Question 1088019
<font color="black" face="times" size="3">Given Info


alpha = 0.10
mu = 45
sigma = 14
n = 44
xbar = 38


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Null Hypothesis
H0: mu >= 45


Alternative Hypothesis
H1: mu < 45 (claim is made here)


This is a left tailed Z test for the mean (mu)
We will reject the null (H0) if the test statistic is in the critical region Z < -1.28
Use a calculator or table to find the critical value -1.28


If you use a table like <a href="http://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf">this one</a>, then look at the column that has "one tail = 0.10" and look at the second to bottom value in this column to see 1.282. I'm going to round that to 1.28 and change it to a negative value to get -1.28. This is so I can ensure I have the left portion. 


This means P(Z < -1.28) = 0.10 approximately


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Compute the Standard Error


SE = sigma/sqrt(n)
SE = 14/sqrt(44)
SE = 2.11057941204434
SE = 2.110579


That allows us to say


Z = (xbar-mu)/(SE)
Z = (38-45)/(2.110579)
Z = -3.31662543785379
Z = -3.32 <font color=red>Z test statistic (aka Z test value)</font>


The value Z = -3.32 is in the critical region. This is because Z < -1.28 is true when Z = -3.32


Visual confirmation is shown below


<img src = "https://i.imgur.com/ZKXQK8y.png">


So we reject H0.


We have enough sufficient evidence to conclude that the population mean mu is less than 45. So we have enough evidence to conclude that inspectors are slower than average.


<font color=red>Answer is choice C) Yes; test value -3.32 falls in the critical (rejection) region</font>


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