research article on cronbach s alpha as the mean of all...

Research ArticleOn Cronbachrsquos Alpha as the Mean of All Possible 119896-Split Alphas

Matthijs J Warrens

Institute of Psychology Unit of Methodology and Statistics Leiden University PO Box 9555 2300 RB Leiden The Netherlands

Correspondence should be addressed to Matthijs J Warrens warrensfswleidenunivnl

Received 13 July 2014 Accepted 17 September 2014 Published 30 September 2014

Academic Editor Steve Su

Copyright copy 2014 Matthijs J Warrens This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Coefficient alpha is the most commonly used internal consistency reliability coefficient Alpha is the mean of all possible 119896-splitalphas if the items are divided into 119896 parts of equal size This result gives proper interpretations of alpha interpretations that alsohold if (some of) its assumptions are not valid Here we consider the cases where the items cannot be split into parts of equal sizeIt is shown that if a 119896-split is made such that the items are divided as evenly as possible the difference between alpha and the meanof all possible 119896-split alphas can be made arbitrarily small by increasing the number of items

1 Introduction

In test theory test scores are used to summarize the perfor-mance of participants on a test An important concept of atest score is its reliability which concerns the precision ofthe administration of a participantrsquos score Reliability can beconceptualized in different ways In laymanrsquos terms a testscore is said to be reliable if it produces similar outcomesfor participants under similar administration conditions Amore formal definition of reliability comes from classicaltest theory namely reliability is defined as the ratio of thetrue score variance and the total score variance [1ndash3] Thetrue score variance cannot be observed directly Thereforeit has to be estimated from the data If there is only onetest administration reliability can be estimated using the so-called internal consistency reliability coefficients [4] Theinternal consistency coefficient that is most commonly usedin psychology and other behavioral sciences is coefficientalpha [5ndash10]

Coefficient alpha was already discussed in Guttman [11]The coefficient was later popularized by Cronbach [6] Sincethen it has been applied in thousands of research studies [59] It has been pointed out that alpha is not a measure of theone-dimensionality of a test score [6 9 12] Furthermore aproblem is that the value of alpha is affected by the length ofthe test Alpha is often high if the test consists of 15 items ormore [5 13 14] Moreover the derivation of alpha is based onseveral assumptions from classical test theory For example

alpha is an estimate of the reliability if the items of the testare essentially tau-equivalent [1 2] These assumptions maynot hold in practice Various authors have therefore studiedalpharsquos robustness to violations of its assumptions [12 15ndash17]

Other authors have presented results that give a meaningto alpha and to its estimate from a sample in case itsassumptions do not hold [6 18] The items of a test can besplit into two parts This will be called a 2-split For eachpart the sum score of the items can be determined If weapply Cronbachrsquos alpha to the two sum scores we obtainthe Flanagan-Rulon split-half estimate of the reliability [1920] The split-half reliability is an alternative approach ofestimating the reliability of a test when there is only oneadministration [3 7 8] The problem with a 2-split of a testis that there are multiple ways to divide the items into twoparts The value of the 2-split alpha therefore depends on theway the 2-splits are made [7] Cronbach [6] showed that ifthe 2-splits are made such that the parts have equal size thenthe overall alpha is the mean of all possible 2-split alphasCronbachrsquos result was generalized by Raju [18] to other typesof splits Let 119896 ge 2 be an integer Raju [18] showed that if the119896-split is such that the 119896 parts have the same number of itemsthen alpha is the mean of all possible 119896-split alphas

The algebraic results by Cronbach [6] and Raju [18]are important since they provide a proper interpretation ofalpha and of its estimate from a sample in the case that itsassumptions do not hold [2 5] Thus if the number of itemsis a composite number 119899 then there exists an interpretation of

Hindawi Publishing CorporationAdvances in StatisticsVolume 2014 Article ID 742863 5 pageshttpdxdoiorg1011552014742863

2 Advances in Statistics

alpha for any proper divisor of 119899 Raju [18] also showed that ifthe 119896 parts do not have equal size then the mean of all 119896-splitalphas is strictly smaller than the overall alpha Moreoverthis difference can be substantial [18] Thus at present it isunknown if there is an (approximate) interpretation of alphaas a mean of 119896-split alphas in case the number of items is aprime number This is the topic of investigation of this paper

