Question 1189992
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Hypothesis:
H0: mu = 2
H1: mu > 2
Based on the inequality sign in the alternate hypothesis, we have a right-tailed test.


n = 32 = sample size
Since n > 30, we can use the standard normal Z distribution (even though we don't know sigma).


From the given data set, compute the following
xbar = sample mean = 2.06375
s = sample standard deviation = 0.19656


The z test statistic is
z = (xbar - mu)/(s/sqrt(n))
z = (2.06375 - 2)/(0.19656/sqrt(32))
z = 1.83468
z = 1.83


Now use a Z table such as this one
<a href = "https://www.ztable.net/">https://www.ztable.net/</a>
to find that
P(Z < 1.83) = 0.96638 approximately
The idea is to locate the row that starts with 1.8 at the far left side. Then locate the column 0.03 up top. The row and column combine to form the z value of z = 1.83; the value in this row and column is the 0.96638 mentioned.
Also, notice that on the link is the subsection titled "How to Read The Z Table" which may clear up any confusion that still lingers. 


Then from here we can say
P(Z > 1.83) = 1 - P(Z < 1.83)
P(Z > 1.83) = 1 - 0.96638
P(Z > 1.83) = 0.03362
This is the approximate p-value.
You can use a p-value calculator to get a more accurate number if you wanted.


The p-value 0.03362 is not smaller than alpha = 0.02, so we fail to reject the null.


To determine the critical z value, we can use a calculator such as this one
<a href="https://www.statology.org/critical-z-value-calculator/">https://www.statology.org/critical-z-value-calculator/</a>
Plug in the significance level of 0.02 to find that 2.054 is the z critical value for the right-tailed test. This result is approximate. 


Using a Z table is an alternative route to finding the critical value, but you'll need to use the table in reverse of the usual method done earlier.
You'll need to locate 0.02 or close to it inside the table itself, then trace backward to the row and column headers and you should get somewhere around 2.05
I recommend using a calculator since it's more straight forward.


The test statistic z = 1.83 is not larger than the critical value z = 2.054; therefore, this test statistic is not in the rejection region and we won't reject the null.
This reinforces the conclusion we came to with the p-value. 


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Answers:<ul><li>test statistic: z = 1.83</li><li>p-value = 0.03362</li><li>critical value: z = 2.054</li><li>Decision: Fail to reject the null</li><li>Interpretation: the mercury levels do not appear to exceed 2 mcg/L</li></ul>Each decimal value is approximate.</font>