I have never been able to clearly understand the relationship between Hypothesis Testing, Central Limit Theorem and the Normal Distribution.

As an example, suppose we have the salaries of randomly sampled people in some town (I made sure that the data does not look normally distributed). Using the R programming language, I simulated some data and plotted the results:

a = rnorm(100000,20000,1000)

a2 = rnorm(100000,40000,10000)

a1 = rnorm(100000,100000,10000)

salary = c(a,a1, a2)


plot_1 =  hist(salary,  100000, ylab = "Number of People" , xlab = " Salary ",   main="Salary of Randomly Selected People in Some Town" )

enter image description here

My Question:

  • A researcher believes that 95% of the town earns less than $105,000.00.

  • However, The 90th percentile of this sampled data is approximately: $105,000.00

     quantile(salary, probs = 0.9)

Can we use hypothesis tests to estimate whether 90% of the population in this town earn less $105,281.00?

enter image description here

Null Hypothesis: Ho = 0.95

Alternate Hypothesis: Ha < 0.95

Test Statistics : Z = (0.95 - 0.90) / sqrt(0.9*(1 - 0.9) / N)) = 0.05 / sqrt(0.09/300000) = 91.28

Based on this extremely large value of the the Test Statistic, I don't think I am doing something right? This large value of the Test Statistic would suggest you accept the Null Hypothesis and conclude with 95% statistical significance that 95% of the people in the town (i.e. true population) earn less than $105,000.00?

I have a feeling the test statistic is so large because I have a really large sample size? Does the non-normal shape of my distribution affect the hypothesis test? Should I be using some other type of hypothesis test (e.g. non-parametric) for this problem?



  1. https://stats.libretexts.org/Courses/Las_Positas_College/Math_40%3A_Statistics_and_Probability/08%3A_Hypothesis_Testing_with_One_Sample/8.04%3A_Hypothesis_Test_Examples_for_Proportions

  2. https://online.stat.psu.edu/statprogram/reviews/statistical-concepts/proportions

  • 1
    $\begingroup$ In your case, the distribution is irrelevant, since the test is based solely on the derived Bernoulli data. $\endgroup$ Oct 20, 2021 at 11:44
  • $\begingroup$ @ bigbendregion: thank you for your reply! Can you please explain this a bit more? Thank you! $\endgroup$
    – stats_noob
    Oct 20, 2021 at 13:54
  • $\begingroup$ The only data used in your test are 0s and 1s. If the income is less than 105k, $Y=1$, else, $Y=0$. It makes no difference what the income distribution looks like, the test is still based on 0s and 1s. $\endgroup$ Oct 20, 2021 at 14:36

1 Answer 1


Generally speaking, the answer is yes. The methodology of hypothesis testing isn't bound only for normal data, but can be used for other types as well (see an example here). However, You might want to formulate your hypotheses differently, e.g.

In my opinion, there are three ways in which you can approach this problem:

  1. Non-parametric, sampling-based methods: Bootstraping percentile confidence intervals (chapter 13 of Efron & Tibshirani) or some kind of permutation testing (intro here).
  2. Non-parametric, rank-based methods: Wilcoxon, Mann-Whitney, there are a few possibilities.
  3. Going with the usual hypothesis testing methodology. Given a series $\hat{\theta}_n$ of consistent and asymptotically-normal estimators (that is, $\forall\theta\in\Theta:\sqrt{n}(\hat{\theta}_n-\theta)\overset{d}{\rightarrow}N\left(0,\frac{1}{I(\theta)}\right)$), according to Wilks' theorem the generalized LRT statistic $\lambda$ for the problem $H_0:\theta=\theta_0,H_a:\theta\neq\theta_0$ follows $2\log(\lambda)\overset{d}{\rightarrow}\chi^2$ under $H_0$. You can play with this idea in order to construct a test for the 95th percentile of your data. You should, however, use this with extreme caution as the assumptions on $\hat{\theta}_n$ are not trivial.
  • $\begingroup$ thank you for your answer! I had an idea about non-parametric sampling methods, i will post it here - I never knew this was a real method, thank you! Note: is it called non-parametric because it doesn't depend on tje distribution of the data? $\endgroup$
    – stats_noob
    Oct 20, 2021 at 14:00
  • $\begingroup$ If possible, can you please explain why hypothesis testing general generally does not require the data to be normally distributed? $\endgroup$
    – stats_noob
    Oct 20, 2021 at 14:09
  • $\begingroup$ Given that the 1st moment exists, we get from The Weak Law of Large Numbers that $\lim_{n\rightarrow\infty}P\left(\left| \bar{X}_n-\mu \right|>\epsilon \right )=0$ for all $\epsilon>0$. Next, if the second moment exists and the variance is finite, we define $Z_n=\sqrt{n}\frac{\bar{X}_n-\mu}{\sigma}$ and The Central Limit Theorem gives us $Z_n\overset{d}{\rightarrow}N(0,1)$. Note that we only require that the data (1) has first two moments and (2) they are finite. $\endgroup$
    – Spätzle
    Oct 21, 2021 at 4:18

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