©Richard Lowry, 1999-
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From the example introduced in Chapter 15, Part 1:

	A [0cps]	B [2cps]	C [6cps]	All groups combined
	Na=18 -Xai=466 -X2ai=    12800 Ma=25.9 SSa=735.8	Nb=18 -Xbi=485 -X2bi=    14021 Mb=26.9 SSb=952.9	Nc=18 -Xci=421 -X2ci=    10443 Mc=23.4 SSc=596.3	NT=54 -XTi=1372 -X2Ti=    37264 MT=25.4 SST=2405.0
	SST = 2405.0 SSwg =TSSa+SSb+SSc = 735.8+952.9+596.3 = 2285.0 SSbg =TSST—SSwg = 2405.0—2285.0 = 120.0

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¶Identification and Removal of SS_subj

   The first five columns of the following table are a somewhat streamlined version of the data table you saw in Part 1. The means for the individual subjects are now highlighted in blue, and next to each subject's mean, in a new column, is a calculation whose general form you will surely recognize from Chapter 14. The difference between each subject's mean and the mean of the total array of data is taken as a deviate:

	deviate = Msubj* — MT		We will use the subscript "subj*" to mean "any particular subject." The first subject would be subj1; the second, subj2; and so on.

The square of that difference is accordingly a squared deviate:

	squared deviate = (Msubj* — MT)2

And that squared deviate is then weighted by k=3, which is the number of measures (X_ai, X_bi, and X_ci) on which the subject's mean is based.

	weighted squared deviate = k(Msubj* — MT)2

   It is the same general structure we used in Chapter 14 to calculate the value of SS_bg from scratch, to measure the aggregate differences among the k groups. Now what we come out with is SS_subj, which is a measure of the aggregate differences among the 18 individual subjects.

Sub- jects	A	B	C		Subject Means
21 22 23 24 25 26 27 28 29 210 211 212 213 214 215 216 217 218	235 232 233 232 231 229 229 227 227 228 227 227 224 224 217 217 214 213	239 235 232 232 233 230 231 229 231 227 227 226 229 225 216 215 215 213	232 231 228 229 226 229 227 227 224 224 223 223 219 219 218 217 212 213		235.3  232.7  231.0  231.0  230.0  229.3  229.0  227.7  227.3  226.3  225.7  225.3  224.0  222.7  217.0  216.3  213.7  213.0	3(35.3—25.4)2=294 3(32.7—25.4)2=159.9 3(31.0—25.4)2=94.1 3(31.0—25.4)2=94.1 3(30.0—25.4)2=63.5 3(29.3—25.4)2=45.6 3(29.0—25.4)2=38.9 3(27.7—25.4)2=15.9 3(27.3—25.4)2=10.8 3(26.3—25.4)2=2.4 3(25.7—25.4)2=0.3 3(25.3—25.4)2=0 3(24.0—25.4)2=5.9 3(22.7—25.4)2=21.9 3(17.0—25.4)2=211.7 3(16.3—25.4)2=248.4 3(13.7—25.4)2=410.7 3(13.0—25.4)2=461.3
MT=25.4						SSsubj=2179.4

   I show you this particular method for calculating SS_subj only because it gives a clear idea of the conceptual structure of the measure. It would not be the method to use for practical computational purposes, as it is likely to end up with a considerable accumulation of rounding error. Also, if you are doing it by hand, it can be rather laborious. The following method is somewhat less laborious and will in any event minimize the risk of rounding errors. You will recognize it as analogous to the computational formula for SS_bg that we worked through in Part 1; namely

SSbg

=

(Xai)2 Na

+

(Xbi)2 Nb

+

(Xci)2 Nc

—

(XTi)2 NT

Except now the string of items to the left of the minus sign is replaced by another string that looks like this:

(Xsubj1)2 k

+

(Xsubj2)2 k

+

(Xsubj3)2 k

+

. . .

and so on, for as many subjects as you have in the analysis.

That is: For each subject in the analysis, take the sum of that subject's scores across all k conditions, square that sum, and then divide the result by k, which is the number of scores on which the sum is based.

