Question

In: Statistics and Probability

1.Reducing scrap of 4-foot planks of hardwood is an important factor in reducing cost at a...

1.Reducing scrap of 4-foot planks of hardwood is an important factor in reducing cost at a wood-flooring manufacturing company. Accordingly, engineers at Lumberworks are investigating a potential new cutting method involving lateral sawing that may reduce the scrap rate. To examine its viability, samples of 500 and 400 planks, respectively, were examined under the old and new methods. Sixty-two of the 500 planks were scrapped under the old method, whereas 33 of the 400 planks were scrapped under the new method.

1a. Construct the 90% confidence interval for the difference between the population scrap rates between the old and new methods, respectively.

1b. Select the null and alternative hypotheses to test for differences in the population scrap rates between the old and new cutting methods, respectively.

  • H0: p1p2 = 0; HA: p1p2 ≠ 0

  • H0: p1p2 ≤ 0; HA: p1p2 > 0

  • H0: p1p2 ≥ 0; HA: p1p2 < 0

1c. Using the part a results, can we conclude at the 10% significance level that the scrap rate of the new method is different than the old method?

we____ H0. At the 10% significance level, we _____ conclude the proportions are different between the old and the new methods.

Solutions

Expert Solution

Calculation:

Critical value:

Z/2 = Z 0.1/2 = 1.645        ...............Using standard Normal table

90% confidence interval:

(0.008, 0.075)

OR

0.008 < P1-P2 < 0.075

1b)

ANSWER: A

A. H0: p1 − p2 = 0; HA: p1 − p2 ≠ 0

1c)

In Above 90% Confidence interval does not include 0. So We Reject Null Hypothesis.

We reject Ho. At the 10% significance level, we Can conclude the proportions are different between the old and the new methods.


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