You haven't specified the significance level or the level of confidence so I'll use 95%. We need the t-distribution table for 5 degrees of freedom (one less than the sample size). The table has 2.571. This tells us how many standard deviations from the mean would be critical, that is, would give us a CI of 95%.
Because we are dealing with a sample and we don't have the population standard deviation we divide the sample SD by √n, where n=6, the sample size. So we use 0.9/√6=0.3674 approx.
So 2.571×0.3674=0.945oz approx on either side of the mean giving us the CI [9.3-0.945,9.3+0.945]=[8.36,10.24] approx. So, 8.36≤μ≤10.24, were μ is the mean. We are 95% sure that the mean lies in this interval.