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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.

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    $\begingroup$ The role of significance is to capture whether or not chance alone is a good explanation for the data. There's some element of randomness in test-taking - even giving the exact same test on different days could result in different scores. Some days the class will score a bit better, some days they'll perform a bit worse, all by chance. By knowing the range of normal variability, you can test whether a 10-point difference is statistically significant, or if it's in the normal range of variation. Basically, is this class truly smarter than average, or did they just do a little better by chance? $\endgroup$
    – Nuclear Hoagie
    Feb 8, 2021 at 20:44
  • $\begingroup$ Although this shows no computations, it does ask a question that often puzzles beginning students; (+1) for asking it pointedly. $\endgroup$
    – BruceET
    Feb 9, 2021 at 3:26

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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.

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  • $\begingroup$ A confidence interval for $\mu$ based on $\bar X = 535$ should be centered at 535. Then you get $(518.47, 551.53),$ which does contain $\mu = 525,$ showing that $H_0: \mu = 525$ should not be rejected in a two-sided 5% level test. (+1) for the interval idea anyhow. $\endgroup$
    – BruceET
    Feb 9, 2021 at 3:34
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@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.

enter image description here

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