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DSE 2012 Ques 39-plzz help

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DSE 2012 Ques 39-plzz help

niki
14 posts
Suppose you have a sample of size one from one of the
following densities
H0 : f(x) = 2x   0 <= x<=1
H1 : f(x) = 2 - 2x 0 <= x <= 1
Let alpha and beta denote type I error and type II error, respectively. Find the test
procedure of the form "Reject H0 if x < k" with alpha = 0:09. Find beta for this
test
(a) k = 0:5; = 0:25
(b) k = 0:5; = 0:36
(c) k = 0:3; = 0:16
(d) k = 0:3; = 0:49
how to do this question..??
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Re: DSE 2012 Ques 39-plzz help

anon_econ
496 posts
You have to find k such that P(X<k)=0.09 if X follows the distribution of the null hypothesis. Once you find k you have to find P(X>k) if X follows the distribution of the alternative hypothesis. This basically follows from the definitions of type I and type II errors.
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Re: DSE 2012 Ques 39-plzz help

niki
14 posts
thanks a lot vasudha..:)
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Re: DSE 2012 Ques 39-plzz help

laracroft
129 posts
Is the answer Option D ??
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Re: DSE 2012 Ques 39-plzz help

niki
14 posts
yup..ans is option D..
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Re: DSE 2012 Ques 39-plzz help

Kuldeep
2 posts
In reply to this post by laracroft
Hi guys can u help me with q9 and q25 of the dse 2012...
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Re: DSE 2012 Ques 39-plzz help

Ayushya Kaul
150 posts
In reply to this post by anon_econ
Which test do we use?
The t-test?
What would be the formula for the test? Could anyone show the calculations?
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Re: DSE 2012 Ques 39-plzz help

Ayushya Kaul
150 posts
In reply to this post by anon_econ
Which test do we use?
The t-test?
What would be the formula for the test? Could anyone show the calculations?
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Re: DSE 2012 Ques 39-plzz help

duck
442 posts
In reply to this post by Ayushya Kaul
Hi Ayushya.. :)

We dont need to use t-test or anything here.
Just use the definition/meaning of type1 and type2 errors.
Type1 error is rejecting the Null given that its true.
It means while finding type1 error, you need to take Null to be true and then, use the test procedure of rejecting null (which is in this case x<k).
So, we've : Type1 error = Integral (2x)dx from 0 to k . And its given Type1 error = 0.09
⇒ Integral (2x) from 0 to k = 0.09
⇒ k =0.3

Now,Type2 error is accepting the null given that alternate hypothesis is true.
So, take Alternate hypothesis to be true and then, use the test procedue of accepting the null (which is x>k).
So, we've: Type2 error = Integral (2-2x)dx from 0.3 to 1
                                = 0.49
:)
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Re: DSE 2012 Ques 39-plzz help

Ayushya Kaul
150 posts
You're a life saver :)
Cheers Duck!
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Re: DSE 2012 Ques 39-plzz help

Abhyudaya
2 posts
In reply to this post by anon_econ
could u pls explain...didnt get it frm ur explaination...I know d theory....