SOLUTION: A student has a coin that she claims is biased towards heads. She believes that it will result in heads 60% of the time. She finally decides to test the coin and see if her claim
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Question 64395: A student has a coin that she claims is biased towards heads. She believes that it will result in heads 60% of the time. She finally decides to test the coin and see if her claim is true, before she takes it to Las Vegas to make her fortune. She flips the coin 30 times, keeping track of S, the number of times it comes up heads. Assuming Ho: p=0, Ha: p>.6, calculate the following, show your work.
a. Theoretical mean for the distribution of S.
b. Theoretical standard deviation of distribution of S
c. The number of heads (S) necessary to reject Ho at the .05 significance level.
m=0,standard deviation= 7.2, S=
this is as far as I got am I on the right track?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A student has a coin that she claims is biased towards heads. She believes that it will result in heads 60% of the time. She finally decides to test the coin and see if her claim is true, before she takes it to Las Vegas to make her fortune. She flips the coin 30 times, keeping track of S, the number of times it comes up heads. Assuming Ho: p=0, Ha: p>.6, calculate the following, show your work.
Comment: If your Ha is p>0.6, your Ho is p<=0.6.
a. Theoretical mean for the distribution of S.
Since S is "number of heads in 30 tosses" the theoretical mean is
(1/2)30=15
Comment: The test is binomial where the mean is np=30(1/2)=15
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b. Theoretical standard deviation of distribution of S
The std = sqrt(npq)=sqrt(30(1/2)(1/2))=(1/2)sqrt30=2.74
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c. The number of heads (S) necessary to reject Ho at the .05 significance level.
The critical value corresponding to a right tail of 0.05 is z=1.645
Using z(S)=[S-15]/2.74 solve for S as follows:
1.645(2.74)=S-15
S=19.5
If S>19.5 reject Ho.
Cheers,
Stan H.
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