1. The problem statement, all variables and given/known data
Hester suspected that a die was biased in favour of a four occuring. She decided to carry out a hypothesis test. When she threw the die 15 times, she obtained a four on 6 occasions. Carry out the test, at the 5% level, stating your conclusion clearly.
2. Relevant equations
None.
3. The attempt at a solution
Ho: p = 1/6
H1: p > 1/6
If Ho is true, then X~B (15, 1/6)
Reject Ho if P (X >= 6) < 5%
P (X >= 6) = 1 - P(X <=5)
= 1- (5/6)^15 - (15C1)(1/6)(5/6)^14 - (15C2)(1/6)^2 (5/6)^13 - (15C3)(1/6)^3 (5/6) ^12 -(15C4)(1/6)^4 (5/6)^11 -(15C5)(1/6)^5 (5/6)^10
=0.0274
=3% (1 s.f.)
Since P (X>=6) < 5%, reject Ho.
The die is biased in favour of 4.
(Answer: no evidence that shows die is biased)
Why am I wrong?
Hester suspected that a die was biased in favour of a four occuring. She decided to carry out a hypothesis test. When she threw the die 15 times, she obtained a four on 6 occasions. Carry out the test, at the 5% level, stating your conclusion clearly.
2. Relevant equations
None.
3. The attempt at a solution
Ho: p = 1/6
H1: p > 1/6
If Ho is true, then X~B (15, 1/6)
Reject Ho if P (X >= 6) < 5%
P (X >= 6) = 1 - P(X <=5)
= 1- (5/6)^15 - (15C1)(1/6)(5/6)^14 - (15C2)(1/6)^2 (5/6)^13 - (15C3)(1/6)^3 (5/6) ^12 -(15C4)(1/6)^4 (5/6)^11 -(15C5)(1/6)^5 (5/6)^10
=0.0274
=3% (1 s.f.)
Since P (X>=6) < 5%, reject Ho.
The die is biased in favour of 4.
(Answer: no evidence that shows die is biased)
Why am I wrong?
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