CHI square test - finding degree of freedom

gxc9800
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Homework Statement


i have problem of finding the degree f freedom for this question. the ans for v is 3 , but my ans is v=n-1 , where n = 6 , so my v=5...

Homework Equations

The Attempt at a Solution

 

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For the Chi Squared test, there is a minimum frequency needed for it to be accurate. I would expect that some of the less frequent observations were grouped or "binned" together to reduce your operational "n".
 
gxc9800 said:

Homework Statement


i have problem of finding the degree f freedom for this question. the ans for v is 3 , but my ans is v=n-1 , where n = 6 , so my v=5...

Homework Equations

The Attempt at a Solution

Please type out the problem; you are not supposed to post thumbnails. (Read Vela's 'pinned' post called 'Guidelines for students and helpers', especially topic 4.) This morning I could not read your thumbnail on the medium I was using then.
 
Ray Vickson said:
Please type out the problem; you are not supposed to post thumbnails. (Read Vela's 'pinned' post called 'Guidelines for students and helpers', especially topic 4.) This morning I could not read your thumbnail on the medium I was using then.
the question contains boxes. if i type out , it would look weird... so i would rather post the image of the original question
 
Did combining the lower frequency observations solve your problem getting to v=3?
I think the general rule is frequency ## \geq## 5.
 
RUber said:
Did combining the lower frequency observations solve your problem getting to v=3?
I think the general rule is frequency ## \geq## 5.
thanks! question solved!
 
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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