1.a. mu=350; sigma=100 by FPDA
1.b. z=(375-350)/100=0.25. Required area is 40.13%.
1.c. n=25; Since development time is normally distributed then:
          s(ybar)=100/sqrt(25)=20. 
     The 95% CI is:
          ybar plus/minus t(24,2.5)*s(ybar)
		350 plus/minus 2.064*20
		[308.72, 391.28]
     Hence, we are 95% confident that the population mean is between
     308.72 and 391.28 days.

2.a. H0: mu=$62,500; Ha: mu<$62,500 
2.b. First, ybar=$60,000 is consistent with Ha. Assume H0 true, and since 
     n is large, determine the z value. s(ybar)=10000/sqrt(100)=1000 hence
     z=(60000-62500)/1000=-2.5. The p-value is 0.62% (or 0.0062 
     expressed as a probability) which is highly significant and so we
     reject the null hypothesis and conclude that the columnist is
     ill-informed.

3.a. H0: mu=$2.10; Ha: mu<$2.10
3.b. First, ybar=$2.04 is consistent with Ha. Assume H0 true, but since 
     n is small, also assume that y is normally distributed in which
     case ybar is Student t with n-1=24df. Since s(ybar)=0.30/sqrt(25)=0.06 
     then t=(2.04-2.10)/0.06=-1. The p-value is between 10% and 25% (or 0.1 
     and 0.25 expressed as a probability) which is non-significant and so we
     have insufficient evidence to reject the null hypothesis. The
     directors claim is therefore reasonable.
3.c. Since the null hypothesis was not rejected then the null hypothesis 
     mu=$2.10 is reasonable. Some other point estimate for the population
     mean would therefore be inappropriate.  

4.   Since n=64 then mu(ybar)=mu(y)=80 and sigma(ybar)=sigma(y)/sqrt(n)=
     24.0/8=3. n is large and so ybars are normally distributed hence
     z=(75-80)/3=-1.67. The desired % is 4.75%.

5.a. i) H(0): mu(y)=250 H(a): mu(y)<250 
    ii) n=25; ybar=245; s(y)=50. Since ybar<250 then ybar consistent with H(a).
   iii) Assume H(0) true (i.e. mu(y)=250
        Since n=25 is small then assume y normally distb and so ybar is student
        t with 24df and mu(ybar)=mu(y)=250 and s(ybar)=s(y)/sqrt(n)=50/5=10
        hence t=-0.5 and so the p-value is between 25% and 50%
    iv) Since p-value>5% then non-signif and insuff evidence to reject H(0).
5.b. Since H(0) is not rejected a point estimate is not appropriate.
5.c. n=(2.6*50/10)^2=169 users needed.

6.   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. Notice that this is a 
     two independent sample problem since the times for the users in 
     each group are independent. 
     i) H0: mu(n)-mu(o)=0
        Ha: mu(n)-mu(o)<0
    ii) a) n(n)=36; ynbar=250; s(n)=71 
           n(o)=36; yobar=300; s(o)=86 
        b) ynbar-yobar=-50 hence consistent with Ha
   iii) a) Assume H0 true: mu(n)-mu(o)=0
        b) Both samples are large so Th'm 1 applies.
           s(nbar-obar)=sqrt((71)^2/36+(86)^2/36)=18.587
           z=-50/18.587=-2.69
           So the p-value is 0.36%
    iv) The p-value is highly significant and so we reject H0 and conclude 
        that the new proposed interface is indeed faster.