Question

Scenario 3: Cereal Boxes ~ A large box of corn flakes claims to contain 510 grams of cereal. Since cereal boxes must contain at least as much product as their packaging claims, the machine that fills the boxes is set to put 513 grams in each box. The machine has a known standard deviation of 3 grams and the distribution of fills is known to be normal. At random intervals throughout the day, workers sample 5 boxes and weigh the cereal in each box. If the average is less than 511 grams, the machine is shut down and adjusted.

Q3A

How often will the workers make a Type I error with this decision rule and the hypotheses  vs.  ? Submit your answer rounded to four decimal places.

Q3B

Suppose the workers are dissatisfied by the Type I error rate in (Q3A). A portion of the group argues the decision rule should be changed such that workers will only shut down the factory if the average is less than 509 grams. Another group believes they should keep the same decision rule, but sample 10 boxes for each quality control check, rather than only 5. Which option will result in the ‘better’ Type I error rate?

Q3C

Using the same decision rule, what is the power of the hypothesis test if the machine is actually filling boxes with an average of 510 grams? That is, calculate the chance of observing a sample mean less than 511 if the true mean is 510.

Answer 1

Null and Alternative Hypothesis:

H₀: µ = 513

H₁ : µ < 513

Q3A:

z = (x̅- µ)/(σ/√n) = (511 - 513)/(3/√5) = -1.4907

p-value = NORM.S.DIST(-1.4907, 1) = 0.0680

Type I error = 0.0680

Q3B:

For sample 5 and sample mean = 509

Test statistic:

z = (x̅- µ)/(σ/√n) = (509 - 513)/(3/√5) = -2.9814

p-value = NORM.S.DIST(-2.9814, 1) = 0.0014

For sample 10 and sample mean = 511

Test statistic:

z = (x̅- µ)/(σ/√n) = (511 - 513)/(3/√10) = -2.1082

p-value = NORM.S.DIST(-2.1082, 1) = 0.0175

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509 grams will give lesser or 'better' Type 1 error.

Q3C:

If true µ₁ = 510, then Power, 1 - β =

1 - β = P(X̅ ≤ 511 | µ₁ = 510)

= P(Z ≤ (511 - 510)/(3/√5) )

= P(Z ≤ 0.75)

Using excel function:

= NORM.S.DIST(0.75, 1)

= 0.772

Power of test = 0.772