COMPARING MORE THAN TWO POPULATION MEANS: AN ANALYSIS OF VARIANCE

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1 COMPARING MORE THAN TWO POPULATION MEANS: AN ANALYSIS OF VARIANCE To see how the piniple behind the analysis of vaiane method woks, let us onside the following simple expeiment. The means ( 1 and ) of two populations ae to be ompaed using independent andom samples of size n 1 = n = 5 fom eah of the populations. The sample obsevations and the sample means ae shown below: Sample fom Population 1 Sample fom Population x 1 = x = 3 Do you think these data povide suffiient evidene to indiate a diffeene between the population means 1 and? Now look at two moe samples of n 1 = n = 5 measuements fom the populations, as shown below. Do these data appea to povide evidene of a diffeene between 1 and? Sample fom Population 1 Sample fom Population x 1 = x = 3 One way to detemine whethe a diffeene exists between the population means 1 and is to examine the spead (o vaiation) between the sample means x 1 and x, and to ompae it to a measue of vaiability within the samples. The geate the diffeene in the vaiations, the geate will be the evidene to indiate a diffeene between 1 and. Fo the data in the fist table, you an see that the diffeene between the sample means is small elative to the vaiability within the sample obsevations. Thus, we think you will agee that the diffeene between x 1 and x is not lage enough to indiate a diffeene between 1 and. Notie that the diffeene between the sample means fo the data in the seond table is idential to the diffeene shown in the fist table. Howeve, sine thee is now no vaiability within the sample obsevations, the diffeene between the sample means is lage ompaed to the vaiability within the sample obsevations. Thus, the data appea to give lea evidene of a diffeene between 1 and. We apply the piniple of this example to the geneal poblem of ompaing k population means: If the vaiability among the k sample means is lage elative to the vaiability within the k samples, then thee is evidene to indiate that a diffeene exists among the k population means. 1

2 ONE-WAY ANALYSIS OF VARIANCE Suppose that we have samples of espetive sizes n 1, n,... n. Let SS b = the between sample sum of squaes, whih measues the vaiation between the sample means (also known as SST, the sum of squaes fo teatments, sine it measues the vaiation due to the diffeenes between teatments, o explained vaiation, sine it measues vaiation that might be due to the inheent diffeenes in the teatments), = n i ( x i x ) = n i x i N x whee x i, the sample mean = 1 n x ij, i = 1,..., i x, the oveall mean = 1 N n i x ij, whee N = n i SS w = the within sample sum of squaes, whih measues the vaiation within the samples (also known as SSE, the sum of squaes fo eos, o unexplained vaiation, sine it measues vaiation due to hane, fo whih no identifiable ause an be found), = n i (x ij x i ) = n i x ij n i n i x i Test statisti F 1,N = SS b / ( 1) with numeato degee of feedom = 1, and SS w / (N ) denominato degee of feedom = N. Example The hin-up soes of 1 hilden taking 3 types of s ae as follow: Chin-up soes nomal 3, 3, 3, 4 x 1 = 3.5 high potein, 4, 4, 5 x = 3.75 x = 3 vegetaian 1,,, 3 x 3 = Does the make a diffeene in the soe obtained? SS b = 4(3.5) + 4(3.75) + 4() 1(3) = = 6.5 SS w = = = 7.5 Test statisti F,9 = 6.5 / 7.5 / 9 = 3.9 Sine 3.9 < F 0.05,,9 = 4.6, we have insuffiient evidene at 5% level of signifiane to onlude that the make a diffeene in the soe obtained.

