Please show work. Thanks in advance.
SOLUTION:
Here are the probabilities they assign to the values x of X
X | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
P0 | 0.1 | 0.21 | 02 | 0.3 | 0.1 | 0.1 | 0.1 |
P1 | 0.1 | 0.3 | 0.2 | 0.1 | 0.1 | 0.1 | 0.1 |
Here H0: P0 is correct and
Ha: P1 is correct.
Also we reject H0 onlty if x1
a)
Here we want to find probability of type-1 error.
i.e., P(reject H0 when H0 is true)
P( x
1/
H0 true)
=P( x=0)+P( x=2)
=0.1+0.1=0.2 (probabilities of P0 i.e., H0 true)
probability of type-1 error=0.2
b)
Now,P(type-2 error)
Type-2 error is accept H0 is false.
i.e., Accept H0 when Ha is true, (if H0 false then Ha true)
P(Accept H0/Ha
true)=1- P(Reject H0/Ha true)
=1-P(x1/Ha
true)
=1-[P(x=0)+P(x=1)]=1-[0.1+0.3] , (for P1 as Ha true)
P(type-2 error)=1-0.4=0.6.
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