Since k is the same for each subject, this portion of the formula to the left of the minus sign can be rewritten as

(Xsubj*)2 k

That is: For each subject in the analysis, take the sum of that subject's scores across all k conditions, square that sum, add up these squared sums across all subjects, and then divide the total of squared sums by k.

and the entire formula can then be written more compactly as

SSsubj

=

(Xsubj*)2 k

—

(XTi)2 NT

The next version of our data table shows the sum for each subject, along with the calculation of the square of each subject's sum.

Sub- jects	A	B	C		Subject Sums	(Xsubj*)2
21 22 23 24 25 26 27 28 29 210 211 212 213 214 215 216 217 218	235 232 233 232 231 229 229 227 227 228 227 227 224 224 217 217 214 213	239 235 232 232 233 230 231 229 231 227 227 226 229 225 216 215 215 213	232 231 228 229 226 229 227 227 224 224 223 223 219 219 218 217 212 213		2106  298  293  293  290  288  287  283  282  279  277  276  272  268  251  249  241  239	1062=11236    982=9604    932=8649    932=8649    902=8100    882=7744    872=7569    832=6889    822=6724    792=6241    772=5929    762=5776    722=5184    682=4624    512=2601    492=2401    412=1681    392=1521
(Xsubj*)2=111122

Recalling that X_Ti=1372 and N_T=54, our calculation of SS_subj is accordingly

	SSsubj	=	(Xsubj*)2 k	—	(XTi)2 NT
		=	111122 3	—	(1372)2 54
		=	2181.7

   As always, one must be careful not to get so lost in the calculations as to lose sight of the concepts. What we have just calculated with SS_subj is the amount of variability within the total array of data that derives from individual differences among the subjects. And now that this amount has been identified, it can be removed. The process of removal is one of simple subtraction. The total amount of within-groups variability is SS_wg=2285.0. Of this amount, SS_subj=2181.7 is attributal not to sheer, cussed random variability, but to extraneous pre-existing individual differences among the 18 subjects. Remove the latter from the former, and what remains is

	SSerror	= SSwg—SSsubj
		= 2285.0—2181.7
		= 103.3

   Here is one of the same diagrams you saw in Part 1, except now the specific numerical values have been included for the several values of SS. The bottom line is that a huge chunk of SS_wg is attributable to individual differences among the subjects, and that chunk is now removed. From this point on, it is all a smooth ride across familiar terrain.

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¶Calculation of MS_bg, MS_error, and F

   We begin by sorting out the several components of degrees of freedom. The first three are the same as we would have with a one-way independent-samples ANOVA:

	dfT = NT—1 = 54—1 = 53		Note that:T dfT = dfbg+dfwg
	dfbg = k—1 = 3—1 = 2
and
	dfwg = NT—k = 54—3 = 51

The only difference is that now we break df_wg into two further components, corresponding respectively to the two components of within-groups variability, SS_subj and SS_error. For df_subj the basic concept is once again "the number of items minus one." In this case the number of items is the number of subjects, which we will designate as N_subj. Thus

	dfsubj = Nsubj—1 = 18—1 = 17

Remove this component from df_wg, and what remains is

	dferror = dfwg—dfsubj = 51—17 = 34

With these, we can then calculate the two relevant MS values as

	MSbg	=	SSbg dfbg	=	120.0 2	= 60.0

and

	MSerror	=	SSerror dferror	=	103.3 34	= 3.0

which in turn give us

	F	=	MSbg MSerror	=	60.0 3.0	= 20.0    with df=2,34

   Figure 15.2 shows the sampling distribution of F for df=2,34, and the adjacent table shows the corresponding portion of Appendix D. As indicated, F=3.28 and F=5.29 mark the points in this distribution beyond which fall 5% and 1%, respectively, of all possible mere-chance outcomes, assuming the null hypothesis to be true.

Figure 15.2. Sampling Distribution of F for df=2,34

	df denomi- nator	df numerator
		1	2	3
	34	4.13 7.44	3.28 5.29	2.88 4.42

As the observed value of F=20.0 falls far to the right of F=5.29, the aggregate differences among the means of the three groups of measures, A|B|C, can be adjudged significant well beyond the .01 level. Our investigator can accordingly reject the null hypothesis, concluding with a high degree of confidence that the aggregate mean differences among the three groups of measures stem from something more than mere chance coincidence.