Quick question on results of Student-t test

Ron Rompen

Trusted Information Resource
:bonk::bonk::bonk::bonk::bonk:

Been quite a few years since I have done ANY real stats analysis, so I'm not completely certain that I am interpreting my results correctly.

I am trying to determine if two sets of data are the same, using Minitab. The results are:

Two-Sample T-Test and CI: Inside, Outside

Two-sample T for Inside vs Outside

N Mean StDev SE Mean
Inside 10 0.730 0.192 0.061
Outside 10 0.906 0.132 0.042


Difference = μ (Inside) - μ (Outside)
Estimate for difference: -0.1760
95% CI for difference: (-0.3308, -0.0212)
T-Test of difference = 0 (vs ≠): T-Value = -2.39 P-Value = 0.028 DF = 18
Both use Pooled StDev = 0.1648


Based on the above, am I correct to assume that since both the lower and upper limit of the 95% CI is < 0, that this supports the null hypothesis (i.e. Inside and Outside are the same)?
 

Steve Prevette

Deming Disciple
Leader
Super Moderator
Actually it is saying you reject the null hypothesis, if your desired confidence level is 95%.

The 95% confidence interval for the difference does not include zero, so the hypothesis test for they are the same fails. I haven't double checked your calculations, just interpreting the final results you gave.
 

Ron Rompen

Trusted Information Resource
Thanks Steve. I wasn't completely sure about the inclusion of '0' in the range, and its always best to double check.
Calculations were done by Minitab v.17, so I will presume that they are correct.

Thanks again
 

Miner

Forum Moderator
Leader
Admin
In addition, look at the P-value. If the P-value is >= your alpha, you do not reject the null hypothesis. If it is < alpha, you do reject the null and conclude that there is a difference. If you desire 95% confidence, alpha is 0.05. The P-value was 0.028, which is less than 0.05, so you would reject the null hypothesis and conclude that there is a difference.

I'm still on Minitab 16, so I don't know if the following is still true. You need to specify whether you want a one or two tailed test under the Options button.
 

Attachments

  • 2T.png
    2T.png
    5.8 KB · Views: 210

Ron Rompen

Trusted Information Resource
Thanks Miner. I guess I need to pull out my old study guides - there was a time when i KNEW this stuff, but it's been so long that it's been filed away somewhere in the dusty corners of my brain :)
 

Ron Rompen

Trusted Information Resource
My brain is rustier than I thought it was :confused:

If I change from a 95% CI to a 99% CI, 0 is now between the two values for the 99% CI.

Is it a correct statement that there is a 99% confidence that both sets of data came from the same population? :confused:



Two-sample T for Inside vs Outside

N Mean StDev SE Mean
Inside 10 0.730 0.192 0.061
Outside 10 0.906 0.132 0.042


Difference = μ (Inside) - μ (Outside)
Estimate for difference: -0.1760
99% CI for difference: (-0.3881, 0.0361)
T-Test of difference = 0 (vs ≠): T-Value = -2.39 P-Value = 0.028 DF = 18
Both use Pooled StDev = 0.1648
 

Miner

Forum Moderator
Leader
Admin
Your p-value was 0.028. If you use an alpha risk of 0.05 then there is evidence of a significant difference. If you "raise the bar" to an alpha risk of 0.01 there is insufficient evidence of a significant difference.

With p-values you are dealing with the risk of making a Type I error (mistakenly calling something significant that is not significant), not with percent confidence. With confidence intervals, when you increase the desired confidence, you widen the interval, so you could simply keep increasing the desired confidence until the interval encompasses zero.
 

Ron Rompen

Trusted Information Resource
Thanks Miner. I think I understand it now (or I will after getting back to the textbooks over the weekend).
 
Top Bottom