So im trying to figure out how this person got this answer. i keep doing it and am getting different answers, Then again i dont really know what im doing.
I get this: ( how to fid the test statistic value)
x_bar is approx. 2.444, μ = 0 under our null hypothesis, s_d = 2.698, n = 9 (x_bar and s_d obtained using calc). Then T = 2.444 / (2.698 / sqrt(9)) = 3 * 2.444 / 2.698 = 2.718
but its here, where they ask the P-value, like in this example i do not understand where the 3.415 is coming from, and what they did to get it and the 0.013.
b) The test statistic p-value is P(T > 3.415) where T is a t-distributed random variable with 8 degrees of freedom (just larger than 3.415 because it's a one-sided test). Using a calc we get p = 0.013.
THANKS SO MUCH !
I get this: ( how to fid the test statistic value)
x_bar is approx. 2.444, μ = 0 under our null hypothesis, s_d = 2.698, n = 9 (x_bar and s_d obtained using calc). Then T = 2.444 / (2.698 / sqrt(9)) = 3 * 2.444 / 2.698 = 2.718
but its here, where they ask the P-value, like in this example i do not understand where the 3.415 is coming from, and what they did to get it and the 0.013.
b) The test statistic p-value is P(T > 3.415) where T is a t-distributed random variable with 8 degrees of freedom (just larger than 3.415 because it's a one-sided test). Using a calc we get p = 0.013.
THANKS SO MUCH !
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You have provided the value
_
x = 2.444 and s(sample standard deviation) = 2.698
and n{sample size} = 9
Now any T or z is calculated
using
.......x - µ
z = ▬▬▬
.........σ
T is calculated with the estimators
......._
.......x - {hypothesized µ}
T = ▬▬▬
.........s {of xbar}
but of course, hypothesized µ = 0 so that is why you
get your
2.44
▬▬▬▬ = 2.717568569
2.698/√9
For a small sample, it has a
students T distribution with 8 degrees of freedom
and I am not sure what resourses you are using
to evaluate T. I studied statistics in the 1970's and
1980's when we had to refer to T tables in the backs of
stats books. Now, there are online calculators such as
the one I linked in the source.
Using that calculator, I determine that
Pr(T< 2.717568569) with 8 degrees of freedom is
0.9868
It turns out that the P value is actually
Pr(T>2.71568569) and that is 1-0.9868 and that
calculates to be 0.0132
Now your part b Test statistic is suggested at 3.415 and
its p value is P(T > 3.415)
That number calculates as 0.004584405
which is less than 1/2 of 1%.
_
x = 2.444 and s(sample standard deviation) = 2.698
and n{sample size} = 9
Now any T or z is calculated
using
.......x - µ
z = ▬▬▬
.........σ
T is calculated with the estimators
......._
.......x - {hypothesized µ}
T = ▬▬▬
.........s {of xbar}
but of course, hypothesized µ = 0 so that is why you
get your
2.44
▬▬▬▬ = 2.717568569
2.698/√9
For a small sample, it has a
students T distribution with 8 degrees of freedom
and I am not sure what resourses you are using
to evaluate T. I studied statistics in the 1970's and
1980's when we had to refer to T tables in the backs of
stats books. Now, there are online calculators such as
the one I linked in the source.
Using that calculator, I determine that
Pr(T< 2.717568569) with 8 degrees of freedom is
0.9868
It turns out that the P value is actually
Pr(T>2.71568569) and that is 1-0.9868 and that
calculates to be 0.0132
Now your part b Test statistic is suggested at 3.415 and
its p value is P(T > 3.415)
That number calculates as 0.004584405
which is less than 1/2 of 1%.
-
but i am still slightly confused, i looked up 2.718 in the z table and found it to be .9966 and 1-.9966 = 0.0034
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and when i look up 3.14 in the z table i get .9997 , 1-.9997= 0.0003.
i figured maybe im lloking in the wrong table so i tried the t table but just by looking at the table it cant be it.
online calc is very useful but wont have during exam !!
what am i not seeing !!! :)
i figured maybe im lloking in the wrong table so i tried the t table but just by looking at the table it cant be it.
online calc is very useful but wont have during exam !!
what am i not seeing !!! :)
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