Question 426246: 1) probability of 59, with Mean=60, and Standard deviation=7?
My answer: X=59
z=(59-60)/7
z= -0.143
p= 0.444
p=0.444, .04<.05, Decision to REJECT, Possible Error Type I
APA Statement: The hypothesis was supported X=45 s, z=0.143, p= 0.44
This is a one tailed hypotheisis.
Is my answer correct?
2)score at z = –1.96
I am not sure how to figure it out
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1) probability of 59, with Mean=60, and Standard deviation=7?
My answer: X=59
z=(59-60)/7
z= -0.143
p= 0.444
p=0.444, .04<.05, Decision to REJECT, Possible Error Type I
APA Statement: The hypothesis was supported X=45 s, z=0.143, p= 0.44
This is a one tailed hypotheisis.
Is my answer correct?
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I don't think you are right.
z(59) = -0.143
Assuming you have a left-tail test the p-value
is P(z < -0.143) = 0.443
That is not less than 5% so the decision is to
not reject Ho.
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Cheers,
Stan H.
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2)score at z = –1.96
I am not sure how to figure it out
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