John Johnson’s gross pay for this week is $520.95. He is single and claims one withholding allowance. What amount will be withhe
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Answer:
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 <em>level of confidence</em> 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
= <em>mean of the sample</em>
= <em>average established by the specification</em>
s = <em>standard deviation of the sample</em>
n = <em>size of the sample</em>
Computing this value of z we get z = 3.42
Since z > we can conclude that the sample has not met the required specification.