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-five of the 500 planks were scrapped under the old method, whereas 30 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; HA: p1 − p2 ≠ 0 H0: p1 − p2 ≤ 0; HA: p1 − p2 > 0 H0: p1 − p2 ≥ 0; HA: 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?

Answer 1

a)

	old	new
x=	65	30
n =	500	400
p̂=x/n=	0.1300	0.0750
estimated diff. in proportion=p̂1-p̂2=		0.0550
Se =√(p̂1*(1-p̂1)/n1+p̂2*(1-p̂2)/n2) =		0.0200

for 95 % CI value of z=		1.960	from excel:normsinv((1+0.95)/2)
margin of error E=z*std error =		0.039182
lower bound=(p̂1-p̂2)-E=		0.0158
Upper bound=(p̂1-p̂2)+E=		0.0942
from above 95% confidence interval for difference in population proportion =(0.02,0.09)

b)

H0: p1 − p2 = 0; HA: p1 − p2 ≠ 0

c)

since confidence interval contains values above 0;

there is significant evidence at 5% level that the scrap rate of the new method is different than the old method