Q1 a) mu(y)=ybar=150 by FPDA
   b) s(ybar)=s(y)/sqrt(n)=30/5=6
   c) Since n is small we assume that y is normally distributed and
      so ybar will be Student t with n-1=24df.
      ybar +/- t(24,5)*6 = 150 +/- 1.71*6 = [139.74,160.26]
      I am 90% confident that mu(y) is between 139.74 and 160.26
   d) Since n is small we assume that y is normally distributed and
      so ybar will be Student t with n-1=24df.
      ybar +/- t(24,2.5)*6 = 150 +/- 2.06*6 = [137.64,162.36]
      I am 95% confident that mu(y) is between 137.64 and 162.36
   e) n=(z(99)*s(y)/delta)^2=(2.6*30/2)^2=1521

Q2 a) H0:mu(t)-mu(c)=0; Ha:mu(t)-mu(c)<0; Note: t-treatment, c-control
   b) Note that this is clearly a two independent sample problem. 
      Notice that the sample sizes are different.
      Step 2)a) n(c)=49; ynbar=56; s(yc)=18 
                n(t)=64; ytbar=50; s(yt)=20
             b) ytbar-ynbar=-6 hence consistent with Ha
      Step 3)a) Assume H0 true: mu(t)-mu(c)=0
             b) Both samples are large so Th'm 1 applies.
                s(ytbar-ynbar)=sqrt((20)^2/64+(18)^2/49)=3.586
                z=-6/3.469=-1.67
                The p-value is 4.75% (Ott) or 4.945% (online table) 
      Step 4) The p-value is significant and so we reject H0 and 
              conclude that training casual users is indeed beneficial.

Q3 a) H0:mu(a)-mu(b)=0; Ha:mu(a)-mu(b)<0; Note: a-method a, b-method b
      Remember that the technique that detects more bugs is better.
   b) Note that this is clearly a two independent sample problem. 
      Notice that you have two distinct groups of programmers.
      Step 2)a) n(a)=36; yabar=18; s(ya)=6 
                n(b)=36; ybbar=23; s(yb)=4
             b) yabar-ybbar=-5 hence consistent with Ha
      Step 3)a) Assume H0 true: mu(a)-mu(b)=0
             b) Both samples are large so Th'm 1 applies.
                s(yabar-ybbar)=sqrt((4)^2/36+(6)^2/36)=1.2
                z=-5/1.2=-4.16
                The p-value is about 0.003167% (Ott) or 0.00165% (online 
                table)
      Step 4) The p-value is highly significant and so we reject H0.
              Since the null hypothesis is rejected then mu(a)<mu(b) 
              and so we conclude that method b is indeed better than 
              method a since method b detects more bugs than method a 
              on average.

Q4 Step1 H0:mu(2)-mu(1)=0; Ha:mu(2)-mu(1)>0; Note: 1-GUI#1, 2-GUI#2 
         Expressed in terms of time difference:
         H0:mu(d)=0; Ha:mu(d)>0
         Note that this is clearly a paired sample problem since each 
         user completes the task suite for each of the GUI's.
   Step 2)a) n=25; dbar=6; s(d)=2.4 
          b) dbar=6>0 hence consistent with Ha
   Step 3)a) Assume H0 true: mu(d)=0
          b) The sample is small so we will assume that d is normally
             distributed and so dbar is Student t with 24df.
             s(dbar)=s(d)/sqrt(n)=2.4/5=0.48
             t=6/0.48=12.5
             The p-value is essentially 0%.
   Step 4) The p-value is highly significant and so we reject H0.
           Since the null hypothesis is rejected then mu(2)>mu(1) 
           and so we conclude that GUI#1 allows users to complete
           tasks in less time (i.e. more quickly) than GUI#2 on 
           average.