1 a. H0: mu(c)=mu(t)
     Ha: mu(c)>mu(t)
  b. ii) ybar(c) > ybar(t) and so the sample statistics are consistent
         with Ha.
    iii) n(c), n(t) are both small and, since the normality p-values for
         y(c), y(t) are 11.24% and 12.08%, respectively, then normality is 
         reasonable and so we may use the ttest output.

         The equal variance p-value is 90.25% and so equality of population
         variances is reasonable and so the p-value for our hypotheses in
         a is 25.17/2=12.585% since our Ha is one-sided.
     iv) We therefore have insufficient evidence to reject H0 and so the
         consultant has no evidence to support her point of view.


2 a. H0: mu(a)=mu(b)
     Ha: mu(a)!=mu(b)
  b. i. Yes. 

        H0: Sample a is selected from a normally distributed population
        Ha: Not so.
        The p-value in this case is 64.55% and so normality is reasonable.

        H0: Sample b is selected from a normally distributed population
        Ha: Not so.
        The p-value in this case is 89.67% and so normality is reasonable.

    ii. The sample sizes are small and so we must also address the equal
        variance requirement.

        H0: sigma(a)=sigma(b)
        Ha: sigma(a)!=sigma(b)
        The p-value may be obtained from the ttest output since normality
        is reasonable. In this case the p-value is 43.25% and so we cannot
        reject H0 and conclude that the population variances are equal.

   iii. The p-value for the hypotheses stated in part a is 2.84% and so we
        reject H0 and conclude that mu(a) is not equal to mu(b).

   c. Technique a is better than technique b since ybar(a)=50.091>ybar(b)
      =39.9 indicating that technique a, on average, detects more bugs.

3 a. Intercept=17.33146 which means that when coverage and experience are
     both zero we can expect quality to be 17.33146 hours. This could be
     reasonable since it may take some time for a program that has not 
     been tested and that has been written by an inexperienced programmer 
     to fail. Since the p-value is 1.72% then this intercept is not
     equal to zero.
     x1=5.42188 which means that for unit increase in all-uses coverage,
     we can expect quality to increase by 5.42188 hours. Since this p-value 
     is 1.8% then this slope parameter is not equal to zero and so 
     coverage is important for this model.
     x2=3.00957 which means that for unit increase in programming experience
     we can expect quality to increase by 3.00957 hours. Since the p-value 
     is 7.2% then this slope parameter may be equal to zero and so 
     experience is not important for this model. 
  b. i) H0: beta1=4 (Note: 240 minutes is equal to 4 hours) 
        Ha: beta1>4
    ii) b1=5.42188>4 hence consistent with Ha
   iii) z=(5.42188-4)/1.87820=0.76 hence the p-value is 22.36%
    iv) The p-value is non-significant and so you have insufficient
        evidence to reject H0 and so your colleagues are correct.