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The model tried to determine if the factors (gender, age, and time of day) and their interaction with each other have any significant effect to the public grooming behavior of the population.

To get the test statistics ANOVA table was used.

Sources of Variation Degrees of Freedom Sum of Squares Mean Square Computed f

Time of Day (A) (a-1) SSA SSA / (a-1) MSA / MSE

Gender (B) (b-1) SSB SSB / (b-1) MSB / MSE

Age (C) (c-1) SSC SSC /(c-1) MSC / MSE

AB (ineteraction) (a-1)(b-1) SSAB SSAB / (a-1)(b-1) MSAB /MSE

AC (interaction) (a-1)(b-1) SSAC SSAC / (a-1)(c-1) MSAC / MSE

BC (interaction) (b-1)(c-1) SSBC SSBC /(b-1)(c-1) MSBC / MSE

ABC (interaction) (a-1)(b-1)(c-1) SSABC SSABC/ (a-1)(b-1)(c-1) MSABC / MSE

Error abc(n-1) SSE SSE / abc(n-1)

TOTAL Abcn - 1 SST

To get the values for the above table, the following formulas were used :

SSA = [(? yI)/bcn] ? [(?y)2 / N ]

SSB = [(? yj)/acn] - [(?y)2 / N ]

SSc = [(? yk)/abn] - [(?y)2 / N ]

SSAB = [(?yij)/cn] - [(?y)2 / N ]

SSAC = [(? yik)/bn] - [(?y)2 / N ]

SSBC = [(? yjk)/an] - [(?y)2 / N ]

SSABC = [(? yijk)/n] - [(?y)2 / N ]

SST = ? yijkl - [(?y)2 / N ]

SSE = SST ? SSA ? SSB ? SSB ? SSAB ? SSAC ¬? SSBC - SSABC

The resulting Sum of Squares values were divided by their corresponding degrees of freedom...

Then the calculated F were compared to the f-distribution to determine if the factor concerned follows the null hypothesis.

6.0 DATA COLLECTION

DATA (See Appendix)

H0 : ?I = 0 for all i

?j = 0 for all j

?k = 0 for all k

(??)ij = 0 for all i and j

(??)ik = 0 for all i and k

(??)jk = 0 for all i and k

(???)ijk = 0 for all I, j and k

? where i = 1,2

j = 1,2

k = 1,2,3

H1 : ?I ? 0 for at least one i

?j ? 0 for at least one j

?k ? 0 for at least one k

(??)ij ? 0 for at least one i and one j

(??)ik ? 0 for at least one i and one k

(??)jk ? 0 for at least one i and k

(???)ijk ? 0 for at least one I, one j and one k

Compute for the ANOVA Table:

SSA = [(774...

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