Question

comcertain motor oil is packed in tins holding 5 litres each. The filling machine can maintain this but with a S.D. of 0.15 litre. Two samples of 36 tins each are taken from the production line. If the sample means are 5.20 and 4.95 litres respectively, can we be 99% sure that the sample have come from a population of 5 litres?

Answer 1

$1) N_1=36, \overline{x}_1=5.2$

$H_0: a=a_0=5, H_1:a>a_0=5\text{ (one-sided)}\\ a \text{ is population mean}\\ \alpha=0.01\\ \sigma=0.15\\ \text{We assume that volume has normal distribution with mean } a\\ \text{and standard deviation } \sigma=0.15.\\ \text{We will use the following random variable:}\\ U=\frac{(\overline{X}-a_0)\sqrt{n}}{\sigma}\\ \overline{X}\text{ --- random value of sample mean}\\ u_{obs}=\frac{(5.2-5)\sqrt{36}}{0.15}=8\\ \Phi(u_{cr})=\frac{1-2\alpha}{2}=0.49\\ u_{cr}=2.33\\ \Phi(x)\text{ --- Laplace function}\\ (2.33,\infty)\text{ --- critical region}\\ u_{obs} \text{ is in the critical region. So we reject } H_0.\\ \text{We cannot be 99 percent sure that the sample have come from}\\ \text{a population of 5 litres}.$ $2) N_2=36, \overline{x}_2=4.95\\ H_0: a=a_0=5, H_1:a<a_0=5\text{ (one-sided)}\\ a \text{ is population mean}\\ \alpha=0.01\\ \sigma=0.15\\ \text{We assume that volume has normal distribution with mean } a\\ \text{and standard deviation } \sigma=0.15.\\ \text{We will use the following random variable:}\\ U=\frac{(\overline{X}-a_0)\sqrt{n}}{\sigma}\\ \overline{X}\text{ --- random value of sample mean}\\ u_{obs}=\frac{(4.95-5)\sqrt{36}}{0.15}=-2\\ (-\infty,-u_{cr})\text{ --- critical region where } u_{cr}=2.33\\ u_{obs} \text{ is not in the critical region. So we accept } H_0.\\ \text{We can be 99 percent sure that the sample have come from}\\ \text{a population of 5 litres}.$