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

Boston 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

(a)

H0: Null Hypothesis: 0.40

HA: Alternative Hypothesis: < 0.40 (Claim)

(b)

= 0.05

ndf = n - 1 = 30 - 1 = 29

One Tail - Left Side Test

From Table, critical value of t = - 1.6991

Rejection Region: Reject null hypothesis if t < - 1.6991

(c)

From the given data, the following statistics are calculated:

n = 30

= 10.5/30 = 0.35

s = 0.1544

SE = s/

= 0.1544/ = 0.0282

Test statstic is given by:

t = (0.35 - 0.40)/0.0282 = - 1.7730

(d)

Since the calculated value of t = - 1.7730 is less than t = - 1.6991, Reject null hypothesis.

So,

There is evidence at the 0.05 level of significance that the mean moisture content isless than 0.40 pound per 100 square feet.

(e)

t = - 1.7730

ndf = 29

By Technology, p-value = 0.043367

(f) Since p = value = 0.043367 is less than = 0.05, Reject null hypothesis.