The paper is organized as follows Notation and Defini-tions are presented in Section 2 We also present a formulaof the mean of all possible 119896-split alphas in Section 2 InSection 3 the nonnegative difference between alpha and themean of all 119896-split alphas is studied It is shown that ifwe divide the items as evenly as possible the difference isstrictly decreasing in the number of items Furthermore thedifference goes to zero if the number of items increasesHence the difference can be made arbitrarily small byincreasing the number of items In Section 4 we determinefor three criteria and for small values of 119896 how many itemsare needed Finally Section 5 contains a conclusion

2 Notation and Definitions

Suppose we have a test that consists of 119899 ge 2 items Let 1205901198941198941015840

for 1 le 119894 1198941015840 le 119899 denote the covariance between items 119894 and1198941015840 and let 1205902

119883denote the variance of the total score Guttmanrsquos

lambda3 or Cronbachrsquos alpha is defined as

120572 =119899

119899 minus 1sdotsum119894 =1198941015840 1205901198941198941015840

1205902119883

(1)

In the summation sum119894 =1198941015840 1205901198941198941015840 both counters 119894 and 1198941015840 run from 1

to 119899 but they are never equalNext suppose we split the test into 119896 parts of sizes1198991 1198992 119899

119896with 1 le 119899

119895lt 119899 for 1 le 119895 le 119896 and

119896

sum

119895=1

119899119895= 119899 (2)

Let 119901119895= 119899119895119899 for 1 le 119895 le 119896 denote the proportion of items

in part 119895 Furthermore let 1205901198951198951015840 denote the covariance between

the sum scores of parts 119895 and 1198951015840 For a 119896-split coefficient alphais defined as [18]

120572119896=119896

119896 minus 1sdotsum119895 =1198951015840 1205901198951198951015840

1205902119883

(3)

We have 120572119899= 120572 Furthermore for fixed 119896 and 119899

1 1198992 119899

119896

the number of different 119896-splits is given by [21 page 823]

119878 =119899

11989911198992 sdot sdot sdot 119899119896 (4)

For each 119896-split there is an associated 120572119896 The mean of all 119896-

split alphas is denoted by 119864(120572119896) An expression for 119864(120572

119896) is

presented in Lemma 1 The result is not derived in Raju [18]and is believed to be new

Lemma 1 It holds that

119864 (120572119896) =119899

119899 minus 1sdot119896

119896 minus 1sdot(sum119895 =1198951015840 1199011198951199011198951015840) (sum119894 =1198941015840 1205901198941198941015840)

1205902119883

(5)

Proof The variance of the total score 1205902119883is not affected by the

119896-split Furthermore 119896(119896minus1) is fixed since 119896 is fixed Hence

119864 (120572119896) =119896

119896 minus 1sdot119864 (sum119895 =1198951015840 1205901198951198951015840)

1205902119883

(6)

where using (4)

119864(sum

119895 =1198951015840

1205901198951198951015840) =1

119878

119878

sum

ℓ=1

(sum

119895 =1198951015840

1205901198951198951015840) (7)

Wewant to determine how often each covariance1205901198941198941015840 between

items 119894 and 1198941015840 occurs in the summation on the right-hand sideof (7) Raju [18 page 559] showed that this amount is givenby

2119878 sdot119899

119899 minus 1sdot sum

119895 =1198951015840

1199011198951199011198951015840 (8)

Since each covariance 1205901198941198941015840 occurs twice in the summation

sum119894 =1198941015840 1205901198941198941015840 (7) is given by

119864(sum

119895 =1198951015840

1205901198951198951015840) =119899

119899 minus 1(sum

119895 =1198951015840

1199011198951199011198951015840)(sum

119894 =1198941015840

1205901198941198941015840) (9)

Using (9) in (6) we obtain (5)

3 Difference between 120572 and 119864(120572119896)

In this section we study the difference between alpha and themean of all 119896-split alphas Using (1) we can write (5) as

119864 (120572119896) = 120572 sdot

119896

119896 minus 1sum

119895 =1198951015840

1199011198951199011198951015840 (10)

Cronbach [6] showed that 120572 = 119864(1205722) if 1199011= 1199012 A proof

can also be found in Lord and Novick [1] Raju [18 page 560]showed that 120572 = 119864(120572

