A fertilizer mixing machine is set to give 12 Kg of nitrate for every 100 kg of fertilizer. Ten bags of 100 kg each are examined. The percentage of nitrate so obtained is: 11,14,13,12,13, 12,13,14,11and 12. Is there reason to believe that the machine is defective, given that the population from where the sample has been drawn assumes normality? ( Test the hypothesis at 1% level of significance)
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Expert's answer
2021-05-07T09:24:08-0400
Given \ sample \ size \ is \ n=10\\
Percentage \ of \ nitrate \ in \ the \ samples \ are \ 11,14,13,12,13, 12,13,14,11and 12 Samplemeanisxˉ=10(11+14+13+12+13+12+13+14+11+12)=10125=12.5∴Samplemeanxˉ=12.5Samplestandarddeviations=(n−1)∑i=1n(xi−xˉ)2Substitutingthevaluesofxiandxˉ,wegets=(10−1)(11−12.5)2+(14−12.5)2+(13−12.5)2+(12−12.5)2+(13−12.5)2+(12−12.5)2+(13−12.5)2+(14−12.5)2+(11−12.5)2+(12−12.5)2=(9)(−1.5)2+(1.5)2+(0.5)2+(−0.5)2+(0.5)2+(−0.5)2+(−0.5)2+(1.5)2+(−1.5)2+(−0.5)2=(9)2.25+2.25+0.25+0.25+0.25+0.25+0.25+2.25+2.25+0.25=(9)10.5=(1.16666666667)≈1.08∴Samplestandarddeviations≈1.08LetusformulatethenullandalternativehypothesesNullHypothesisH0:Machineisnotdefective(μ=μ0=12)⇒μ=12AlternativeHypothesisH1:Machineisdefective(μ=μ0=12)⇒μ=12Theteststatisticisz=((n)s)xˉ−μSubstitutingxˉ=12.5μ=12s=1.08andn=10wegetz=(101.08)12.5−12=1.080.5(10)=1.464≈1.46⇒Teststatisticz=1.46Forthe1%siginificancelevelthecriticalvalueisz2α=2.58(Fromthestandardnormaltables)Since,Theteststatisticz<z2α=2.58,thereisnoreasontorejecttheNullHypothesis.Therefore,Machineisnotdefectiveat1%significancelevel,bytwotailedtest.
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