Answer:


And the confidence interval for this case 
Step-by-step explanation:
We know the following info from the problem
sample mean for the group 1
the standard deviation for the group 1
the sample size for group 1
sample mean for the group 2
the standard deviation for the group 2
the sample size for group 2
We have all the conditions satisifed since we have random samples.
We want to construct a confidence interval for the true difference of means and the correct formula for this case is:

The degrees of freedom are given :

The confidence level is 0.9 or 90% and the significance level is
and
and the critical value for this case is:

And replacing the info given we got:


And the confidence interval for this case 