SOLUTION: A sample of 105 workers in a given factory earns an average of $ 24,375 per year. The average salary for all workers is $ 24,230 with a standard deviation of $ 523. Are workers in

Question 1202042: A sample of 105 workers in a given factory earns an average of $ 24,375 per year. The average salary for all workers is $ 24,230 with a standard deviation of $ 523. Are workers in the factory overpaid? Use α = 0.05
Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!

Hypotheses
Null: mu = 24230
Alternate: mu > 24230
This is a right tailed test because of the "greater than" sign in the alternate hypothesis.

Parameters:
mu = population mean = 24230
sigma = population standard deviation = 523

Statistics
n = sample size = 105
xbar = sample mean = 24375

SE = standard error
SE = sigma/sqrt(n)
SE = 523/sqrt(105)
SE = 51.0395738 (approximate)

Compute the z score
z = (xbar - mu)/SE
z = (24375 - 24230)/51.0395738
z = 2.84 (approximate)

Now use a table such as this one
https://www.ztable.net/
Similar tables are found in the back of your stats textbook.

That table says
P(Z < 2.84) = 0.99774
which leads to
P(Z > 2.84) = 1-P(Z < 2.84)
P(Z > 2.84) = 1-0.99774
P(Z > 2.84) = 0.00226
This is the approximate area under the Z curve to the right of z = 2.84
It is the approximate p-value (due to the right tailed test).

p-value = 0.00226
alpha = 0.05

The p-value is smaller than alpha, so we reject the null.
We conclude that mu > 24230

Answer: The workers from this particular factory appear to be overpaid compared to their peers from other similar factories.

A very similar question has been asked and answered at this link
https://courses.washington.edu/urbdp520/UDP520/T_TEST_exerise_example.pdf
Scroll down to the last page to see problem 8.6
But I would advise caution. I don't know why the professor used xbar in the hypothesis when it sould have been mu. We don't need to test xbar since it is easily computed in the sample. It appears that professor made a typo. Everything else looks fine.

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