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A student researcher developed a novel analytical method for determining the lead ( Pb2+ ) content in waste water. To demonstrate the efficacy of the new method, the researcher measured the Pb2+ concentration in an unknown sample using the novel method (Method 1). She then measured the same sample using the standard, conventional analytical method (Method 2). The results of this experiment are shown as the mean (

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Answer #1

Answer: _ Date: ---31/5/2019 Here to check if standard deviations are signi ficantly di fferent are not, we have test to check significance H0 : Null hypothesis : σ H 1: Alternative hypothesis : σ2 σ 2 Test statistics S2 0.240A2/0.190A2-0.0576/0.0361- 1.5955- 1.60 We need to use F-test for this: Decision criteria : Reject H0 if F(cal) > F(tab) degrees of freedom-n-l using decision criteria : F(cal) = 1.60 and F(tab) =5.05 F(cal) F(tab) Hence we do not reject Ho, we may say that standard deviations are not significantly di fferent Part B)To compare data sets from method 1 and method 2, We will test if their population ,means are equal, Hence, Null hypothesis :H0 : μ1 142 Alternative hypothesis: HI: μι 142 Level of significance 190(As it is being checked on 99%) s2 s2 Test statistics Vnín2-3.590-3.740/sqrt(0.24012/6+0.190л2/6)--1.2003

t(cal) -1.20 Table value (tab)+a-2/2 10,0005 3.169 l)> t(tab) са Decision rule Reiect Ho ifIca But 1.20 < 3.169 Hence we do not reject Ho. as both have no significant difference from test, we may say that new method is replacement for old method.

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Answer 1

Answer: _ Date: ---31/5/2019 Here to check if standard deviations are signi ficantly di fferent are not, we have test to check significance H0 : Null hypothesis : σ H 1: Alternative hypothesis : σ2 σ 2 Test statistics S2 0.240A2/0.190A2-0.0576/0.0361- 1.5955- 1.60 We need to use F-test for this: Decision criteria : Reject H0 if F(cal) > F(tab) degrees of freedom-n-l using decision criteria : F(cal) = 1.60 and F(tab) =5.05 F(cal) F(tab) Hence we do not reject Ho, we may say that standard deviations are not significantly di fferent Part B)To compare data sets from method 1 and method 2, We will test if their population ,means are equal, Hence, Null hypothesis :H0 : μ1 142 Alternative hypothesis: HI: μι 142 Level of significance 190(As it is being checked on 99%) s2 s2 Test statistics Vnín2-3.590-3.740/sqrt(0.24012/6+0.190л2/6)--1.2003

t(cal) -1.20 Table value (tab)+a-2/2 10,0005 3.169 l)> t(tab) са Decision rule Reiect Ho ifIca But 1.20 < 3.169 Hence we do not reject Ho. as both have no significant difference from test, we may say that new method is replacement for old method.