Hypothesis testing

I'm not for sure how to get started. The question reads...The average yearly earnings of male college grads are $58500. for men aged 25 to 34. The average yearly earnins of female college grads withthe same qualifications are $49339. based on the results below, can it be concluded that there is a difference in mean earnings between male and female college graduates? Us the 0.01 level of sig. Sample mean m+59,235 f=$52,487
Pop Std Dev m=8,945 f= 10,125
Sample Size m=40 f=35 Where do I begin? I am so not a math person and I just need this class to get into Grad School for Nurse Pract. Help...My whole class wraps up Dec 18

For guys, the population mean is $58,500.
The standard deviation of their population is $8,945
(a figure for how their earnings deviate above and below the mean on average).

For girls with the same qualifications, their overall average earnings is $49,339.
Their standard deviation is $10,125.

These figures come from the populace.

What you are being asked to work with is a group taken from that population,
for one thing because it's way easier and faster, but we need to allow for the fact
that we are examining a pair of small samples of the entire population.

We can assume that the earnings are normally distributed, meaning we use the maths of the
Standard Normal Distribution.

When calculating z-values for the Standard Normal Distribution, convert with z = (x-mean)/(SD).
When dealing with a sample of size n, use z = (sample mean-population mean)/(SD/[sqrt(n)]).
n for guys is 40.
n for girls is 35.

You calculate those values of z for guys and girls and compare them with the value z that corresponds to
the 0.01 level, which is the 1% level.

As you are checking for a "difference", then if your readings lie between 0.5% and 99.5% of the graph,
then the samples are accepted as being representative of the population to 99% confidence,
ie 99% of the time samples from the population will lie between those z values.

Therefore you need to find the z that corresponds to 99.5% or 0.995, or 0.495 if your table of z starts from 0.
You only need to check for positive z since both sample means are above the population means.
The z values you calculate for both guys ang girls would need to be less than about 2.57,
in order that those samples have a 99% confidence of belonging to the overall population.

If you perform those calculations, you will find both z values are <2,
meaning that at the 0.01 level, those samples are taken as representative samples of that population.

Since this is clear as mud to me...can you help me plug in the numbers? It seems that I am having a hard time on the bottom half of the equation. so far...(59235-58500)/8945/sqrt40 what am i doing wrong?

It was mud to me when I started seeing this stuff too!
Persistence pays off eventually.

The z conversion for the population is z = (x-mean) divided by "standard deviation for the population".

For a sample, it's (sample mean)-(population mean) divided by {standard deviation/(sqrt[n])}.

You're doing that correctly, though you have no parenthesis.
First calculate the numerator, then calculate 8945 divided by the square root of 40.
Divide the resulting numerator by the resulting denominator.

Now you have the z-value for the sample of blokes.
This is the value you look up in the z-tables.

Now do the same thing for the girls to get the second z-value.
You will need to look that up too.

Just go step by step.
If it's not clear, just go slowly, step by step.

Try that and if you don't know what to do about the two z-values, then drop us a line again.

for an entire population, you calculate z using z = (x-mean)/(SD),
where SD is the standard deviation.

For a sample of a population,
which is a little group (like a class of students out of all the schools in the country),
you calculate z in a different way.

For a sample z = (x-mean)/(Q), where Q is (SD)/{sqrt(n)}.

Therefore, first you calculate the square root on "n" for both guys and girls,
since there are different figures for both, including different "n".

Then you divide the standard deviation by sqrt(n).
This is your new denominator Q.

Use the values of x, mean, SD and n for guys to get a z-value for guys,
then do the same procedure for girls.

You will now have two z-values.

For guys, Q = 8945/{sqrt40} = 8945/(6.3) = 1414.3,
so their z-value is (59235-58500)/1414.3 = 735/1414.3 = 0.52.

For girls Q = 10125/{sqrt35} = 1711.4,
so the girls' z-value is (52487-49339)/1711.4 = 1.84.

Now you need to correctly interpret what these two z-values mean.
For the guys, z = 0.52 gives a reading of 0.6985.
This means that 0.6985 or 69.85% of the graph lies to the left of that z-value.

The 0.01 level means the 1% level, meaning the first 0.5% and last 0.5% of the graph,
which is the first 0 to 0.5% and last 99.5% to 100%.
Reading off those values corresponding to 0.005 and 0.995, we get z = -0.258 and 2.58 approximately.
Z would need to be less than or equal to -2.58 or greater than or equal to 2.58,
to be in the extremes mentioned.
99% of the time, z will be between those limits.
If outside those limits, we would say there was only a 1% chance of that, so chances are we can't trust these readings.

As both values of z are within the limits, we can say, chances are the boys and girls' distributions have been
accurately formulated and there is a discrepancy between their wages.

I'm sorry this is late, but you learn if you try.