Counting degree of freedom

[Sophie H.]

Case I: If given some distribution and provide parameters, df=n-1

Case II: If given some distribution and provide parameters, fitted from the data, df= n-r-1
From here,
given binomial n and q, df=n-2-1
given Pareto alpha and theta, df= n-2-1
given negative binomial, df= n-2-1

what if uniform distribution, so Ej will be total and equally divided, then df= n-??-1....?
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Let's assume 40.14 and 40.10 are really the case II. Then, what will their df's be?

40.14 is as follow:

The observed # of claims for a group of 50 risks has been recorded as:
# of claims // # of risks
0 // 7
1 // 10
2 // 12
3 // 17
4 // 4
Null hypothesis: the # of claims per risk follows a uniform distribution on 0,1,2,3 and 4.
So 50 risk equally divided into 5 groups is 10 each group. Can I assume 10 is a given parameter? So, df=5-1-1=3?

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40.10
100 observed losses have been recorded in thousands of dollars and are grouped as follow:
Interval // # of Losses
(0,1) // 15
[1,5) // 40
[5,10) // 20
[10,15) // 15
[15, infinity)// 10
Random variable X underlying observed losses, in thousands, is believed to have density function
f(x) = 0.2 e^(-0.2x)

Can I assume it's given as an exponential with theta=1/.2=5,
so df= 5-1-1=3?

[Actuarialsuck]

How many parameters are fitted for uniform?

Maybe you should post 40.14

[Sophie H.]

Another confusion -> 40.19.

"null hypothesis, H0, is the number of claims per risk follows a Poisson Distribution.
.........
min chi-square estimate of the mean of the Poisson distribution is .3055."

Case I: This is what I got. 4 groups down to 3 groups. df=3-1=2 because it doesn't said it's fitted from the data so it doesn't minus the given parameter (Poisson mean).

Case II: which is the solution. Is it because of this sentence? "min chi-square estimate of the mean of the Poisson distribution is .3055"

So, can I draw a conclustion that whenever a distribution parameter is given along with key words such as "chi-square estimate", "fitted by data", ...,etc, then we use df=n-r-1.

[Actuarialsuck]

Have you had a chance to read this http://www.actuarialoutpost.com/actu...hlight=freedom

[Actuarialsuck]

Originally Posted by Sophie H.

Another confusion -> 40.19.

"null hypothesis, H0, is the number of claims per risk follows a Poisson Distribution.
.........
min chi-square estimate of the mean of the Poisson distribution is .3055."

Case I: This is what I got. 4 groups down to 3 groups. df=3-1=2 because it doesn't said it's fitted from the data so it doesn't minus the given parameter (Poisson mean).

Case II: which is the solution. Is it because of this sentence? "min chi-square estimate of the mean of the Poisson distribution is .3055"

So, can I draw a conclustion that whenever a distribution parameter is given along with key words such as "chi-square estimate", "fitted by data", ...,etc, then we use df=n-r-1.

You are told that a parameter is fitted! "chi square estimate of the mean..." and how is Poisson dist. determined? By lambda. And what does the mean = ?

[Sophie H.]

Originally Posted by Actuarialsuck

Have you had a chance to read this http://www.actuarialoutpost.com/actu...hlight=freedom

I read that post already, and it doesn't add much to what I know.

I'm still not getting everything for chi-square!