Understanding and interpreting t-value in order to get p-value from t-distribution table

#1
Hi there,

Im currently doing an assignment for college involving the paired samples t-test using SPSS. I am having some trouble understanding why or what I am doing wrong.
I have run the pair sample t test on for pre cortisol levels and post cortisol levels, however, I have gotten a t value of -26.394 and a two-tailed significance value of 0.000. Of course when I look at the t table, my degrees of freedom is 99, however, the values do not reach such a high number that i have gotten for my t-value from the test. Have I done something wrong or am I using the wrong test or should I be possible using a non-parametric test?

Asking for a friend as I am in dire need of help!!
 

obh

Active Member
#2
Hi Grainne,

No one uses tables any more ... but you may need it to your assignment
When I'm using a simple t distribution calculator:

p(t ≤ -26.394) = 6.73943e-47. {df=99}

Since you use 2-tailed test, 2*p(t ≤ -26.394) = 2*6.73943e-47 = 1.347886e-46 rounded to 0.000

Now if you still need it, please show me how did you get a different number. (please attach a screenshot of the table you use
how did you get to other the value ...)
 
#3
Hi obh,

Thank you so much for replying to my post, I really appreciate it!
Oh okay, I'm still quite new to statistics and as per my lecture notes, I am required to use the t distribution table that is found online when searched for. I am sorry but I do not understand how you calculated you two-tailed test p-value as I have not been taught to calculate it, only to use to t-table to find the p-value corresponding to my t-value and df.

Here is a screenshot of the t-table i am required to use.

Thank you so much again for helping me and I apologise for being rather beginners level at statistics.



1578223557035.png
 
#4
Hi Grainne,

No one uses tables any more ... but you may need it to your assignment
When I'm using a simple t distribution calculator:

p(t ≤ -26.394) = 6.73943e-47. {df=99}

Since you use 2-tailed test, 2*p(t ≤ -26.394) = 2*6.73943e-47 = 1.347886e-46 rounded to 0.000

Now if you still need it, please show me how did you get a different number. (please attach a screenshot of the table you use
how did you get to other the value ...)

Hi obh,

Thank you so much for replying to my post, I really appreciate it!
Oh okay, I'm still quite new to statistics and as per my lecture notes, I am required to use the t distribution table that is found online when searched for. I am sorry but I do not understand how you calculated you two-tailed test p-value as I have not been taught to calculate it, only to use to t-table to find the p-value corresponding to my t-value and df.

Here is a screenshot of the t-table i am required to use.

Thank you so much again for helping me and I apologise for being rather beginners level at statistics.

1578223988095.png
 

obh

Active Member
#5
Hi Grainne,

please let me know what number did you get and how ?

You should look at df=100

Per the table - row 2 and row 3 of the header:

P( x > 3.390 ) = 0.0005
P( x < -3.390 ) = 0.0005 (symetrical)
or for 2 tailes P( x < 3.39 ) + P( x > 3.39 ) = 0.001

but your number is smaller then -3.39 so 1 tail will be smaller than 0.0005 or 2 tails will be smaller than 0.001
 
#7
Maybe a chart will be a better explanation

Example for -3.39

now imagine -26.394


View attachment 1663
http://www.statskingdom.com/p_value.html

Hi obh,

Yes, I understand exactly what you mean now, thank you so much for explaining it to me.
Hence, p-value of 0.001 would suggest there is a statistical significance? Hence, I can reject the null hypothesis and accept the alternative hypothesis??

Thank you so much for the graph, it makes a lot of sense when it is visually represented. I am truly grateful for your help!!
 
#8
Maybe a chart will be a better explanation

Example for -3.39

now imagine -26.394


View attachment 1663
http://www.statskingdom.com/p_value.html

Hi obh,

Yes, I understand exactly what you mean now, thank you so much for explaining it to me.
Hence, p-value of 0.001 would suggest there is a statistical significance? Hence, I can reject the null hypothesis and accept the alternative hypothesis??

Thank you so much for the graph, it makes a lot of sense when it is visually represented. I am truly grateful for your help!!
 
#9
Hi obh,

Yes, I understand exactly what you mean now, thank you so much for explaining it to me.
Hence, p-value of 0.001 would suggest there is a statistical significance? Hence, I can reject the null hypothesis and accept the alternative hypothesis??

Thank you so much for the graph, it makes a lot of sense when it is visually represented. I am truly grateful for your help!!


Also, apologies for only getting back to you now, I did not see the notification in my emails until now.
 

obh

Active Member
#10
Hi obh,

Yes, I understand exactly what you mean now, thank you so much for explaining it to me.
Hence, p-value of 0.001 would suggest there is a statistical significance? Hence, I can reject the null hypothesis and accept the alternative hypothesis??

Thank you so much for the graph, it makes a lot of sense when it is visually represented. I am truly grateful for your help!!
Great Grainne :)

Correct, you reject the null assumption