To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil for a 2-year period. the maximum penetration (in mils) for each specimen is then measured, yielding a sample average penetration of x = 52.4 and a sample standard deviation of s = 4.7. the conduits were manufactured with the specification that true average penetration be at most 50 mils. they will be used unless it can be demonstrated conclusively that the specification has not been met. what would you conclude? (use α = 0.05.) state the appropriate null and alternative hypotheses.

The sample has not met the required specification.

Step-by-step explanation:

As the average of the sample suggests that the true average penetration of the sample could be greater than the 50 mils established, we formulate our hypothesis as follow

: The true average penetration is 50 mils

: The true average penetration is > 50 mils

Since we are trying to see if the true average is greater than 50, this is a right-tailed test.

If the level of confidence is α = 0.05 then the score against we are comparing with, is 1.64 (this is because the area under the normal curve N(0;1) to the right of 1.64 is 0.05)

The z-score associated with this test is

where

= mean of the sample

= average established by the specification

s = standard deviation of the sample

n = size of the sample

Computing this value of z we get z = 3.42

Since z > we can conclude that the sample has not met the required specification.

54

step-by-step explanation:

answer:

c. the third graph