119896) if 1199011= 1199012= sdot sdot sdot = 119901

119896and that 120572 gt

119864(120572119896) if 1199011 1199012 119901

119896are not all equal Hence the difference

120572 minus 119864 (120572119896) = 120572(1 minus

119896

119896 minus 1sum

119895 =1198951015840

1199011198951199011198951015840) (11)

is nonnegative Difference (11) can be used to study how closethe values of alpha and themean of all 119896-split alphas are Since0 le 120572 le 1 we have the inequality

120572 minus 119864 (120572119896) le 1 minus

119896

119896 minus 1sum

119895 =1198951015840

1199011198951199011198951015840 (12)

The right-hand side of (12) is an upper bound of 120572 minus 119864(120572119896)

for all values of 120572 (0 le 120572 le 1) It only depends on 119896 and theproportions of items in the parts

It turns out that difference (11) can be quite large [18]This is the case when the variance between 119901

1 1199012 119901

119896is

large that is if 1199011 1199012 119901

119896are very different To illustrate

this property of difference (11) suppose we have119898+ 119896 items

Advances in Statistics 3

where 119898 ge 3 is an integer Furthermore suppose wedistribute the items such that we have 119896 minus 1 parts with 1 itemand 1 part with119898+1 itemsThe proportions of items per partare given by

1199011=119898 + 1

119898 + 119896 119901

2= sdot sdot sdot = 119901

119896=1

119898 + 119896 (13)

Using the proportions in (13) we have

sum

119894 =119895

119901119894119901119895=(119896 minus 1) (119896 minus 2)

(119898 + 119896)2+2 (119896 minus 1) (119898 + 1)

(119898 + 119896)2 (14)

Due to the1198982-term in the denominator of the right-hand sideof (14) we have for fixed 119896 lt 119898 that sum

119894 =119895119901119894119901119895rarr 0 as119898 rarr

infinIn this paper we are not interested in an arbitrary 119896-split

of the items but in a 119896-split in which the items are distributedas evenly as possible over the parts Suppose we have 119896119898 + 119888items where 119898 119888 ge 1 are integers with 0 lt 119888 lt 119896 If we wantto distribute the items as evenly as possible over the 119896 partsan optimal 119896-split is to have 119896 minus 119888 parts with 119898 items and 119888parts with 119898 + 1 items In this case the proportions of itemsper part are given by

1199011= sdot sdot sdot = 119901

119888=119898 + 1

119896119898 + 119888 119901

119888+1= sdot sdot sdot = 119901

119896=119898

119896119898 + 119888

(15)

Lemma 2 shows that for a split of the type in (15) the quantitysum119895 =1198951015840 1199011198951199011198951015840 approaches the ratio (119896 minus 1)119896 as the counter 119898

goes to infinity


sum

119894 =119895

119901119894119901119895997888rarr119896 minus 1

119896as 119898 rarr infin (16)

Proof Using the proportions in (15) we have the identity

sum

119894 =119895

119901119894119901119895=119888 (119888 minus 1) (119898 + 1)

2

(119896119898 + 119888)2+2119888 (119896 minus 119888)119898 (119898 + 1)

(119896119898 + 119888)2

+(119896 minus 119888 minus 1) (119896 minus 119888)119898

2

(119896119898 + 119888)2

(17)

Using the identity

119888 (119888 minus 1) + 2119888 (119896 minus 119888) + (119896 minus 119888 minus 1) (119896 minus 119888) = 119896 (119896 minus 1)

(18)

we can write (17) as

sum

119894 =119895

119901119894119901119895=119888 (119888 minus 1) (2119898 + 1) + 2119888 (119896 minus 119888)119898 + 119896 (119896 minus 1)119898

2

(119896119898 + 119888)2

=119888 (119888 minus 1) + 2119888 (119896 minus 1)119898 + 119896 (119896 minus 1)119898

2

(119896119898 + 119888)2

=119888 (119888 minus 1) + (119896 minus 1) (2119888119898 + 119896119898

2)

(119896119898 + 119888)2

(19)

For119898 rarr infin we have

sum

119894 =119895

119901119894119901119895997888rarr119896 (119896 minus 1)

1198962=119896 minus 1

119896 (20)

It follows from Lemma 2 that the right-hand-side of (12)and thus the difference 120572minus119864(120572

119896) goes to zero as119898 increases

In other words if the 119896-split is of the type in (15) thedifference between alpha and the mean of the best 119896-splitsvanishes as the number of items becomes large It is not clearfromLemma 2 thatsum

