Some sports that involve a significant amount of running, jumping, or hopping put participants at risk of Achilles tendinopathy (AT), an inflammation and thickening of the Achilles tendon. A study in the American Journal of Sports Medicine looked at the diameter (in mm) of the affected tendons for patients who participated in these types of sports activities. Suppose that the Achilles tendon diameters in the general population have a mean of 5.97 millimetres. When the diameters of the affected tendon were measured for a random sample of 41 patients, the average diameter was 9.58 with a standard deviation of 1.95 mm. Is there sufficient evidence to indicate that the average diameter of the tendon for patients with AT is greater than 5.97 mm? Test at the 5% level of significance.

Can someone explain how the person got 1.684 from the critical value test? I'm absolutely lost even after looking at youtube tutorials. I think I might be reading the t table wrong.


Since the population standard deviation is NOT known, t-test is used

Null hypothesis: $H_o: \mu \leq 5.97 $

Alternate hypothesis: $H_a: \mu > 5.97$ (Right tailed/one tailed test)

$$t = \frac{(\bar X - \mu)}{ (\frac{s}{\sqrt n}) }= \frac{(9.58 - 5.97) }{(\frac{1.95}{\sqrt 41})} = 11.854 $$

Degrees of freedom = ddf = n-1 = 41-1 = 40

Level of significance = Alpha = 5% or 0.05

Critical (table) value of 't' for 40 df at alpha = 0.05 applicable to right tailed test is 1.684

Since the calculated 't' value (11.854) > the critical value of 't'(1.684), the Ho will be REJECTED.

It can be said that there is sufficient evidence to indicate ....is > 5.97 mm


1 Answer 1


The "critical value" they are referring to ($1.684$) is the $t$-value such that $\alpha=0.05=5\%$ of the t-distribution with $df=40$ is to the right of that value. Since the $t$-value of the test ($11.854$) is far greater than this, we can infer that the probability of the event happening under random conditions is far less than $\alpha=0.05,$ so the null hypothesis was rejected.

This is equivalent to deriving the P-value based on $11.854$ and noting that it's less than $0.05$ (and therefore rejecting the null, as with the solution's method).


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