Answer to Question #317081 in Statistics and Probability for Cheska

Question

A ramdom sample of 18 bottles of 500-ml C-Softdrink had a mean content of 480 ml with a standard deviation of 4 ml, while sample of 14 bottles of 500-ml P-Softdrink had a mean content of 490 ml with standard deviation of 5 ml. Using 0.01 level of significance, can you say that C-Softdrink has lesser mean content than P-Softdrink

Answer 1

$H_0:\mu _1\geqslant \mu _2\\H_1:\mu _1<\mu _2\\\nu =\frac{\left( \frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2} \right) ^2}{\frac{1}{n_1-1}\left( \frac{{s_1}^2}{n_1} \right) ^2+\frac{1}{n_2-1}\left( \frac{{s_2}^2}{n_2} \right) ^2}=\frac{\left( \frac{4^2}{18}+\frac{5^2}{14} \right) ^2}{\frac{1}{17}\left( \frac{4^2}{18} \right) ^2+\frac{1}{13}\left( \frac{5^2}{14} \right) ^2}=24.5178\approx 25\\T=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}}}=\frac{480-490}{\sqrt{\frac{4^2}{18}+\frac{5^2}{14}}}=-6.11463\\P-value:\\P\left( T<-6.11463 \right) =F_{t,25}\left( -6.11463 \right) =1.08\cdot 10^{-6}<\alpha \Rightarrow \\\Rightarrow \mu _1<\mu _2$