Mathematical Statistics- two sample t-test

[Roni1985]

Homework Statement

At the 5% level, will the two means be equal ?

Homework Equations

The Attempt at a Solution

I tested the variances and found out that it's very likely that the variances are equal, so this is an assumption we make when we do the second test.
Now, I'm trying to test the means and see if they are equal.
I'm getting confused because this is what I get:

Two Sample t-test

data: Var1 and Var2
t = -2.1372 df=3998 p-value = 0.03264
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.128821801 -0.005552541
sample estimates:
mean of x mean of y
-0.0670763877 0.0001107835

zero is not included in the but my p-value is greater than 2.5% (for a two sided test)...
what do I do in such case ?
my t value is more extreme than at the 2.5%.
I think the p-value I'm given is wrong, is it possible ?

Thanks,
Roni.

 

[Mark44]

It seems very odd to me that you confidence interval doesn't include 0. Also, since df = 3998, your sample size must be about 4000. With that large a sample, you don't need to use the Student's t test, but could instead use a normal distribution.

There is a lot you don't show, but I infer that you have a binomial distribution that you can approximate with a normal distribution.

 

[Roni1985]

It seems very odd to me that you confidence interval doesn't include 0. Also, since df = 3998, your sample size must be about 4000. With that large a sample, you don't need to use the Student's t test, but could instead use a normal distribution.

There is a lot you don't show, but I infer that you have a binomial distribution that you can approximate with a normal distribution.

You are right, I have 2 different sets of data with 2000 r.v's in each of them.

I think if it works with the normal dist test, it should also work with the t-test (the df is large)

here is some of the data:

1 -1.67695968151127 0.577070066857405
2 -0.642441727404665 0.53974331907915
3 1.03286440763869 -1.31999608111349
4 2.02667778667435 -0.0361000625995362
5 0.417407406923095 1.23635840784727
6 0.338215656007098 0.295867822353842
7 -0.831289368212899 -0.419309627121533
8 0.90774100682453 1.09654263070674
9 -1.22308033810869 0.474269912619356
10 0.201242609350191 0.793994235426449
11 -0.349209983009311 -0.620406980101651
12 0.303036015889208 -1.01786836015372
13 0.946824776561656 1.61722911478799
14 1.04405267789264 0.95021579495309
15 -0.5631803909092 -0.773840047711912
16 -0.618697005188519 0.877219620279467
17 -1.14813286274261 0.774274810083378
18 0.361139276781852 1.29857639982538
19 -1.69816166131597 -0.132765129167876
20 -1.85142475578048 2.01112325343722
21 -0.75348529082793 0.125903773730686
22 0.373202072550282 0.210826733939696

I don't know which distribution is used to generate these r.v's...

 

[Mark44]

Well, something is not working, it seems. If your null hypothesis is that the difference of the two means is zero (i.e. they are equal), your confidence interval should straddle 0.

As you have things, 0 is outside the confidence interval, so if your test statistic came out as 0 you would reject the null hypothesis! That doesn't seem reasonable at all.

 

[Roni1985]

Well, something is not working, it seems. If your null hypothesis is that the difference of the two means is zero (i.e. they are equal), your confidence interval should straddle 0.

As you have things, 0 is outside the confidence interval, so if your test statistic came out as 0 you would reject the null hypothesis! That doesn't seem reasonable at all.

I think the p-value that's given to me is already multiplied by two.
I just did the z-test for two samples with excel and I got the same p-value.
Now, that the p-value is less than 5%, we can reject the null hypothesis.

Thanks for the help.