Question 761781: Hi, I have a Stats question. For Student's t distribution with degrees of freedom=16 and t=-1.830 find the corresponding p-value for a one and two tailed test. I need a formula to do this, and am not clear on how to start, because I feel like I need a sample size to determine this. Thank you so much!
Found 2 solutions by stanbon, rothauserc:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Hi, I have a Stats question. For Student's t distribution with degrees of freedom=16 and t=-1.830 find the corresponding p-value for a one and two tailed test. I need a formula to do this, and am not clear on how to start, because I feel like I need a sample size to determine this.
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For left-tail test:
p-value = P(t < -1.830) = tcdf(-100,-1.830,16) = 0.0430
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For 2-tail test:
p-value = 2*P(t < -1.830) = 2*0.0430 = 0.0859
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Cheers,
Stan H.
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Answer by rothauserc(4718)  (Show Source):
You can put this solution on YOUR website! My answer uses the t-table.
First notice that the given t value is negative, the reason it is negative is two sample population means were compared and in this case the second sample population's mean is larger. Without loss of generality, we can drop the negative sign.
Now, referencing the t table for 16 degrees of freedom and t value 1.830, we find the following:
1.830 lies between 1.746 and 2.120 which corresponds to the following probabilities:
one tailed test = .05 > P(t=1.830) > .025
two tailed test = .10 > P(t=1.830) > .05
sometimes the t table will only list one tailed or two tailed probabilities - note that the two tailed value is just twice the one tailed value.
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