Q1 a) mu(t)-mu(c) is approx ytbar-ycbar=50-56=-6
      Note: t-treatment, c-control
   b) Since n(c), n(t) are both large then Definition 1 applies. That
      is:
           (ytbar-ycbar) +/- z(80)*s(ytbar-ycbar)
      Since s(ytbar-ycbar)=sqrt((20)^2/64+(18)^2/49)=3.586 then the
      80% CI is:
           -6 +/- 1.3*3.586 = [-10.6618,-1.3382]
      I am 80% confident that mu(t)-mu(c) is between -10.6618 and
      -1.3382.   
   c) Since n(c), n(t) are both large then Definition 1 applies. That
      is:
           (ytbar-ycbar) +/- z(95)*s(ytbar-ycbar)
      Since s(ytbar-ycbar)=sqrt((20)^2/64+(18)^2/49)=3.586 then the
      95% CI is:
           -6 +/- 1.95*3.586 = [-12.9927, 0.9927]
      I am 95% confident that mu(t)-mu(c) is between -12.9927 and
       0.9927.   
   d) As before, this is clearly a two independent sample problem. 
      Notice that the sample sizes are different.
      Step 1) H0:mu(t)-mu(c)=0; Ha:mu(t)-mu(c)<0
      Step 2)a) n(c)=20; ycbar=56; s(yc)=18 
                n(t)=16; ytbar=50; s(yt)=20
             b) ytbar-ycbar=-6 hence consistent with Ha
      Step 3)a) Assume H0 true: mu(t)-mu(c)=0
             b) Both samples are small so let us assume that yt
                and yc are normally distributed and that the
                poulation standard deviations are equal then Th'm 2a
                applies. That is, ytbar-ycbar is Student t with 34df.
                   s(ytbar-ycbar)=s(p)*sqrt(1/20+1/16)
                Since s(p)=sqrt(19*(18)^2+15*(20)^2/34)=18.908 then:
                   s(ytbar-ycbar)=18.908*sqrt(1/20+1/16)=6.342
                Thus:
                   t=-6/6.342=-0.946
                The p-value is between 10% and 25% (Ott) or 17.105% 
                (using Normal approx) 
      Step 4) The p-value is non-significant and so we have insufficient
              evidence to reject H0 and conclude that training casual users 
              is indeed ineffective.

Q2 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)=20; yabar=18; s(ya)=6 
                n(b)=20; 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 small so let us assume that ya
                and yb are normally distributed and that the
                poulation standard deviations are equal then Th'm 2a
                applies. That is, yabar-ybbar is Student t with 38df.
                   s(yabar-ybbar)=s(p)*sqrt(1/20+1/20)
                Since s(p)=sqrt(19*(6)^2+19*(4)^2/38)=5.099 then:
                   s(yabar-ybbar)=5.099*sqrt(1/20+1/20)=1.612
                Thus:
                   t=-5/1.612=-3.1
                The p-value is between 0.1% and 0.5% (Ott) or 0.097% 
                (using Normal approx) 
      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.
   c) Since we reject H0 then a point estimate is appropriate.
      Since n(a), n(b) are both small then Definition 2a applies. That
      is:
           (yabar-ybbar) +/- t(38;5)*s(yabar-ybbar)
      Since s(yabar-ybbar)=5.099*sqrt(1/20+1/20)=1.612 then the
      90% CI is:
           -5 +/- 1.684*1.612 = [-7.715,-2.285]
      I am 90% confident that mu(a)-mu(b) is between -7.715 and
      -2.285.