Answer:
There is enough statistical evidence to say that the means are different.
Step-by-step explanation:
<em>The question is incomplete:</em>
<em>"The two-sample t statistic for comparing the population means has value (±0.001)."</em>
<em />
The null and alternative hypothesis are:

The level of significance is assumed to be α=0.01.
The difference of the sample means is:

The standard deviation for the difference of the means is:

Then, the t-statistic is:

The degrees of freedom are:

With 40 degrees of freedom the t-critial for a significance of α=0.01 is t=±2.705.
As the t-statistic lies in the rejection region, the effect is significant and the null hypothesis is rejected.
There is enough statistical evidence to say that the means are different.