p-value! need done by 11:30

According to the NY Times, college students spend on average 50 hrs p/week on the internet. Most college students think it is significantly different. It is known the standard deviation for the amount of time spent on the internet is 17.5. A sample of 100 students was taken and the avg. time spent on the net was 45 hrs. At the a=0.05, does this show a significant difference in the amount of time spent of the net from what the NY Times reported?

A. pval = .9979 and a=0.05; do not reject H0

B. p val = .0021 and a=.05; reject H0

C. pval = .0021 and a= .025; reject H0

D. 0.0025< p val < .001, reject H0

E. none of the above

Re: p-value! need done by 11:30

I think option D looks good.