1. a) Conduct a test of hypotheses.

      i) H0: mu(SE)-mu(CS)=0
         Ha: mu(SE)-mu(CS)<0
     ii) a) n(SE)=35; ySEbar=4.6; s(SE)=1.08 
            n(CS)=40; yCSbar=5.0; s(CS)=1.1 
         b) ySEbar-yCSbar=-0.4 hence consistent with Ha
    iii) a) mu(SE)-mu(CS)=0
         b) Both samples are large so Th'm 1 applies.
            z=-0.4/sqrt((1.1)^2/40+(1.08)^2/35)=-1.59
            p-value is 5.59%
     iv) Insufficient evidence to reject H0 and so the CTI student
         cannot justify his belief.

   b) Since H0 was not rejected, a point estimate of the difference is
      not appropriate.


2. a) Conduct a test of hypotheses.

      Let mu(n) denote the mean for the new interface and mu(o) the
      mean for the old interface. Similarly n, o will be used to
      distinguish the relevant statistics.

      i) H0: mu(n)-mu(o)=0
         Ha: mu(n)-mu(o)<0
     ii) a) n(n)=14; ynbar=250; s(n)=71 
            n(o)=14; yobar=315; s(o)=86 
         b) ynbar-yobar=-65 hence consistent with Ha
    iii) a) mu(n)-mu(o)=0
         b) Both samples are small so if we assume y(n) and y(o)
            normally distributed and also assume that sigma(n)=sigma(o)
            then Th'm 2a applies.
            s(p)^2=((13)(71)^2+(13)(86)^2)/(14+14-2)=6218.5
            t=-65/sqrt(6218.5(1/14 + 1/14))=-2.18
            So since df=26 then the p-value is between 1% and 2.5%
     iv) Reject H0 and conclude that the proposed interface is indeed 
         faster.

   b) Since H0 is rejected then a point estimate is appropriate. That is:
             mu(n)-mu(o)=ynbar-yobar=-65
      A 90% CI is:
            -65 +/- (t(26;5))(29.81)=-65 +/- (1.706)(29.81)=[-115.85,-14.15]

   c) We must assume that the sample sizes are the same and the standard
      deviations the same. In this case we will use s(p) as our standard
      deviation. Since s(p)^2=6218.5 then s(p)=78.86
                     n=(2.6*78.86*2/10)^2=1681.6 
      Hence each sample would need 1682 users to achieve a level of accuracy
      of 10 seconds with 99% confidence.