Answer to Question #184429 in Statistics and Probability for Sunny Paul Paulose

Question

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)

Answer 1

$Given \ sample \ size \ is \ n=10\\ Percentage \ of \ nitrate \ in \ the \ samples \ are \$  11,14,13,12,13, 12,13,14,11and 12 $Sample \ mean \ is \ \bar x = \frac{(11+14+13+12+13+ 12+13+14+11+12)}{10}\\ = \frac{125}{10}\\ =12.5\\ \therefore Sample \ mean \ \bar x=12.5\\ Sample\ standard \ deviation \ s=\sqrt\frac{\sum_{i=1}^{n}(x_i-\bar x)^2}{(n-1)} \\ Substituting\ the \ values \ of \ x_i \ and \ \bar x, \ we get \\ s=\sqrt{\frac{(11-12.5)^2+(14-12.5)^2+(13-12.5)^2+(12-12.5)^2+(13-12.5)^2+(12-12.5)^2+(13-12.5)^2+(14-12.5)^2+(11-12.5)^2+(12-12.5)^2}{(10-1)}} \\ =\sqrt{\frac{(-1.5)^2+(1.5)^2+(0.5)^2+(-0.5)^2+(0.5)^2+(-0.5)^2+(-0.5)^2+(1.5)^2+(-1.5)^2+(-0.5)^2}{(9)}} \\ =\sqrt{\frac{2.25+2.25+0.25+0.25+0.25+0.25+0.25+2.25+2.25+0.25}{(9)}} \\ =\sqrt{\frac{10.5}{(9)}} \\ =\sqrt{(1.16666666667)}\\ \approx 1.08\\ \therefore Sample \ standard \ deviation \ s\approx1.08\\ Let \ us \ formulate \ the \ null \ and \ alternative \ hypotheses\\ Null\ Hypothesis \ H_0: Machine \ is \ not \ defective\ (\mu \ = \ \mu_{0}=12)\Rightarrow \mu=12 \\ Alternative \ Hypothesis \ H_1: Machine \ is \ defective\ (\mu \ \ne \ \mu_{0}=12)\Rightarrow \mu \ne 12 \\ The \ test \ statistic \ is \ z=\frac{\bar x-\mu}{(\frac{s}{\sqrt(n)})}\\ Substituting \ \bar x=12.5 \ \mu=12 \ s=1.08 \ and \ n= 10 \ we \ get \\ z=\frac{12.5-12}{(\frac{1.08}{\sqrt{10}})}\\ =\frac{0.5(\sqrt{10})}{1.08}\\ =1.464\\ \approx1.46\\ \Rightarrow Test \ statistic \ z= 1.46\\ For \ the \ 1\% \ siginificance \ level\ the \ critical \ value\ is \ z_{\frac{\alpha}{2}}=2.58\\ (From \ the \ standard \ normal \ tables)\\ Since, \ The \ test \ statistic \ z< z_{\frac{\alpha}{2}}=2.58, there \ is \ no \ reason \ to \\ reject \ the \ Null \ Hypothesis.\\ Therefore, Machine \ is \ not \ defective \ at 1\% \ significance \ level, \ by\ two \ tailed \ test.$