2. This is a two-independent sample situation.(a) LetYAandYBbe the concentrations onA and B respectively and letμAandμBbe the respective population means. Theneeded summary statistics are¯yA= 6.1,s2A= 0.8733,¯yB= 4.9, ands2B= 0.74.Then, due to the balanced data,s2p= (s2A+s2B)/2 =.8067.The CI forμA-μBis¯yA-¯yB±t*s2p(1/nA+ 1/nB). Sinces2phas 6 df, the appropriate t-value is 1.943.Thus, (-0.03< μA-μB<2.43).(b) The null hypothesis is written:H0:μA-μB= 0.3. We notice that the value of 0.3iscontained within the CI above.Thus,we can conclude that thep-value >0.10for the given test.3. The appropriate formula for the pooled varianceis:s2p= (∑ki=1(ni-1)s2i)/(N-k) wherekis thenumber of treatments andNis the total numberof observations. You are given the standard de-viations so you must square them.s2p= 157.02.There areN-k= 46 degrees of freedom asso-ciated with this variance. This can be thoughtof as the addition of the number of df for eachvariety (9 + 9 + 6 + 5 + 8 + 9).
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