home *** CD-ROM | disk | FTP | other *** search
- .MCD 20000 0
- .CMD PLOTFORMAT logs=0,0 subdivs=1,1 size=5,15 type=l
- .CMD FORMAT rd=d ct=10 im=i et=3 zt=15 pr=3 mass length time charge
- .CMD SET ORIGIN 0
- .CMD SET TOL 0.001000
- .CMD MARGIN 0
- .CMD LINELENGTH 78
- .CMD SET PRNCOLWIDTH 8
- .CMD SET PRNPRECISION 4
- .TXT 0 39 1 39
- a1,38,39,37
- Copyright (c) 1988 by MathSoft, Inc.
- .TXT 1 43 1 41
- a1,40,78,39
- /EQUATIONS FOR BUILDING VARIANCE TABLE
- .TXT 1 -57 1 31
- a1,30,78,29
- ONE-WAY ANALYSIS OF VARIANCE
- .TXT 2 -23 2 74
- a2,73,76,117
- This application calculates within and among group sums of squares for
- carrying out a one-way analysis of variance.
- .TXT 3 -1 1 16
- a1,15,78,14
- read in data:
- .EQN 0 16 1 20
- D~READPRN(VDATA)
- .TXT 0 43 1 8
- a1,7,26,6
- group
- .TXT 0 8 1 13
- a1,12,17,11
- group mean
- .EQN 0 15 1 13
- N~cols(D)
- .EQN 0 18 1 13
- M~rows(D)
- .EQN 0 18 1 15
- n~0;N-1
- .EQN 0 20 1 15
- m~0;M-1
- .EQN 1 -70 2 8
- µ[n=
- .TXT 1 -68 1 20
- a1,19,78,18
- number of groups:
- .EQN 0 20 1 9
- N=?
- .TXT 0 10 1 14
- a1,13,78,12
- group size:
- .EQN 0 14 1 9
- M=?
- .EQN 0 16 1 7
- n=
- .TXT 0 22 1 18
- a1,17,78,16
- sum of elements
- .TXT 0 31 1 14
- a1,13,78,12
- group means
- .TXT 0 24 1 21
- a1,20,78,19
- degrees of freedom
- .TXT 2 -117 1 29
- a1,28,63,27
- analysis of variance table
- .EQN 0 62 4 20
- s~m$n$D[(m,n)
- .EQN 0 31 3 17
- µ[n~mean(D{52}n)
- .EQN 0 24 3 18
- df~({3,1}÷M*N-1÷N*(M-1)÷N-1)
- .TXT 1 -120 1 35
- a1,34,58,33
- --------------------------------
- .TXT 1 0 2 14
- a2,13,61,20
- degrees of
- freedom
- .TXT 0 13 2 11
- a2,10,47,17
- sum of
- squares
- .TXT 0 11 2 11
- a2,10,35,16
- mean
- square
- .EQN 2 -22 2 9
- df[i=
- .EQN 0 10 2 9
- ss[i=
- .EQN 0 12 2 9
- ms[j=
- .TXT 1 41 1 17
- a1,16,78,15
- sum of squares
- .TXT 0 30 1 23
- a1,22,78,21
- sum of group squares
- .TXT 0 26 1 16
- a1,15,78,14
- table indices
- .TXT 1 -136 3 18
- a3,17,76,38
- among groups
- within groups
- totals
- .EQN 0 110 5 23
- g~n$(m$D[(m,n))^2
- .EQN 1 -29 4 20
- q~m$n$D[(m,n)^2
- .EQN 0 56 1 11
- i~0;2
- .TXT 1 -77 1 11
- a1,10,25,9
- F ratio:
- .EQN 2 0 1 9
- F={18994}?
- .TXT 1 -60 1 15
- a1,14,75,13
- [Ctrl][PgDn]
- .TXT 1 79 1 15
- a1,14,78,13
- [Ctrl][PgDn]
- .TXT 2 -80 1 12
- a1,11,77,10
- choose α:
- .EQN 0 12 1 10
- α~.025
- .TXT 0 47 2 18
- a2,17,17,29
- approximate
- critical value:
- .TXT 0 20 1 63
- a1,62,78,61
- These equations calculate the table entries and the F ratio.
- .TXT 2 -79 2 52
- a2,51,53,88
- If F > CV then there is a significant difference
- among the group means at the α level.
- .EQN 0 79 1 11
- j~0;1
- .EQN 1 -19 1 10
- CV={18994}?
- .EQN 1 22 11 16
- ss~({3,1}÷q-s^2/(M*N)÷q-(g/M)÷g/M-s^2/(M*N))
- .TXT 1 -82 5 79
- a5,78,76,371
- The data used in the example above are contained in the matrix B. You can
- change the data by redefining B and pressing F9 to write it to the file
- VDATA. To use data from a different file, first delete the following two
- equations, so that your data will not be overwritten. Then use the filename
- command to assign the name of your file to the file variable VDATA.
- .EQN 0 106 8 18
- ms~({2,1}÷ss[1/(N*(M-1))÷ss[0/(N-1))
- .EQN 2 26 5 9
- F~ms[0/ms[1
- .EQN 4 -133 5 27
- B:({5,4}÷5.9÷5.8÷6.0÷6.1÷5.3÷4.5÷5.4÷4.2÷5.2÷3.9÷6.8÷6.5÷7.1÷6.3÷5.2÷3.9÷4.7÷4.7÷3.2÷3.1)
- .TXT 5 80 1 15
- a1,14,78,13
- [Ctrl][PgDn]
- .EQN 1 -80 1 22
- WRITEPRN(VDATA):B
- .TXT 1 81 2 74
- a2,73,78,125
- Approximation for incomplete beta function (see Abramowitz and Stegun,
- Handbook of Mathematical Functions, number 26.5.21):
- .EQN 4 1 13 65
- I(a,b,x)~cnorm((3*((b*x)^(1/3)*(1-1/(9*b))-(a*(1-x))^(1/3)*(1-1/(9*a))))/\((b*x)^(2/3)/b+(a*(1-x))^(2/3)/a))
- .TXT 14 -1 1 15
- a1,14,78,13
- [Ctrl][PgDn]
- .TXT 2 0 5 79
- a5,78,76,369
- These equations calculate CV, an approximation for the critical F value
- corresponding to α. For α between .01 and .1, CV is within 1% of the true
- value for three or more groups with four or more observations in each group,
- and within 2.5% of the true value for two groups with six or more
- observations in each. For α = .005, CV is within 2.5% of the true value.
- .EQN 7 0 1 14
- TOL~.00001
- .EQN 2 0 5 17
- x~df[1/(df[1+2*df[0)
- .EQN 0 22 4 30
- y~root(I(df[1/2,df[0/2,x)-α,x)
- .EQN 0 33 5 18
- CV~(df[1*(1-y))/(df[0*y)
-