119895 =1198951015840 1199011198951199011198951015840 is strictly increasing in119898This

is shown in Lemma 3

Lemma 3 For split (15)sum119895 =1198951015840 1199011198951199011198951015840 is strictly increasing in119898

Proof Using (19) the first order partial derivative is

120597sum119894 =119895119901119894119901119895

120597119898

=2 (119896 minus 1) (119896119898 + 119888)

2minus[119888 (119888 minus 1) +(119896 minus 1) (2119888119898 + 119896119898

2)] 2119896

(119896119898 + 119888)3

=2119888 (119896 minus 119888)

(119896119898 + 119888)3

(21)

Since 119896 minus 119888 gt 0 (21) is strictly positive and it follows thatsum119895 =1198951015840 1199011198951199011198951015840 is strictly increasing in119898

4 Numerical Results

Lemmas 2 and 3 from the previous section show that thenonnegative difference 120572 minus 119864(120572

119896) decreases strictly to zero

as 119898 increases if the 119896-split is of the type in (15) Hencefor sufficiently large 119898 the difference 120572 minus 119864(120572

119896) can be

made smaller than the specified criteria In this section wedetermine the numbers of items that are needed so that thedifference 120572 minus 119864(120572

119896) is less than 0010 0005 and 0001 Since

a difference of 0010 is negligible for most practical purposeswe may say that 120572 is approximately equal to 119864(120572

119896) if the

difference is less than 0010 that is if

120572 minus 119864 (120572119896) le 0010 (22)

The values 0005 and 0001 can be used if one wants a strongerdefinition of approximately equal

If we are interested in a 119896-split but the number of itemsis not a multiple of 119896 there are 119896 minus 1 different scenarios toconsider For example if 119896 = 3 but the number of items is nota multiple of 3 we either have 3119898 + 1 items or have 3119898 + 2items Inequality (22) may hold for different119898 depending onwhether we have 3119898 + 1 items or 3119898 + 2 items We thereforeconsider the number of items of all 119896 minus 1 variants of a 119896-splitseparately For all possibilities of 119896 and 119888 with 2 le 119896 le 7 and1 le 119888 lt 119896 we applied the following numerical strategy Forfixed 119896 and 119888 we increased 119898 in steps of 1 starting at 1 untilinequality (22) was satisfiedThe results for 119896 = 2 3 7 are


Table 1 The minimum number of items per split variant for threecriteria

Split type Split variant Minimum number of itemsle0010 le0005 le0001

2-split 2119898 + 1 11 15 33

3-split 3119898 + 1 10 16 343119898 + 2 11 17 32

4-split4119898 + 1 13 17 334119898 + 2 14 18 384119898 + 3 11 15 35

5-split

5119898 + 1 11 16 365119898 + 2 17 22 425119898 + 3 13 18 435119898 + 4 14 19 34

6-split

6119898 + 1 13 19 376119898 + 2 14 20 446119898 + 3 15 21 456119898 + 4 16 22 466119898 + 5 11 17 35

7-split

7119898 + 1 15 15 367119898 + 2 16 23 447119898 + 3 17 24 457119898 + 4 18 25 467119898 + 5 19 19 477119898 + 6 13 20 34

presented in Table 1The table presents for each variant of a 119896-split the minimum number of items for which the difference120572minus119864(120572

119896) isle0010 (column 3)le0005 (column 4) andle0001

(column 5)Table 2 summarizes for each 119896-split the minimum num-

ber of items required per criteria For example consider thecase 119896 = 2 If the number of items is even we have a perfect2-split and 120572 = 119864(120572

2) If the number of items is odd and we

distribute the items as evenly as possible Table 1 shows thatthe difference 120572 minus 119864(120572

2) is no more than 0010 if we have at

least 11 items no more than 0005 if we have at least 15 itemsand no more than 0001 if we have at least 33 items Hencewe may conclude that in the case that we distribute the itemsas evenly as possible the difference 120572 minus 119864(120572

2) is no more than

0010 if we have at least 10 items no more than 0005 if wehave at least 14 items and no more than 0001 if we have atleast 32 items These numbers are presented in Table 2

As another example consider the case 119896 = 3 If we have3119898 items we have a perfect 3-split and 120572 = 119864(120572

3) If we have

3119898 + 1 items Table 1 shows that the difference 120572 minus 119864(1205723) is

no more than 0010 if we have at least 10 items no more than0005 if we have at least 16 items and nomore than 0001 if wehave at least 34 items If we have 3119898 + 2 items Table 1 showsthat the difference120572minus119864(120572

