Question

1. nment standards limit the airborne amount of a known cancer causing gas. A company using a known cancer causing gas in its production process limits the exposure of its employees to the gas by shutting down production when the amount of the gas reaches 3 ppm with a standard deviation of 0.5 ppm.

1. (2) If you are the plant manager and you are told management wants as little down time as possible, would you choose a larger or a smaller value of alpha to test whether the amount of gas is dangerous? Explain.
2. (2) If you were an employee in the plant, would you want the choice of alpha to be larger or smaller? Explain.
3. (5) A random sample of 50 air specimens from various sensors in the plant produced a mean of 3.1 ppm. Is this sample data sufficient to shut down production? Use alpha = 0.01.
4. (2) Describe both Type I and Type II errors in the context of this problem.
5. (6) Calculate the probability of Type II error if in fact the true mean plant level if 3.1 ppm. Use alpha = 0.01.
6. (1) What is the value of the Power of the test in this case?

Answer 1

ANSWER:

Given that,

a)

Herewe would choose a lower amount of alpha to test whether the amount of gas is dangerous or not as we are needing a little downtown so decreasing the alpha would give us more confidence and then it will reduce the down time.

b)

Here if i am an employyee of plant, i want a higher value of alpha so that if there is slight increase the concetration of gas, it would be out of limit and the system would be shut down

c)

Here

Test statistic

Z = (3.1 - 3.0)/ (0.5/sqrt(50) = 0.1/ . 0.0707 = 1.4142

so here alpha = 0.01

so Here Z(Critical) = 2.326

so here Z < Z(critical) so we will not shut down producion.

d)

Here type I error means that we will reject the null hypothesis even if it is true that means we will shut down the production even if the gas concentration is under standard limit.

Here type II error means that we fail to reject the null hypothesis even if it is false that means we will not shut down the productioneven if the gas concentration is over standard limit.

e)

Here Pr(Type II error) = Pr( > 3 + 2.576 * 0.5/sqrt(50) ; 3.1 ; 0.5/sqrt(50) )

Z = (3 + 2.326* 0.5/sqrt(50) - 3.1)/ (0.5/sqrt(50)

Z = 0.9121

Pr(Type II error) = Pr( < 3 + 2.236 * 0.5/sqrt(50) ; 3.1 ; 0.5/sqrt(50) ) = Pr(Z < 0.9121) = 0.8191

f)

Power = 1 - Pr(Type II error) = 1 - 0.8191 = 0.1809