Answer to Question #328655 in Statistics and Probability for Red

Question

A company producing lubricating oil that the average content of the containers is 20 litres. Test this claim if a random sample of ten containers are 20.5, 19.4, 20.2, 19.8, 20.6, and 19.6 litres .assume normal distribution and use 1% level of significance

Answer 1

$\bar{x}=20.017\\s=0.4916\\H_0:\mu =20\\H_1:\mu \ne 20\\T=\sqrt{n}\frac{\bar{x}-\mu}{s}=\sqrt{6}\frac{20.017-20}{0.4916}=0.0847057\sim t_{n-1}=t_5\\P\left( \left| T \right|>0.0847 \right) =2F_{t_5}\left( -0.0847 \right) =2\cdot 0.4679=0.9358>0.01\Rightarrow \\\Rightarrow H_0\,\,not\,\,rejected$