3) is nomore than 0010 if we have at

least 11 items no more than 0005 if we have at least 17 itemsand no more than 0001 if we have at least 32 items Hencewe may conclude that in the case that we distribute the itemsas evenly as possible the difference 120572minus119864(120572

3) is no more than

0010 if we have at least 9 items nomore than 0005 if we have

Table 2 The minimum number of items per split type for threecriteria

Split type Minimum number of itemsle0010 le0005 le0001

2-split 10 14 323-split 9 15 324-split 11 15 355-split 13 18 396-split 11 17 417-split 13 19 41

at least 15 items and no more than 0001 if we have at least 32items These numbers are presented in Table 2

5 Conclusion

Alpha is the most commonly used coefficient for estimatingreliability of a test score if there is only one test administration[5ndash9] Cronbach [6] and Raju [18] have presented results thatgive a meaning to alpha and to its estimate from a sampleeven if its assumptions do not hold Cronbach [6] showed thatif the 2-splits are made such that the parts have equal sizethen alpha is the mean of all possible 2-split alphas Raju [18]showed that if a 119896-split for 119896 ge 2 is made such that the 119896 partshave the same number of items then alpha is the mean of allpossible 119896-split alphas The importance of these results lies inthe fact that they provide a proper interpretation of alpha if(some of) its assumptions do not hold [2 5]

In this paper we consider the cases where the itemscannot be split into parts of equal size In these cases thevalue of alpha exceeds the mean of all possible 119896-split alphas[18] However it was shown that if a 119896-split is made suchthat the items are divided as evenly as possible the differencebetween alpha and the mean of all possible 119896-split alphasis negligible for sufficiently many items (Lemmas 2 and 3)Using numerical simulations it was shown in the case of a2-split that the difference is no more than 0010 if we have atleast 10 items no more than 0005 if we have at least 14 itemsand nomore than 0001 if we have at least 32 items and in thecase of a 3-split that the difference is nomore than 0010 if wehave at least 9 items no more than 0005 if we have at least 15items and no more than 0001 if we have at least 32 items

The results in Cronbach [6] and Raju [18] provide inter-pretations of alpha if the number of items is a compositenumber 119899 for any proper divisor of 119899 The results andsimulations in this paper provide interpretations of alpha ifthe number of items is prime

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This research was done while the author was funded bythe Netherlands Organisation for Scientific Research VeniProject 451-11-026


References

[1] F M Lord and M R Novick Statistical Theories of Mental TestScores Addison-Wesley Reading Mass USA 1968

[2] R P McDonald Test Theory A Unified Treatment ErlbaumMahwah NJ USA 1999

[3] W Revelle and R E Zinbarg ldquoCoefficients alpha beta omegaand the glb comments on Sijtsmardquo Psychometrika vol 74 no1 pp 145ndash154 2009

[4] H G Osburn ldquoCoefficient alpha and related internal consis-tency reliability coefficientsrdquo Psychological Methods vol 5 no3 pp 343ndash355 2000

[5] J M Cortina ldquoWhat is coefficient alpha An examination oftheory and applicationsrdquo Journal of Applied Psychology vol 78no 1 pp 98ndash104 1993

[6] L J Cronbach ldquoCoefficient alpha and the internal structure oftestsrdquo Psychometrika vol 16 no 3 pp 297ndash334 1951

[7] A Field Discovering Statistics Using SPSS Sage Los AngelesCalif USA 3rd edition 2009

[8] RM Furr and V R Bacharach Psychometrics An IntroductionSage Los Angeles Calif USA 2008

[9] K Sijtsma ldquoOn the use the misuse and the very limitedusefulness of Cronbachrsquos alphardquoPsychometrika vol 74 no 1 pp107ndash120 2009

[10] M J Warrens ldquoSome relationships between Cronbachrsquos alphaand the Spearman-Brown formulardquo Journal of Classification Inpress

[11] L Guttman ldquoA basis for analyzing test-retest reliabilityrdquo Psy-chometrika vol 10 pp 255ndash282 1945

[12] D Grayson ldquoSome myths and legends in quantitative psychol-ogyrdquo Understanding Statistics vol 3 no 2 pp 101ndash134 2004

[13] A P Keszei M Novak and D L Streiner ldquoIntroduction tohealth measurement scalesrdquo Journal of Psychosomatic Researchvol 68 no 4 pp 319ndash323 2010

