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Critical value

The critical value is the threshold in our hypothesis test. It marks the limit of the rejection area and is determined by the significance level (α). Values that fall beyond the critical value will lead to rejection of the null hypothesis.

The critical value answers to the question: “What is the critical value that our sample mean should be for us to be able to be able to reject the null hypothesis? It is determined by the significance level (α).



Critical value scenarios

We can calculate the critical value for one-tailed (left/right) and for two tailed tests:


Critical value for right tailed test

Say that our assumed mean (µ) = 10 and that we conduct a sample of 15 (n). Our sample statistics return a sample mean (x̄) of 11.5 with a sample standard deviation (s) = 4.0 which result in a t-score of 1.45. Looking up in the t-table, we find that the critical t-score at one-tailed test with α=0.05 and degrees of freedom (df) = 14, is 1.7613.

So, our t-score falls within the non-rejection area, as 1.45 < 1.7613, and we therefore fail to reject the null hypothesis:

Critical value visualized in density curve

If 11.5 is not enough in order to reject the null hypothesis, how much should the sample mean be in order to reject the null hypothesis? How much is this critical value?

We can run this calculation with a bit of algebra knowing the critical value is 1.7616:

Critical value calculation

The critical value of 11.81 which can be visualized in the density curve:

Critical value in density curve

The calculation shows that our critical value is 11.81, so we were “close”, but not close enough in order to reject the null hypothesis. The p-value is 0.0842 meaning there is approximately 8.4% probability of finding a result at least as extreme as ours (the 11.5), assuming that the mean is 10.



Critical value for left tailed test

Let’s take the example from above, but now with a sample mean of 8.5:

  • Assumed mean (µ) = 10
  • Sample mean (x̄) = 8.5
  • Sample standard deviation (s) = 4.0
  • Critical t = -1.7613

Say we wish to test, if the mean is less than 10 at α = 0.05. We will then conduct a one sided left tailed test. To find the critical value the calculations are the same as we did for the right tailed test (above), only that the critical t-score is negative 1.7613. Therefore, we will get the same absolute value of 1.814, only in negative:

Critical value calculation

The critical value is 8.18. Visualized in the curve:

Critical value in density curve


Critical value for two-tailed test


Now, the same example (as above) but as a two-tailed test stating the our hypotheses as to whether the mean is different from the assumed mean. Thus, we will be looking both to the right and to the left of the assumed mean.

Looking up the critical t-value for a two-tailed test at α = 0.05 and df=14, returns 2.145, and the calculations for the critical value therefore becomes:

Critical value calculation

The critical value to the left is (10-2.215) 7.785 and to the right (10+2.215) 12.215. Visualized in the curve:

Critical value in density curve


Parrot example

Let’s use the same example as in the chapter Hypothesis testing where we are testing for a potential weight loss amongst an African parrot specie:

We are working with an assumed mean weight (µ) of the parrots of 420 grams, but that we now have the intuition that they are losing weight. We therefore conduct a simple random sample of 15 (n) and find a sample mean (x̄) of 416 grams with a sample standard deviation (s) of 12.21.

This is a t-test as the sample size (n) < 30 and σ is unknown. We calculate the t-score and find that this new finding is not significant. So, we fail to get support for our alternative hypothesis. We do not have enough prove to support that we could be right about a potential decrease in mean.

Now that our finding, our sample mean, of 416 is not enough to give prove for our intuition or believe, what result should we have had in order to get support for our alternative hypothesis assuming that n and s are unchanged (n=15 and s=12.21)? Or, just how low should our sample mean have been in order for there to be support for rejection of H0? These questions are asking for the critical value.

The critical t-value for our test can be looked up in the t-table or simply taken from the statistical software. Our n=15, so our degrees of freedom (df) is 14. We are running a left-tailed test, because our alternative hypothesis says less than and our null hypothesis equal to or greater than, because we are suspecting a decrease.

The calculations for the critical value:

Critical value calculation

Visualizing the critical value in the density curve:

Critical value in density curve


Critical values in Excel

To obtain the critical value in Excel, I calculate the as to the actual formula, meaning that I do the algebra as in the examples above. =T.INV for the t-distribution and the =NORM.S.INV return the critical values.

Critical value in Excel

Learning statistics

Some of my preferred places to learn about the critical value:

Carsten Grube

Carsten Grube

Freelance Data Analyst


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