Question

A production line operation is tested for filling weight accuracy using the following hypotheses. Hypothesis Conclusion and Action H 0: = 16 Filling okay, keep running H a:       16 Filling off standard; stop and adjust machine The sample size is 33 and the population standard deviation is = 1. Use = .05. Do not round intermediate calculations.

What would a Type II error mean in this situation?

What is the probability of making a Type II error when the machine is overfilling by .5 ounces (to 4 decimals)?

What is the power of the statistical test when the machine is overfilling by .5 ounces (to 4 decimals)?

Answer 1

Answer:

Given data

n=33

population standard deviation () = 1

significance level () =0.05

To find the Type II error mean in this situation:

Concluding that the mean filling weight is 16 ounces when it actually is n't

To find the probability of making a Type II error when the machine is overfilling by .5 ounces:

lower cut off sample mean=16-1.96(1/(33))=15.659

upper cut off sample mean=16+1.96(1/(33))=16.341

when true mean is 16.5

z(15.659)=(15.659-16.5)/(1/33))=-4.83

z(16.341)=(16.341-16.5)/(1/33)) = -0.91

P(type II error)=P(-4.83<z<-0.91)

=P(z<-0.91)-P(z<-4.83)

= (1−P ( Z<0.91 )) - (1-p(Z<4.83)

= (1-0.8186) -( 1-1)

= 0.1814

P(type II error) = 0.1814

To find the power of the statistical test when the machine is overfilling by .5 ounces:

Power=1-P(type II error)

=1-0.1814

=0.8186

Therefore the power of the statistical test when the machine is overfilling by .5 ounces is 0.8186