Fisheries example - H o μ x = μ y H a μ x<> μ y We...

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Fisheries example The goal of this example is to calculate the p-value, or lowest possible α-value at which it is possible to reject H o . The following data are given: Fisheries example x = 5.190 y = 5.100 s x  = 0.32 s y  = 0.45 n x  = 100 n y  = 140 From context, we know that the application is: CASE II:   σ x  and  σ y   unknown  but  n x   and  n y   LARGE   (Devore § 9.1) Given H o , the test statistic  Z  =   X Y 2 2 X Y X Y (X Y) ( ) S S n n - - μ + We also were given that the test is a two-tailed test, because we want to know if X is different from Y (it could be larger or smaller). The significance of why this is important to marine biologists is beyond the scope of the course. Stating the null and alternative hypotheses:
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Unformatted text preview: H o : μ x = μ y H a : μ x <> μ y We then proceed to calculate our test statistics z . Because in the null hypothesis μ x = μ y , μ x - μ y = 0. Thus this term goes away, leaving ( 29 ( 29 80 . 1 140 45 . 100 32 . ) 10 . 5 19 . 5 ( 2 2 = +-= z We now proceed with the formula for the two-tail test from fig.8.7 in Devore Phi(abs(z)) = Phi(1.80) = 0.964 from the z-table Then p = 2[1 – 0.964] = 0.072 = 7.2% Interpretation: the lowest possible alpha for rejecting H o is 7.2%. If the test were given at the 10% level, we would reject H o , but if it were at the 5% level, we would accept it....
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