Answer to Question #171762 in Statistics and Probability for Moon Shah

Question

WCans of black olives are filled by two machines (A and B) in a food processing factory. The distribution of the gross mass is known to be normal with mean 500 g and standard deviation 5.3 g for machine A, and normal with mean 504 g and standard deviation 4.8 g for machine B.

A batch of canned black olives have been filled by the same machine, but it is not known which machine was used. As some substandard olives have accidently been used by machine A, it is decided to test

H0: batch is from machine A, against H1: batch is from machine B

By weighing a random sample of 6 cans and rejecting H0 if the mean mass exceeds a pre- determined mass of k g.

(a) Determine constant k such that the risk of type I error is 5%. What is the corresponding

risk of type II error?

(b) Apply and carry out the test for a sample of 6 cans with masses 511, 499, 500, 498, 507,

495 g.

Answer 1

$a)\ H_0:a=a_0=500\\ H_1:a>a_0=500\\ a\text{ --- population mean}\\ \alpha=0.05\\ \Phi(u_{cr})=\frac{1-2\alpha}{2}\\ \Phi(u_{cr})=0.45\\ u_{cr}\approx 1.64\text{ --- critical value}\\ \text{We will use the following random variable:}\\ U=\frac{(\overline{X}-a_0)\sqrt{n}}{\sigma}\\ \frac{(k-500)\sqrt{6}}{5.3}>1.64\\ k\approx 503.55\\ \text{Type II Error:}\\ \beta=\frac{1}{\sqrt{2\pi}(4.8)}\int_{-\infty}^{503.55}e^{-\frac{(x-504)^2}{2(4.8)^2}}dx=\\ =(t=\frac{x-504}{4.8},\ dt=\frac{1}{4.8}dx)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{-0.09375}e^{-\frac{t^2}{2}}dt=\\ =0.5-\Phi(0.09375)\approx 0.4627.\\ b)\ \overline{x}\approx 501.7\\ u_{obs}=\frac{(501.7-500)\sqrt{6}}{5.3}\approx 0.79\\ u_{obs}<u_{cr}\\ \text{So we accept }H_0:a=a_0=500.$