A 99% confidence interval when n=33? Solution In each case, use df = n ?1; if that number is not in Table D, drop to the lower degrees of freedom df=n-1=33-1=32 For 99% con?dence and df = 32, we drop to df = 30 and use t*= 2.750 Software gives t*=2.7385 For 90% con?dence and df = 249, we drop to df = 100 and use t*= 1.660 Read more: 7.2 http://www.marin.edu/~npsomas/Math115/Resources/IPS7e_ISM_ch07.pdf.