Question

Reducing scrap of 4-foot planks of hardwood is an important factor in reducing cost at a wood-floor 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, two independent, random, representative samples of planks were examined. One sample contained 200 planks which were sawed using the old method. The other sample contained 400 planks which were sawed using the new method. Sixty-two of the 200 planks were scrapped under the old method of sawing, whereas 36 of the 400 planks were scrapped under the new method.

Required:
a. Construct the 90% confidence interval for the difference between the population scrap rates between the old and new methods, respectively.
b. Write the null and alternative hypotheses to test for differences in the population scrap rates between the old and new cutting methods, respectively.
c. 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?

Answer 1

Answer:

The critical value for two tailed test at alpha=0.1 is ± 1.645

The calculated  z= 9.406

Step-by-step explanation:

Formulate the hypotheses as

H0: p1= p2 there is no difference between the population scrap rates between the old and new cutting methods

Ha : p1≠ p2

Choose the significance level ∝= 0.1

The critical value for two tailed test at alpha=0.1 is ± 1.645

The test statistic is

Z = $\frac{p_1- p_2}\sqrt pq(\frac{1}{n_1} + \frac{1}{n_2})$

p1= scrap rate of old method = 62/200=0.31

p2= scrap rate of new method = 36/400= 0.09

p = an estimate of the common scrap rate on the assumption that the two rates are same.

p = n1p1+ n2p2/ n1 + n2

p =200 (0.31) + 400 (0.09) / 600

p= 62+ 36/600= 98/600 =0.1633

now q = 1-p= 1- 0.1633= 0.8367

Thus

z= 0.31- 0.09/ √0.1633*0.8367( 1/200 + 1/400)

z= 0.301/√ 0.13663( 3/400)

z= 0.301/0.0320

z= 9.406

The calculated value of z falls in the critical region therefore we reject the null hypothesis and conclude that the 10% significance level that the scrap rate of the new method is different from the old method.