[14] M Tavakol and R Dennick ldquoMaking sense of Cronbachrsquosalphardquo International Journal ofMedical Education vol 2 pp 53ndash55 2011

[15] S B Green and S L Hershberger ldquoCorrelated errors in truescore models and their effect on coefficient alphardquo StructuralEquation Modeling vol 7 no 2 pp 251ndash270 2000

[16] S B Green and Y Yang ldquoCommentary on coefficient alpha acautionary talerdquo Psychometrika vol 74 no 1 pp 121ndash135 2009

[17] Y Sheng and Z Sheng ldquoIs coefficient alpha robust to non-normal datardquo Frontiers in Psychology vol 3 article 34 2012

[18] N S Raju ldquoA generalization of coefficient alphardquo Psychome-trika vol 42 no 4 pp 549ndash565 1977

[19] J C Flanagan ldquoA proposed procedure for increasing theefficiency of objective testsrdquo Journal of Educational Psychologyvol 28 no 1 pp 17ndash21 1937

[20] P J Rulon ldquoA simplified procedure for determining the reliabil-ity of a test by split-halvesrdquoHarvard Educational Review vol 9pp 99ndash103 1939

[21] M Abramowitz and I A Stegun Handbook of MathematicalFunctions with Formulas Graphs and Mathematical TablesDover New York NY USA 1970

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alpha for any proper divisor of 119899 Raju [18] also showed that ifthe 119896 parts do not have equal size then the mean of all 119896-splitalphas is strictly smaller than the overall alpha Moreoverthis difference can be substantial [18] Thus at present it isunknown if there is an (approximate) interpretation of alphaas a mean of 119896-split alphas in case the number of items is aprime number This is the topic of investigation of this paper

The paper is organized as follows Notation and Defini-tions are presented in Section 2 We also present a formulaof the mean of all possible 119896-split alphas in Section 2 InSection 3 the nonnegative difference between alpha and themean of all 119896-split alphas is studied It is shown that ifwe divide the items as evenly as possible the difference isstrictly decreasing in the number of items Furthermore thedifference goes to zero if the number of items increasesHence the difference can be made arbitrarily small byincreasing the number of items In Section 4 we determinefor three criteria and for small values of 119896 how many itemsare needed Finally Section 5 contains a conclusion

2 Notation and Definitions

Suppose we have a test that consists of 119899 ge 2 items Let 1205901198941198941015840

for 1 le 119894 1198941015840 le 119899 denote the covariance between items 119894 and1198941015840 and let 1205902

119883denote the variance of the total score Guttmanrsquos

lambda3 or Cronbachrsquos alpha is defined as

120572 =119899

119899 minus 1sdotsum119894 =1198941015840 1205901198941198941015840

1205902119883

(1)

In the summation sum119894 =1198941015840 1205901198941198941015840 both counters 119894 and 1198941015840 run from 1

to 119899 but they are never equalNext suppose we split the test into 119896 parts of sizes1198991 1198992 119899

119896with 1 le 119899

119895lt 119899 for 1 le 119895 le 119896 and

119896

sum

119895=1

119899119895= 119899 (2)

Let 119901119895= 119899119895119899 for 1 le 119895 le 119896 denote the proportion of items

in part 119895 Furthermore let 1205901198951198951015840 denote the covariance between

the sum scores of parts 119895 and 1198951015840 For a 119896-split coefficient alphais defined as [18]

120572119896=119896

119896 minus 1sdotsum119895 =1198951015840 1205901198951198951015840

1205902119883

(3)

We have 120572119899= 120572 Furthermore for fixed 119896 and 119899

1 1198992 119899

119896

the number of different 119896-splits is given by [21 page 823]

119878 =119899

11989911198992 sdot sdot sdot 119899119896 (4)

For each 119896-split there is an associated 120572119896 The mean of all 119896-

split alphas is denoted by 119864(120572119896) An expression for 119864(120572

119896) is

presented in Lemma 1 The result is not derived in Raju [18]and is believed to be new


119864 (120572119896) =119899

119899 minus 1sdot119896

119896 minus 1sdot(sum119895 =1198951015840 1199011198951199011198951015840) (sum119894 =1198941015840 1205901198941198941015840)

1205902119883

(5)

Proof The variance of the total score 1205902119883is not affected by the

119896-split Furthermore 119896(119896minus1) is fixed since 119896 is fixed Hence

