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 = 0.7. Use = .05. Do not round
intermediate calculations.
Answer:
Given,
sample n = 33
standard deviation = 0.7
significance level = 0.05
Here at 95% CI, z value is +/- 1.96
Interval = mean +/- z*(s/sqrt(n))
substitute values
= 16 +/- 1.96*0.7/sqrt(33)
= 16 +/- 0.2388
Interval = (15.7612 , 16.2388)
P(Type II error) = P((15.7612 - 16.5)/(0.7/sqrt(33)) < (x-u)/(s/sqrt(n)) < (16.2388 - 16.5)/(0.7/sqrt(33)))
= P(-6.06 < z < -2.14)
= P(z < - 2.14) - P(z < - 6.06)
= 0.0161773 - 0 [since from z table]
= 0.0162
Now to give power overfilling by 0.5
Power = 1 - P(type II error)
= 1 - 0.0162
Power = 0.9838
Get Answers For Free
Most questions answered within 1 hours.