Find difference between sample mean of a year and population mean of the previous year?

Question

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?

Answer 1

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.

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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.