119864 (120572119896) =119896

119896 minus 1sdot119864 (sum119895 =1198951015840 1205901198951198951015840)

1205902119883

(6)

where using (4)

119864(sum

119895 =1198951015840

1205901198951198951015840) =1

119878

119878

sum

ℓ=1

(sum

119895 =1198951015840

1205901198951198951015840) (7)

Wewant to determine how often each covariance1205901198941198941015840 between

items 119894 and 1198941015840 occurs in the summation on the right-hand sideof (7) Raju [18 page 559] showed that this amount is givenby

2119878 sdot119899

119899 minus 1sdot sum

119895 =1198951015840

1199011198951199011198951015840 (8)

Since each covariance 1205901198941198941015840 occurs twice in the summation

sum119894 =1198941015840 1205901198941198941015840 (7) is given by

119864(sum

119895 =1198951015840

1205901198951198951015840) =119899

119899 minus 1(sum

119895 =1198951015840

1199011198951199011198951015840)(sum

119894 =1198941015840

1205901198941198941015840) (9)

Using (9) in (6) we obtain (5)

3 Difference between 120572 and 119864(120572119896)

In this section we study the difference between alpha and themean of all 119896-split alphas Using (1) we can write (5) as

119864 (120572119896) = 120572 sdot

119896

119896 minus 1sum

119895 =1198951015840

1199011198951199011198951015840 (10)

Cronbach [6] showed that 120572 = 119864(1205722) if 1199011= 1199012 A proof

can also be found in Lord and Novick [1] Raju [18 page 560]showed that 120572 = 119864(120572

119896) if 1199011= 1199012= sdot sdot sdot = 119901

119896and that 120572 gt

119864(120572119896) if 1199011 1199012 119901

119896are not all equal Hence the difference

120572 minus 119864 (120572119896) = 120572(1 minus

119896

119896 minus 1sum

119895 =1198951015840

1199011198951199011198951015840) (11)

is nonnegative Difference (11) can be used to study how closethe values of alpha and themean of all 119896-split alphas are Since0 le 120572 le 1 we have the inequality

120572 minus 119864 (120572119896) le 1 minus

119896

119896 minus 1sum

119895 =1198951015840

1199011198951199011198951015840 (12)

The right-hand side of (12) is an upper bound of 120572 minus 119864(120572119896)

for all values of 120572 (0 le 120572 le 1) It only depends on 119896 and theproportions of items in the parts

It turns out that difference (11) can be quite large [18]This is the case when the variance between 119901

1 1199012 119901

119896is

large that is if 1199011 1199012 119901

119896are very different To illustrate

this property of difference (11) suppose we have119898+ 119896 items



1199011=119898 + 1

119898 + 119896 119901


119896=1

119898 + 119896 (13)


sum

119894 =119895

119901119894119901119895=(119896 minus 1) (119896 minus 2)

(119898 + 119896)2+2 (119896 minus 1) (119898 + 1)

(119898 + 119896)2 (14)


119894 =119895119901119894119901119895rarr 0 as119898 rarr



1199011= sdot sdot sdot = 119901

119888=119898 + 1

119896119898 + 119888 119901

119888+1= sdot sdot sdot = 119901

119896=119898

119896119898 + 119888

(15)


goes to infinity


sum

119894 =119895

119901119894119901119895997888rarr119896 minus 1

119896as 119898 rarr infin (16)


sum

119894 =119895

119901119894119901119895=119888 (119888 minus 1) (119898 + 1)

2

(119896119898 + 119888)2+2119888 (119896 minus 119888)119898 (119898 + 1)

(119896119898 + 119888)2


2

(119896119898 + 119888)2

(17)

Using the identity


(18)


sum

119894 =119895


2

(119896119898 + 119888)2


2

(119896119898 + 119888)2

=119888 (119888 minus 1) + (119896 minus 1) (2119888119898 + 119896119898

2)

(119896119898 + 119888)2

(19)


sum

119894 =119895

119901119894119901119895997888rarr119896 (119896 minus 1)

1198962=119896 minus 1

119896 (20)





is shown in Lemma 3



120597sum119894 =119895119901119894119901119895

120597119898

=2 (119896 minus 1) (119896119898 + 119888)


2)] 2119896

(119896119898 + 119888)3

=2119888 (119896 minus 119888)

(119896119898 + 119888)3

(21)


