A software training consultant disagrees with the widely held view that training is ineffective for casual users. She believes that a short training session can improve the efficiency of casual users and particularly so for users of products like Microsoft Word. She decides to conduct an experiment to investigate this issue and randomly selects two groups of casual Microsoft Word users from her organization. She decides to use one group as a Control group and the other group as a Treatment group. That is, the Control group will not receive training but each member in the Treatment group will receive a short training session on various Microsoft Word features. She then assigns a suite of tasks to each individual in each group and records the time required to complete the suite.
1. Given the problem statement above, identify and state the null and alternative hypotheses (i.e. the primary hypotheses).
2. Examine this SAS output. Given the SAS output and your hypotheses, conduct a test of hypotheses. Remember to address the following issues:
  1. What can you say about normality for each population.
  2. Comment on the significance of the p-value for your primary hypotheses and comment on the consultants point of view.
You are interested in comparing two different techniques (say method a and method b) that may be used to conduct software inspections. You would like to determine if one is better than the other. To evaluate the methodologies you decide to randomly select two groups of programmers from your software development team. You train each group on one of the techniques and then give them a few days to get comfortable with the technique. Each programmer is then given a suite of modules to inspect and the number of errors detected is recorded.
1. Given the problem statement above, identify and state the null and alternative hypotheses (i.e. the primary hypotheses).
2. Examine this SAS output. Given the SAS output and your hypotheses, conduct a thorough analysis. That is, address the following issues:
  1. Normality: Do you need to establish normality for each population? If so, state the normality hypotheses and provide a p-value to address them.
  2. Equal Variance: Do you need to establish this requirement? Can you establish it by way of hypotheses? If so, state the hypotheses and provide a p-value to address them or state the mechanism you will use to establish this requirement and comment on the requirement.
  3. Comment on the significance of the p-value for the primary hypotheses.
3. Given your findings above, comment on the inspection techniques. Can you say which is better? If so, identify the better technique and, if not, explain why not.
Consider the quality score prediction problem. That is, you are interested in deriving a multiple regression model to predict quality score from two independent variables. Quality score is the time to the first failure in hours. However, in this case, the independent variables are all-uses coverage, x₁ (i.e. a coverage measure that is similar to decision coverage), and programmer experience, x₂ (i.e. programming experience in years).
You obtain a large random sample of 3-tuples and use proc reg to obtain the output below. Given this output, complete the following:
1. Comment on the estimates of the slope parameters. That is, interpret the estimates and comment on the importance of each parameter to the model.
2. Your colleagues argue that a unit increase in all-uses coverage will lead to an increase in execution time to first failure of 240 minutes. You believe that the benefit of this additional coverage is better than they contend. Formulate the necessary hypotheses to resolve this issue. Given the proc reg output, conduct a test of hypotheses.
```
                         Parameter Estimates

                 Parameter       Standard
    Variable     Estimate          Error    t Value    Pr > |t|

    Intercept     17.33146       5.94571        8.50     0.0172
    x1             5.42188       1.87820        8.33     0.0180
    x2             3.00957       1.47685        4.15     0.0720
```