P-values and statistical significance The mean for the result of a regional test is $525$ points with a deviation of $80$.
A $90$ students class did the test and got a mean score of $535$.
Is the mean score of the class different from the regional mean score for a statistical significance of $0.05$?
I'm trying to understand the role of statistical significance and just can't wrap my head around how could these means not be different, would love to get some help understanding this problem.
 A: You should specify, whether you are interested in a single sided or a two sided test. Suppose we take the latter, we know that the 95% confidence interval is approx. spanned by $\mu \pm 2 \hat \sigma_{\bar x}$, if we assume a normal distribution. In addition, we know that $\hat\sigma_{\bar x} = \hat\sigma_{x}/\sqrt{n} = 80/\sqrt{90}\approx 8.43$. Hence, the $2\sigma$ interval is $525 \pm 2\cdot 8.43 \approx [508.1, 541.8]$. Since the found value $535$ lies within this interval the result is not significant at a 5% value.
A: @Semoi's Answer (+1) in terms of a 95% confidence interval is a correct
approach for this two-sided test at the 5% level (but it should be centered at the class average 535). Here I show the test in
what I hope is a familiar format.
You need to think of $\mu = 525$ as a population mean and $\sigma = 80$ as the population standard deviation (for the 'region').  You need to think of
$\bar X = 535$ as the mean of a sample of size $n = 90$ (this one class) as
explained in @NuclearHoagie's Comment.
"Different from" implies a 2-sided alterative for your z-test. The test statistic is
$z = \frac{535 - 525}{80/\sqrt{90}}= 1.18585.$ The two-sided P-value is
$$P(|Z| >  1.18585) =  0.2357 > 0.05 = 5\%,$$ where $Z$ is standard normal.
So do not reject
$H_0:\mu = 525$ against the two-sided alternative at the 5% level.
In R:
z = (535-525)/(80/sqrt(90));  z
[1] 1.185854
2*pnorm(-z)
[1] 0.2356799


From a recent release of Minitab statistical software (which has a one-sample z-test procedure for summarized data):
One-Sample Z 

Test of μ = 525 vs ≠ 525
The assumed standard deviation = 80

 N    Mean  SE Mean       95% CI          Z      P
90  535.00     8.43  (518.47, 551.53)  1.19  0.236


Below is a simulation of class averages a from a thousand
classes of size $n = 90.$ The summary shows class averages were
as small as 479 and as large as 557. Half of the 1000 classes
had scores between 518 and 532. So an average score of 535 in a class of
90 students is not surprising. [Such a class with an average score of 555 would be surprising.]
set.seed(2021)
a = replicate(1000, mean(rnorm(80, 525, 90)))
summary(a)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  479.1   518.0   525.3   525.4   532.3   557.7 

A histogram of the 1000 simulated class averages is shown below. The vertical dashed line is at 535.

