Understanding P-values in Statistical Analysis

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3. I solved for t distribution = (.03 - 0)/(.1/√25) = .15
P - Value = 2P(T ≥ .15) = 2(.441) = .882. So I got No but instead we fail to reject H0 because 0.882 > .05 (.005 is the typical significance level). Is this correct and would it be a correct interpretation?
 
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I am surprised that s does not have a unit.

0.882 > 0.05 (or whatever you want to use), so we don't reject H0.
The large p-value shows that the result is very close to the "prediction" based on H0.
 
mfb said:
I am surprised that s does not have a unit.

0.882 > 0.05 (or whatever you want to use), so we don't reject H0.
The large p-value shows that the result is very close to the "prediction" based on H0.

Thank You!
 
And do you know in words what that p-value means?
 
abeliando said:
And do you know in words what that p-value means?

The probability of attaining a test statistic.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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