Question

5. An important quality characteristic used by the manufacturer of Boston asphalt shingles is the amount of moisture the shingles contain when they are packaged. Customers may feel that they have purchased a product lacking in quality if they find moisture and wet shingles inside the packaging. In some cases, excessive moisture can cause the granules attached to the shingles for texture and coloring purposes to fall off the shingles, resulting in appearance problems. To monitor the amount of moisture present, the company conducts moisture tests. A shingle is weighed and then dried. The shingle is then reweighed, and, based on the amount of moisture taken out of the product, the pounds of moisture per 100 square feet are calculated. The 30 measurements (in pounds per 100 square feet) were collected by the manufacturer (moisture2.xlsx).

0.20, 0.26, 0.14, 0.33, 0.13, 0.72, 0.51, 0.28, 0.39, 0.39, 0.25, 0.16, 0.20, 0.22, 0.42, 0.24, 0.21, 0.49, 0.34, 0.36, 0.29, 0.27, 0.40, 0.29, 0.43, 0.34, 0.37, 0.55, 0.67, 0.65

a) The company would like to show that the mean moisture content is less than 0.40 pound per 100 square feet. For conducting a (one-sided) test to support the company’s claim, build null and alternative hypotheses.

(b) To conduct the one-tail test based on the hypotheses in (a), identify the rejection region given the 95% critical level.

(c) Using the given data, compute the test statistic for the t test in (b).

(d) Is there evidence at the 0.05 level of significance that the population mean moisture content is less than 0.40 pound per 100 square feet? In the critical-value approach, what is your conclusion based on (b) and (c)? Explain.

(e) Using the test statistic in (c), determine the p-value for the t test.

(f) In the p-value approach, what is your conclusion based on (e)? Explain.

Answer 1

The given sample as

0.20, 0.26, 0.14, 0.33, 0.13, 0.72, 0.51, 0.28, 0.39, 0.39, 0.25, 0.16, 0.20, 0.22, 0.42, 0.24, 0.21, 0.49, 0.34, 0.36, 0.29, 0.27, 0.40, 0.29, 0.43, 0.34, 0.37, 0.55, 0.67, 0.65

and the test mean as 0.40

Hence the hypotheses are

b) Here left tail test is applied hence rejection region as

Reject Ho if t calculated is less than T critical at 95% which is -1.69913

c) the t statistics calculated as

t=-1.7736

d) Sincw the t value calculated is grater than t critical hence we reject the null hypothesis and heence have enough evidence to support the claim.

e) the p value computed as

The P-Value is 0.0433

f) For p valueapproach if the p value is less than 0.05 here which is lwvwl of significance , hence we can reject the null hypothesis. Here also the p value is less than 0.05 hence we follow the previous conclusion as well.