Question 1197695
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n = 34 = sample size
xbar = 53.8 = sample mean
s = 6 = sample standard deviation


mu = population mean
sigma = population standard deviation
Sigma is unknown, which is a common occurrence in stats. 
Despite sigma being unknown, we can use the Z distribution because n > 30. 
The T distribution is approximately fairly close to the standard Z distribution for these large values of n.


If sigma were unknown and n < 30, then we'd have to use the T distribution.


Hypothesis:
{{{H[0]}}}: mu = 48.8
{{{H[A]}}}: mu > 48.8
The claim is in the alternative hypothesis.
This gives us a right tailed test.


Computing the test statistic:
z = (xbar - mu)/(s/sqrt(n))
z = (53.8 - 48.8)/(6/sqrt(34))
z = 4.85912657903776
z = 4.86


Use a Z calculator to find that 
P(Z > 4.86) = 0.000000587
which is the approximate p-value.
I don't know how the tutor @ewatrrr got P(z ≤ 4.859) = .000014 as that is not correct.


This p-value 0.000000587 is very small. 
Depending how you round this, the p-value is effectively 0.


Whenever the p-value is smaller than alpha, we reject the null.
We conclude that mu > 48.8 appears to be the case.


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If you wanted to go the critical value route, then at 10% significance, the critical value is roughly z = 1.28 since P(Z > 1.28) = 0.10 approximately


The test statistic (z = 4.86) is to the right of the critical value (z = 1.28), which places the test statistic in the rejection region.
This is another way to see why we reject the null in favor of the alternative hypothesis.
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