Module 9 Assignment
NOTE: Refer to Module 9b for a similar example
How heavy a load (pounds) is needed to pull apart pieces of Douglas fir 4 inches long and 1.5 inches square? Students doing a laboratory exercise sample 20 pieces of wood and find an average load capacity of 30,841 pounds.
We are willing to regard the wood pieces prepared for the lab session as an SRS of all similar pieces of Douglas fir. Engineers also commonly assume the characteristics of materials vary normally. Suppose that the strength of pieces of wood like these follow a normal distribution with a population standard deviation of 3,000 pounds (As you can see all three necessary assumptions are met).
a. Assuming that all evergreen wood has a known “load” capacity average of 30,000 pounds. Make a two-sided hypothesis (null and alternative statement) about Douglas Fir “load” capacity compared to the overall average.
b. Apply
the formula for finding our test statistic (show your work or describe the
process)
Specifically, use the formula for the z-statistic (pasted below and on page 401) since the population standard deviation is provided in the problem (3,000 lbs). Also, remember that the sample average as 30,841 lbs.
This result will tell us how many “standard errors” the Douglas Fir sample average is away from the overall capacity of 30,000.
c. Using
table “C” on page 701, if our two tailed alpha level threshold is .05, can we
reject the null hypothesis? Why or why not? (Hint: the very bottom row of Table
C is used for two-tailed tests. Find the z-score associated with two-sided
P=.05. This is your "critical value." Is your result from part b greater or less than that z-score?)