3 TWO-WAY ANALYSIS OF VARIANCE (WITHOUT INTERACTION) o TWO-WAY ANALYSIS OF VARIANCE FOR A RANDOMISED BLOCK DESIGN A Randomised Blok Design is a design in whih teatments ae ompaed within eah of bloks. Eah blok ontains mathed expeimental units and the teatments ae andomly assigned, one to eah of the units within eah blok. Let SS = the between ow sum of squaes whih measues the vaiation between the ow means (also known as SST, the sum of squaes of teatments, sine it measues the vaiation due to the diffeenes between teatments), = ( x i x ) = x i x whee x i, the mean of obsevations in ow i = 1 x ij, i = 1,..., x, the oveall mean = 1 x ij = 1 x i = 1 SS = the between olumn sum of squaes whih measues the vaiation between the olumn means (also known as SSB, the sum of squaes of bloks, sine it measues the vaiation due to the diffeenes between bloks), = ( x j x ) = x j x whee x j, the mean of obsevations in olumn j = 1 x ij, j = 1,..., x j SS total = the total sum of squaes = (x ij x ) = x ij x SS e = the eo sum of squaes whih measues the vaiation within the samples = (x ij x i x j + x ) = SS total SS SS SS / ( 1) Test statisti fo omputing ow effets F 1,( 1) ( 1) = SS e / ( 1)( 1) with numeato degee of feedom = 1, and denominato degee of feedom = ( 1)( 1). SS / ( 1) Test statisti fo omputing olumn effets F 1,( 1) ( 1) = SS e / ( 1)( 1) with numeato degee of feedom = 1, and denominato degee of feedom = ( 1)( 1). 3

4 Example The hin-up soes of fou hilden when taking diffeent types of s ae as follow: bloks Alan Bob Cal Dave teatments nomal x 1 = 3.5 high potein x = 3.75 vegetaian 1 3 x 3 = x 1 = x = 3 x 3 = 3 x 4 = 4 x = 3 SS = 4 ( ) 1 (3) = = 6.5 SS = 3 ( ) 1 (3) = = 6 SS total = (3) = = 14 SS e = SS total SS SS = = 1.5 Test statisti fo ompaing effets of F,6 = Test statisti fo ompaing effets of hild F 3,6 = 6.5 / 1.5 / 6 6 / / 6 = 13 = 8 Sine 13 > F 0.05,,6 = 5.14, we have suffiient evidene at 5% level of signifiane to onlude that the does make a diffeene in the soe obtained. Sine 8 > F 0.05,3,6 = 4.76, we have suffiient evidene at 5% level of signifiane to onlude that the hild does make a diffeene in the soe obtained. Note: (1) Although it is possible to alulate a value of F fo the olumns (bloks), this vaiable is usually of a seonday inteest and is used pimaily to allow fine disimination between ows (teatments). () SS e is smalle than in the ealie one-way analysis, beause muh of the fomely unexplained vaiation in the sample obsevations an now be attibuted to diffeenes in the hilden. We obtain a muh highe value of F fo ows (teatments) than befoe, allowing us to ejet the null hypothesis (equal ow/teatment population means) at a smalle level of signifiane than we ould in the one-way analysis. Two-way analysis of vaiane is thus moe effiient than one-way analysis. (3) This method assumes that the ow and olumn effets ae additive. That is, eah affeted all hilden equally. We did not allow fo the possibility that a patiula may inteat stongly with a patiula hild. 4

