Question 1041531
The standard error is t(0.975, df=34)s/sqrt (n)
=2.0322*1.6/sqrt(35)
=0.5496, or 0.55
(69.65,70.75), which I would round to (69.7,70.7)
I would say D
BUT, I am bothered by the sd of the population and not the sample. I think what the question wants you to do is to use a z-test because the sample is larger than 30 AND you know the population sd.  I don't have evidence of normality, which is another assumption needed, but if I accept that, then I would use a z-test but fortunately getting the same answer.  
This would make it a z-test where the SE is 1.96*1.6/sqrt(35)=0,53, slightly different from above.