Question #184429

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 Sample mean is xˉ=(11+14+13+12+13+12+13+14+11+12)10=12510=12.5Sample mean xˉ=12.5Sample standard deviation s=i=1n(xixˉ)2(n1)Substituting the values of xi and xˉ, wegets=(1112.5)2+(1412.5)2+(1312.5)2+(1212.5)2+(1312.5)2+(1212.5)2+(1312.5)2+(1412.5)2+(1112.5)2+(1212.5)2(101)=(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(9)=(1.16666666667)1.08Sample standard deviation s1.08Let us formulate the null and alternative hypothesesNull Hypothesis H0:Machine is not defective (μ = μ0=12)μ=12Alternative Hypothesis H1:Machine is defective (μ  μ0=12)μ12The test statistic is z=xˉμ(s(n))Substituting xˉ=12.5 μ=12 s=1.08 and n=10 we getz=12.512(1.0810)=0.5(10)1.08=1.4641.46Test statistic z=1.46For the 1% siginificance level the critical value is zα2=2.58(From the standard normal tables)Since, The test statistic z<zα2=2.58,there is no reason toreject the Null Hypothesis.Therefore,Machine is not defective at1% significance level, by two tailed test.Sample \ mean \ is \ \bar x = \frac{(11+14+13+12+13+ 12+13+14+11+12)}{10}\\ = \frac{125}{10}\\ =12.5\\ \therefore Sample \ mean \ \bar x=12.5\\ Sample\ standard \ deviation \ s=\sqrt\frac{\sum_{i=1}^{n}(x_i-\bar x)^2}{(n-1)} \\ Substituting\ the \ values \ of \ x_i \ and \ \bar x, \ we get \\ s=\sqrt{\frac{(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}{(10-1)}} \\ =\sqrt{\frac{(-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)}} \\ =\sqrt{\frac{2.25+2.25+0.25+0.25+0.25+0.25+0.25+2.25+2.25+0.25}{(9)}} \\ =\sqrt{\frac{10.5}{(9)}} \\ =\sqrt{(1.16666666667)}\\ \approx 1.08\\ \therefore Sample \ standard \ deviation \ s\approx1.08\\ Let \ us \ formulate \ the \ null \ and \ alternative \ hypotheses\\ Null\ Hypothesis \ H_0: Machine \ is \ not \ defective\ (\mu \ = \ \mu_{0}=12)\Rightarrow \mu=12 \\ Alternative \ Hypothesis \ H_1: Machine \ is \ defective\ (\mu \ \ne \ \mu_{0}=12)\Rightarrow \mu \ne 12 \\ The \ test \ statistic \ is \ z=\frac{\bar x-\mu}{(\frac{s}{\sqrt(n)})}\\ Substituting \ \bar x=12.5 \ \mu=12 \ s=1.08 \ and \ n= 10 \ we \ get \\ z=\frac{12.5-12}{(\frac{1.08}{\sqrt{10}})}\\ =\frac{0.5(\sqrt{10})}{1.08}\\ =1.464\\ \approx1.46\\ \Rightarrow Test \ statistic \ z= 1.46\\ For \ the \ 1\% \ siginificance \ level\ the \ critical \ value\ is \ z_{\frac{\alpha}{2}}=2.58\\ (From \ the \ standard \ normal \ tables)\\ Since, \ The \ test \ statistic \ z< z_{\frac{\alpha}{2}}=2.58, there \ is \ no \ reason \ to \\ reject \ the \ Null \ Hypothesis.\\ Therefore, Machine \ is \ not \ defective \ at 1\% \ significance \ level, \ by\ two \ tailed \ test.


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