against the alternative bi using the test statistic tb bi /ku> xk rse ( where rse is the residual squared errorthat is rse rss/( pthis test statistic has tn- distribution under to see thiswrite zv/( )with zb bi sku> xk / and thenby theorem ( under kh - and and are independent the result now follows directly from corollary comparing two normal linear models suppose we have the following normal linear model for data [ yn ] + { ( in )( xb where and are unknown vectors of dimension and krespectivelyand and are full-rank model matrices of dimensions and ( )respectively above we implicitly defined [ and [ > > suppose we wish to test the hypothesis against following section the idea is to compare the residual sum of squares for both modelsexpressed as ky ( and ky ( using pythagorastheorem we saw that ky ( ky ( ky ( ( and so it makes sense to base the decision whether to retain or reject on the basis of the quotient of ky ( ( and ky ( this leads to the following test statistics theorem test statistic for comparing two normal linear models for the model ( )let ( and ( be the projections of onto the space spanned by the columns of and the columns of respectively then under the test statistic ky ( ( /( kt( ky ( /( phas an ( kn pdistribution proofdefine : / with expectation :xb/sand : / with expectation kp note that andunder uk we can directly apply theorem to find that ky ( / kx kh - andunder ky ( ( / kx xk kh - moreoverthese random variables are independent of each other the proof is completed by applying theorem