5 TWO-WAY ANALYSIS OF VARIANCE WITH INTERACTION o TWO-WAY ANALYSIS OF VARIANCE FOR A FACTORIAL DESIGN A Fatoial Design is one onduted to investigate the effet of two fatos on the mean value of a esponse vaiable. One fato has levels and the othe fato has levels, and n (the numbe of epliations) esponses ae measued fo eah of the fato level ombinations. Let SS = the ow sum of squaes = n ( x i x ) = n x i whee x i, the mean of obsevations in ow i = 1 n x, the oveall mean = SS = the olumn sum of squaes = n ( x j x ) = n x j n x 1 n n x ijk, i = 1,..., k=1 n x ijk k=1 n x whee x j, the mean of obsevations in olumn j = 1 n n x ijk, j = 1,..., k=1 SS int = the inteation sum of squaes = n ( x ij x i x j + x ) = n ( x ij x ) SS SS = n x ij n x SS SS whee x ij, the mean of obsevations in ow i, olumn j = 1 n n x ijk k=1 SS total = the total sum of squaes n n = (x ijk x ) = x ijk n x k=1 k=1 SS e = the eo sum of squaes = n k=1 (x ijk x ij ) = SS total SS SS SS int Test statisti fo deteting inteation F ( 1)( 1), (n 1) = SS int / ( 1)( 1) with numeato SS e / (n 1) degee of feedom = ( 1)( 1), and denominato degee of feedom = (n 1). SS / ( 1) Test statisti fo omputing ow effets F 1, (n 1) = with numeato SS e / (n 1) degee of feedom = 1, and denominato degee of feedom = (n 1). Test statisti fo omputing olumn effets F 1, (n 1) = SS / ( 1) with numeato SS e / (n 1) degee of feedom = 1, and denominato degee of feedom = (n 1). 5

6 Example The following epesents the hin-up soes of 6 gils and 6 boys taking 3 types of s: gils boys 1 3 vegetaian 4 x 11 = 1.5 x 1 = 3.5 x 1 = high potein 6 9 x = 5.5 x 1 = 8 x = nomal 9 6 x 31 = 8 x 3 = 5 x 3 = 6.5 x 1 = 5 x 1 = 5.5 x = 5.5 Is thee any evidene of an inteation effet? Do the data indiate that the and the sex of the hild do indeed affet the soe? SS = 4( ) 1(5.5) = 45.5 SS = 6( ) 1(5.5) = 0.75 SS int = ( ) 1(5.5) = 18.5 SS total = (5.5) = 7.5 SS e = = / Test statisti fo deteting inteation F,6 = 7.5 / 6 = 7.4 Sine 7.4 > F 0.05,,6 = we have suffiient evidene at 5% level of signifiane to onlude that thee is inteation between the and the sex of the hild boys gils 0 vegetaian high potein nomal Notie that the mean soes fo gils ae less than those fo boys taking vegetaian o high potein s. But the means ae evesed fo hilden taking a nomal. Sine the mean depends on the ombination of the fato levels, we say that the two fatos inteat. Note: The F tests fo fato effets ae usually elevant only when fato inteation is insignifiant. If inteation is deteted, we do not pefom the F tests fo fato effets. In fat, one of the most impotant objetives of a two-way analysis of vaiane with inteation is to detet fato inteation if it exists. 6

7 Example Suppose now the soes ae as shown below: gils boys 1 3 vegetaian 4 x 11 = 1.5 x 1 = 3.5 x 1 = high potein 6 9 x = 5.5 x 1 = 8 x = nomal 6 9 x 31 = 5 x 3 = 8 x 3 = 6.5 x 1 = 4 x 1 = 6.5 x = 5.5 Is thee any evidene of an inteation effet? Do the data indiate that the and the sex of the hild do indeed affet the soe? SS = 4( ) 1(5.5) = 45.5 SS = 6( ) 1(5.5) = SS int = ( ) 1(5.5) = 0.5 SS total = (5.5) = 7.5 SS e = = 7.5 Test statisti fo deteting inteation F,6 = 0.5 / 7.5 / 6 = 0. Sine 0. < F 0.05,,6 = we have insuffiient evidene at 5% level of signifiane to onlude that thee is inteation between the and the sex of the hild boys gils 0 vegetaian high potein nomal We now fous on testing the effets of the and the sex of the hild. Test statisti fo ompaing effets of F,6 = Test statisti fo ompaing effets of sex of hild F 1,6 = 45.5 / 7.5 / / / 6 = 18. = 15 Sine 18. > F 0.05,,6 = we have suffiient evidene at 5% level of signifiane to onlude that the does make a diffeene in the soe obtained. Sine 15 > F 0.05,1,6 = we have suffiient evidene at 5% level of signifiane to onlude that the sex of the hild does make a diffeene in the soe obtained. 7