Question

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 36 of the 400 planks were scrapped under the new method. (You may find it useful to reference the appropriate table: z table or t table)

a. Construct the 95% confidence interval for the difference between the population scrap rates between the old and new methods, respectively. (Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answers to 2 decimal places.)

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

• H₀: p₁ − p₂ = 0; H_A: p₁ − p₂ ≠ 0

• H₀: p₁ − p₂ ≤ 0; H_A: p₁ − p₂ > 0

• H₀: p₁ − p₂ ≥ 0; H_A: p₁ − p₂ < 0

c. Using the part a results, can we conclude at the 5% significance level that the scrap rate of the new method is different than the old method?

We		H0. At the 5% significance level, we		conclude the proportions
are different between the old and new methods.

Answer 1

a)

Here, , n1 = 500 , n2 = 400
p1cap = 0.124 , p2cap = 0.09

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.124 * (1-0.124)/500 + 0.09*(1-0.09)/400)
SE = 0.0205

For 0.95 CI, z-value = 1.96
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.124 - 0.09 - 1.96*0.0205, 0.124 - 0.09 + 1.96*0.0205)
CI = (-0.01 , 0.07)

b)

H0: p1 − p2 = 0; HA: p1 − p2 ≠ 0

c)

We do not reject H0. At the 5% significance level, we cannot       conclude the proportions
are different between the old and new methods