Page 3 of 3 2. The second section prints the chisquare test statistic value, the degrees
of freedom, and the cumulative distribution 'l'unction cdl value of the chis uare test statistic. The chisquare test statistic cdf value is an
alternative way of expressing the critical value. This cdf value is compared to the acceptance intervals printed in section three. For an
upper onetailed test. the alternative hypothesis acceptance interval is (1  (1,1), the alternative hypothesis acceptance interval for a lower onetailed
test is (0.0). and the alternative hypothesis acceptance interval for a two tailed test is (1  [Jr/2,1) or (0.11/2). Note that accepting the alternative
hypothesis is equivalent to rejecting the null hypothesis. 3. The third section prints the conclusions for a 95% test since this is the most common case. Results are given in terms of the alternative
hypothesis for the twotailed test and for the onetailed test in both directions. The alternative hypothesis acceptance interval column is
stated in terms of the cdf value printed in section two. The last column speciﬁes whether the alternative hypothesis is accepted or rejected. For a
different signiﬁcance level, the appropriate conclusion can be drawn from the chisquare test statistic cdf value printed in section two. For
example. for a signiﬁcance level of 0.10, the corresponding alternative
hypothesis acceptance intervals are (0.0.05) and (0.95.1), (O. 0.10), and (0.90.1). Output from other statistical software may look somewhat different from the
above output. Questions The chisquare test can be used to answer the following questions: 1. Is the standard deviation equal to some predetermined threshold value?
2. Is the standard deviation greater than some predetermined threshold value?
3. Is the standard deviation less than some predetermined threshold value? Related Test / Techniques Bartlett "licst ﬁrflﬂgsy
l...evene Test SofMare The chisquare test for the standard deviation is available in many general
purpose statistical software programs. including Dataplot. N 1 WWW ......... __.,_ W... W.. MaiEm lHOME toms 3. mos {SEARCH lance; want
5 E M AT E C H http://www.itl.nistgov/divS98/handbook/eda/section3/eda3 5 8 .htm 1/28/2012 1.3.5.9. F—Test for Equality of Two Standard Deviations . Ex aloratorv Data Anal =sis l
[.3. EDA Tochni nos
1 uantitative Technit 1.3.5.9. F—Test for Equality of Two Standard Deviations Purpose: Test if standard
deviations from two
populations
are equal Deﬁnition An Ftest ( Suedecor and Cochran 1983) is used to test if the standard
deviations of two populations are equal. This test can be a twotailed test or a onetailed test. The twotailed version tests against the alternative that the standard deviations are not equal. The one—tailed version only tests in one direction. that is the standard deviation from the first
pOpulation is either greater than or less than (but not both) the second population standard deviation . The choice is determined by the problem. For example, if we are testing a new process. we may only be interested
in knowing if the new process is less variable than the old process. The F hypothesis test is deﬁned as: H0 01 I 02 H3: ‘71 "C ‘72 for a lower one tailed test
0 1 3’ t1'2 for an upper one tailed test
‘71 3'5 ‘72 for a two tailed test Test F = / Statlstlc population variances. Signiﬁcance [1
Level: Critical Region: The hypothesis that the two standard deviations are equal is rejected if I" 32> Fta,N1_1,N2.—1) for an upper onetailed test F <2 F[1_.;1’N1—1,N2—1) for a lower one—tailed test Page 1 of3 1/28/2012 1.3.5.9. FTest for Equality of Two Standard Deviations Page 2 of 3 f" 3‘" ﬁlIIIZM—IJVZ—ll where F[¢,ﬁ_1,1v_ﬁ} is the critical value of the E
distribution with “I and be degrees of freedom and a
signiﬁcance level of or. In the above formulas for the critical regions, the
Handbook follows the convention that F”. is the upper
critical value from the F distribution and F141 is the lower
critical value from the F distribution. Note that this is the opposite of the designation used by some texts and software programs. In particular. Dataplot uses the
opposite convention. Sample Dataplot generated the following output for an Ftest from the
Output JAHAN MllDAT data set: F TEST NULL HYPOTHESIS UNDER TESTSIGMA1 = STGMA2
ALTERNATIVE HYPOTHESTS UNDER TESTSIGMA1 NOT EQUAL SIGMAZ SAMPLE 1:
NUMBER OF OBSERVATIONS = 240
MEAN = 688.9987
STANDARD DEVIATION = 65.54909
SAMPLE 2:
NUMBER OF OBSERVATIONS = 240
MEAN = 611.1559
STANDARD DEVIATION = 61.85425
TEST:
STANDARD DEv. (NUMERATOR) = 65 54909
STANDARD DEv. (DENOMINATOR) = 61.85425
E TEST STATISTIC VALUE = 1.123037
DEG. OF FREEDOM (NUMER.) = 239.0000
DEG. OF FREEDOM (DENOM ) = 1239400001 .
E TEST STATISTIC CDF VALUE = ;10.814808 ‘\
. __ a _ﬂ_,L
NULL NULL HYPOTHESIéi NULL HYPOTHESIS
HYPOTHESIS ACCEPTANCE INTERVAL CONCLUSION
SIGMAl = SIGMAZ (0.000.03950) ACCEPT
Interpretation
of Sample We are testing the hypothesis that the standard deviations for sample one
Output and sample two are equal. The output is divided into four sections. 1. The ﬁrst section prints the sample statistics for sample one used in
the computation of the Ftest. 2. The second section prints the sample statistics for sample two used
in the computation of the Ftest. http://www.itl.nist.gov/div898/handbook/eda/section3/eda3 59.htm 1/28/2012 1.3.5.9. FTest for Equality of Two Standard Deviations Questions Related Techniques Case Study Sofmare HIST
SEMATECH the cumulative distribution
statistic. The F—test statistic cdf value is an alternative way of expressing the critical valuc. This cdf value is compared to the acceptance interval printed in section four. The acceptance interval
for a twotailed test is (0.1  0:). The fourth section prints the conclusions for a 95% test since this is the most common case. Results are printed for an upper one—
tailed test. The acceptance interval column is stated in terms of the cdf value printed in section three. The last column speciﬁes whether the null hypothesis is accepted or rejected. For a different
signiﬁcance level, the appropriate conclusion can be drawn from the Ftest statistic cdf value printed in section four. For example,
for a signiﬁcance level of 0.10, the corresponding acceptance
interval become (0.000,0.9000). Output from other statistical software may look somewhat different from
the above output. The F—test can be used to answer the following questions: 1. 2. uantilc Do two samples come from populations with equal standard deviations?
Does a new process. treatment... or test reduce the variability of the current process? " t‘tantilc P lot Bihistogram
Chi—Sq uare Test
Bartlett's Test
Levcnc il‘est Ceramic stren 3th data. The Ftest for equality of two Standard deviations is available in many
general purpose statistical software programs, including D2an lot. mom Wm M Warn«mail; {Tics mam treats a. 'Alos‘ {SEMEF Page 3 of 3 1/28/2012 ...
View
Full
Document
This note was uploaded on 02/02/2012 for the course EMA 3800 taught by Professor Elshall during the Spring '10 term at University of Florida.
 Spring '10
 ElShall

Click to edit the document details