Find difference between sample mean of a year and population mean of the previous year? I have the data set of the salaries of a current year and the population mean and std deviation of the previous year. I want to test if there is sufficient evidence to prove that the population mean of the current year is at least more than 5 of the previous year. How would one go about this?
 A: This makes sense only if the data for this year
is a random sample from the same large population for which you have $\mu_{last}$ and $\sigma_{last}$ for the previous year.
If the salary sample for this year is roughly normal, and you assume $\sigma_{cur}$ is unchanged from last year, you might use a one-sample normal
test based on the statistic $Z = \frac{\bar X - (\mu_{last}+5)}{\sigma_{last}/\sqrt{n}}.$ where $\bar X$ is
the average of the $n$ salaries sampled this year.
Then you would reject
$H_0:\mu_{cur} \le (\mu_{last} + 5)$
in favor of $H_a:\mu_{cur} > (\mu_{last} + 5)$
at the 5% level of significance if $Z > 1.645.$
If you think the standard deviation may have
changed, then you should estimate $\sigma_{cur}$
by $S,$ the sample standard deviation of the $n$ observations from the current year and use a
one-sample t test. The t statistic is
$T = \frac{\bar X - (\mu_{last}+5)}{S/\sqrt{n}}.$
Here is an example of the appropriate one sided t test in R. Suppose $\mu_{last} = 40$ (per hour) for
the employees of interest and you have the $n=400$
preliminary randomly sampled salaries from the
same population of workers this year, in the vector x.
mean(x);  sd(x)
[1] 48.1549
[1] 20.29432

Then a t test in R is as shown below. You should be able to use $n = 400, \bar X = 48.1549, S = 20.2953$ to
get the t statistic shown in the output. With these
fictitious data the small P-value $0.001$ indicates
that you can reject $H_0$ at the 5% level (also, at the 1%
level, and almost at the 0.1% level).
t.test(x, mu = 45, alt="greater")

            One Sample t-test

data:  x
t = 3.1091, df = 399, p-value = 0.001005
alternative hypothesis: true mean is greater than 45
95 percent confidence interval:
 46.48196      Inf
sample estimates:
mean of x 
  48.1549 

Also, you can reject because $T > c = 1.649,$ where approximate value of $c$
can be found by looking at the bottom of a printed
table of t distributions. (Often, the bottom row is for
the standard normal distribution; the t distribution
with 399 degrees of freedom is very nearly standard normal.)  In R:
qt(.95, 399)
[1] 1.648682

The bottom line is that, with my assumptions and fictitious data, the mean salary $48.15$ for this year is large enough to reject $H_0,$ providing strong evidence that this year's salaries
average above $40 + 45 = 45.$
In a mathematical sense, there is no "proof" here, so it is not
appropriate to use the word "prove" in your question.

Notes: (1) I used the following R code to obtain the sample x used in the example above.
set.seed(1234)
x = rnorm(400, 48, 20)

(2) See @whuber's comment below this answer for
additional issues concerning your question.
