HELP: un paired t test what do these results mean?
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 This topic has 8 replies, 4 voices, and was last updated 18 years, 8 months ago by DP.

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December 9, 2003 at 9:56 pm #34062
Forgive me for being a newby. trying to work through a certification project. I have two sets of timeline variable data. My MBB tells me after reviewing my project that I must add a 2 sample t test to compare the means. The data is inventory levels. The results are below. What conclusions can I draw from them?:
Year N Mean St Dev SE Mean
2001 7 8.900 0.816 0.31
2003 7 3.686 0.598 0.23
Difference
=mu 2001 mu 2003 estimate for difference :5.214 95% CI for difference (4.372, 6.056) TTest of difference =0 (vs not=): TValue= 13.63 PValue= 0.000 DF=11
Plain english for low level guy appreciated
Thanks in advance
0December 10, 2003 at 2:01 pm #93413I’m just a Yellow Belt candidate and I’m trying to absorb al this stuff which is new to me and difficult for me to understand.
I was really hoping someone was going to respond to this thread.
Reg
0December 10, 2003 at 2:24 pm #93415
Robert ButlerParticipant@rbutler Include @rbutler in your post and this person will
be notified via email.A two sample ttest with equal variances and sample sizes (which is what you have) has a t value of 13.63. The only problem that I can see is that the DF for the test should be 12 and not 11 (n1 +n2 2) = 7+72=12. The results of your ttest, for the samples taken, is that there is a significant difference between the two means. An examination of the two standard deviations – using the F test, indicates there is not a significant difference between the variances.
If your data is representative, this would suggest that there really is a significant difference in the inventory levels and that the “noise” in both inventory systems is about the same. Assuming that these average differences matter and that you wish to make inventory levels the same, this would suggest that you look at things in your inventory process that impact the quantity of inventory and not the changes (variability) in the inventory quantity.0December 10, 2003 at 3:17 pm #93428Robert does it make a difference that one month is missing as far as continuous data? Both sets of data are 7 month levels of inventory. & months each but one set has one month missing in the middle.
I have Jan through Aug of this year but may is missing and I have May through Nov of 2001.
Does the interval difference in the one set of data make a difference?
Thanks
Reg0December 10, 2003 at 4:04 pm #93430
Robert ButlerParticipant@rbutler Include @rbutler in your post and this person will
be notified via email.No, it shouldn’t make any difference. However, the missing August data does explain why you have DF = 11 and not 12 since you now have (7+62) = 11.
Since you have monthly data I would also recommend that you plot both inventories on the same graph by month and take a look at the pattern. You may find something of interest with respect to trending over time – i.e. similar trends but different levels, inverse trending – in a month when one goes up the other goes down, etc. Such trends my give you even more insight into your process and may be far more valuable than the simple fact of the statistically significant difference of their means.0December 10, 2003 at 6:03 pm #93434Robert –
The dfs are as they are due to the way MINITAB calculates the pooled sd for the test when equal variances are not assumed. When they are assumed it is simple pooling. When they are not, the degrees of freedom are adjusted by a formula that would be difficult to type in here.
The help for MINITAB V13 gives the formula. I have a demo for V14.0 and they appear to have left out the formula although the technique is the same. Seems somewhere they provide a reference for the methodology , but could not find it in my short search.
So there are 14 data points (look at the n for each set in the printout the original poster gave), but the degrees of freedom have been adjusted due to actual unequal variances. The formula reduces to n(1) plus n(2) 2 when variances are equal.
0December 10, 2003 at 8:11 pm #93442With such small sample sizes (n=7 for each variable), wouldn’t a nonparametric test (like MannWhitney) be more appropriate?
0December 10, 2003 at 8:49 pm #93444
Robert ButlerParticipant@rbutler Include @rbutler in your post and this person will
be notified via email.I’m not aware of any sample size restriction with respect to the use of a ttest. If you are concerned about the issue of the normality of a sample size of 7 you should remember that, like ANOVA, the ttest is quite robust with respect to nonnormality.
The bigger issues with respect to tests of means are:
1 Lack of constancy of variance
2. Lack of independence
According to the original post, the data consists of monthly readings from two separate inventories. It is probably reasonable to assume independence of measurements both within and between the inventories. The original post provided the standard deviations of the two samples and they are not significantly different. I guess you could use a nonparametric test if you wanted to but I don’t see that there would be any advantage to doing so.0December 11, 2003 at 2:00 pm #93477Right or wrong, here are the assumtions I would check before using a ttest:
1. Data stability (if it is time ordered, which is not the case in this example)
2. Data normality – I know that ttests can be robust against nonnormality, but I was under the impression that this robustness was linked to sample size (with larger samples sizes being more robust).
3. Test for equal variances
Given that we only have 7 data points for each time period, I’d be jumpy about the validity of the normality test and the test for equal variances. Instead of straining to use a tool where I can’t really tell if its assumptions are met, I would take the “easy way out” and use a different tool where I KNOW the assumptions are met.
With all that being said, based on the output from the ttest I think a nonparametric test would have given the same result (“is there a significant difference – why, yes, there is!”).
0 
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