4 Numerical Results




119896) can be




119896) if the


120572 minus 119864 (120572119896) le 0010 (22)






2-split 2119898 + 1 11 15 33

3-split 3119898 + 1 10 16 343119898 + 2 11 17 32

4-split4119898 + 1 13 17 334119898 + 2 14 18 384119898 + 3 11 15 35

5-split

5119898 + 1 11 16 365119898 + 2 17 22 425119898 + 3 13 18 435119898 + 4 14 19 34

6-split

6119898 + 1 13 19 376119898 + 2 14 20 446119898 + 3 15 21 456119898 + 4 16 22 466119898 + 5 11 17 35

7-split

7119898 + 1 15 15 367119898 + 2 16 23 447119898 + 3 17 24 457119898 + 4 18 25 467119898 + 5 19 19 477119898 + 6 13 20 34









2) is no more than



3) If we have





3) is no more than






5 Conclusion






Acknowledgment



References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











1199011=119898 + 1

119898 + 119896 119901


119896=1

119898 + 119896 (13)


sum

119894 =119895

119901119894119901119895=(119896 minus 1) (119896 minus 2)

(119898 + 119896)2+2 (119896 minus 1) (119898 + 1)

(119898 + 119896)2 (14)


119894 =119895119901119894119901119895rarr 0 as119898 rarr



1199011= sdot sdot sdot = 119901

119888=119898 + 1

119896119898 + 119888 119901

119888+1= sdot sdot sdot = 119901

119896=119898

119896119898 + 119888

(15)


goes to infinity


sum

119894 =119895

119901119894119901119895997888rarr119896 minus 1

119896as 119898 rarr infin (16)


sum

119894 =119895

119901119894119901119895=119888 (119888 minus 1) (119898 + 1)

2

(119896119898 + 119888)2+2119888 (119896 minus 119888)119898 (119898 + 1)

(119896119898 + 119888)2


2

(119896119898 + 119888)2

(17)

Using the identity


(18)


sum

119894 =119895


2

(119896119898 + 119888)2


2

(119896119898 + 119888)2

=119888 (119888 minus 1) + (119896 minus 1) (2119888119898 + 119896119898

2)

(119896119898 + 119888)2

(19)


sum

119894 =119895

119901119894119901119895997888rarr119896 (119896 minus 1)

1198962=119896 minus 1

119896 (20)





is shown in Lemma 3



120597sum119894 =119895119901119894119901119895

120597119898

=2 (119896 minus 1) (119896119898 + 119888)


2)] 2119896

(119896119898 + 119888)3

=2119888 (119896 minus 119888)

(119896119898 + 119888)3

(21)


4 Numerical Results




119896) can be




119896) if the


120572 minus 119864 (120572119896) le 0010 (22)






2-split 2119898 + 1 11 15 33

3-split 3119898 + 1 10 16 343119898 + 2 11 17 32

4-split4119898 + 1 13 17 334119898 + 2 14 18 384119898 + 3 11 15 35

5-split

5119898 + 1 11 16 365119898 + 2 17 22 425119898 + 3 13 18 435119898 + 4 14 19 34

6-split

6119898 + 1 13 19 376119898 + 2 14 20 446119898 + 3 15 21 456119898 + 4 16 22 466119898 + 5 11 17 35

7-split

7119898 + 1 15 15 367119898 + 2 16 23 447119898 + 3 17 24 457119898 + 4 18 25 467119898 + 5 19 19 477119898 + 6 13 20 34









2) is no more than



3) If we have





3) is no more than






5 Conclusion






Acknowledgment



References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












2-split 2119898 + 1 11 15 33

3-split 3119898 + 1 10 16 343119898 + 2 11 17 32

4-split4119898 + 1 13 17 334119898 + 2 14 18 384119898 + 3 11 15 35

5-split

5119898 + 1 11 16 365119898 + 2 17 22 425119898 + 3 13 18 435119898 + 4 14 19 34

6-split

6119898 + 1 13 19 376119898 + 2 14 20 446119898 + 3 15 21 456119898 + 4 16 22 466119898 + 5 11 17 35

7-split

7119898 + 1 15 15 367119898 + 2 16 23 447119898 + 3 17 24 457119898 + 4 18 25 467119898 + 5 19 19 477119898 + 6 13 20 34









2) is no more than



3) If we have





3) is no more than






5 Conclusion






Acknowledgment



References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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