This example was taken from Bartholomew and Knott (1999) (Example 3.27)

Data are taken from a study by Smith and Stanley (1983)"Clocking g:
relating intelligence and measures of timed performance"
Intelligence 7, 353-368.  Here only the facor anlayis of the ability
variables are considered.  Scores were available for 112 individuals
on the following six variables:
1. a nonverbal measure of general intelligence (Spearman's g) using
Cattell's culture-fair test
2. picture completion test
3. block design
4. mazes
5. reading comprehension
6. vocabulary

1.  Run the SAS program found under SAS program 

	a. How many factors does the scree plot indicate?

        b. What is the Hypothesis being tested by the (second)
           Chi-square test?

	c. What is the value and p-value of the Chi-square test?

        d. What are the estimated values of the standardized factor
           loadings?

2.  To the proc factor statement add the option rotate = promax and
    rerun the program.

        a. What are the resulting factor loadings after the oblique
           rotation. Look at (Std Reg coeff)

        b. Which observed values appear to be strong indicators of
           factor 1 and factor 2?

        c. What is the correlation between the two factors?        
 
3. To the proc factor statement add the option n=1 (this tells SAS
   to fit only 1 factor).

       a.  What is the value and p-value of the Chi-square test?  

       b.  Does this model fit better or worse than the model with
           n=2 (fit in part 1 above)

       c.  How do the rotated loadings compare to the non-rotated
           loadings.

4. To the proc factor statement change the factor option to n=3?

       a.  What error/warning message does SAS give?

       b.  How many degrees of freedom are there for the Chi-square
           test?


5. In BK, the author tried fitting 2 factors to this data with X1
removed.  Remove X1 from the analysis and try fitting the 2 factor
model.

       a.  What error is given?

6.  The error in 5a. is the result of a variance being estimated to
be negative.  This is called a heywood case.  By default since SAS
knows that negative variances are usually not a good thing it just
stops and gives an error. To tell SAS to keep running despite the
negative variance you can add the option heywood to the proc factor
statement.  Add the option "heywood" to the proc factor line and
rerun what you did in 5 above.

       a.  Which variable has the communality greater than 1.

       b.  How do the rotated loadings and correlation between the
           factors compare to your answer in 2b and 2c.