How to read a 2 sample t interval in statistics I calculated a 99% confidence interval for the mean difference between men and women's average body temperature using the 2 sample t interval function. The interval I got was (-0.2825, 1.3025).
After completing that, the problem wants me to state if one of the body temperatures (man or woman) is higher than the other.
The information I was given at the beginning of the problem was:
Men sample mean: 98.586
Men sample standard deviation: 0.655
Men sample size: 18
Women sample mean: 98.076 
Women sample standard deviation: 1.129
Women sample size: 18 
So I am not sure how to know which genders temperature is higher or if they are generally the same, based on the confidence interval I constructed.
 A: First, I used Minitab software to verify your CI.
I tried both the pooled and Welch (separate variances)
procedures, and the pooled procedure came the closest
to the CI you got.
 Two-Sample T-Test and CI 

 Sample   N    Mean  StDev  SE Mean
 Men     18  98.586  0.655     0.15
 Wome    18  98.076  1.129     0.27

 Difference = mu (1) - mu (2)
 Estimate for difference:  0.510
 99% CI for difference:  (-0.329, 1.349)
 T-Test of difference = 0 (vs not =): 
    T-Value = 1.66  P-Value = 0.107  DF = 34
Both use Pooled StDev = 0.9229

The difference between your 99% CI and the one from Minitab
is not enough to change the interpretation, but you might
want to check your computations for a mistake (or for rounding
errors from rounding too much too soon in the computation).
Because your CI covers (includes) 0, you can say that a
two-sided pooled t test would not reject the null hypothesis
$H_0: \mu_M = \mu_W$ against the alternative $H_A: \mu_M \ne \mu_W.$
The CI tells you that "zero difference" is a reasonable 
value, not to be rejected (1% level). I think this is the answer you are
expected to give.
But if your instructor wants you to do a t test in addition to
looking at the CI, you will get the test statistic $T = 1.66$
with degrees of freedom $DF = 18+18-2 = 34$ and P-value 0.107.
This P-value means that you would not reject at the 1%, 5%, or
even 10% level. (You reject at level $\alpha$ if the P-value
is smaller than $\alpha.$ If you are using a t table instead
of software, you will be able to see that the P-value slightly
exceeds 0.10, but not to find its exact value.)
Note: I showed the pooled procedure instead of the Welch
procedure because I guessed that is what you used. My own
preference is to use the Welch procedure, but figuring out the
DF without software is a little messy.
