\magnification=\magstep1
\centerline{MONT 106N -- Identifying Patterns}
\centerline{Selected Solutions -- Problem Set 5}
\centerline{November 16, 2009}
\bigskip
\noindent
Chapter 10 Review Exercises (p. 176 -- 178) 
\bigskip
\noindent
4.  (a) {\it Method 1}:  If the husband has completed 18 years of
schooling, the value of $x$ is 2 SD's above the average of 12.
The regression line predicts that the wife's education level
will be $(r)(2) = (.5)(2) = 1$ SD above the average for the wives.
Thus, the wife's educational level is predicted to be $12 + 3 = 15$
years.  
\bigskip
\noindent
{\it Method 2}:  We can also solve problems like this by writing down
the $xy$-equation of the regression line.  In this case, it 
is $y - {\rm ave}_y = {r\cdot SD_y\over SD_x}(x - {\rm ave}_x)$
or $y - 12 = {(.5)(3)\over 3}(x - 12)$, or $y = .5x + 6$.  With 
$x = 18$, we get predicted value $y = (.5)(18) + 6 = 15$.  
\bigskip
\noindent
(b) This is similar to part (a), except that the roles of 
$x$ and $y$ are reversed in the above.  A man with $15$
years of schooling is $1$ SD above the average, so 
the woman is predicted to be $(r)(1) = .5$ SD above the 
average for the women.  This gives an educational level
of $12 + (.5)(3) = 13.5$ years.  
\bigskip
\noindent
(c)  This is possible since we are actually talking 
about {\it two different} regression lines here, one
for predicting values of $y = $ wife's education level
from values of $x = $ husband's education level, and 
the other for predicting values of $x$ from values of 
$y$.  ({\it Technical Note:}  The first is the line $y = .5x + 6$ found in 
the solution of (a) by Method 2 above.  The other would
be the line $x = .5 y + 6$ in this example.  The general equation
of the 2nd line would be 
$(x - {\rm ave}_x) = {r\cdot SD_x\over SD_y}(y - {\rm ave}_y)$.)  
The
two lines are located as in the figure from problem 6
on page 177.  If we look at the intersection of 
the vertical line $x = 18$ with $y = .5x + 6$, then 
we get the point $(18,15)$ from part (a).  Then the horizontal
line $y = 15$ gives $(13.5,15)$ on the second line $x = .5y + 6$.
Since the line for predicting $x$ from $y$ lies above
the first line, this point is farther to the left.  ({\it Draw
the picture to make sure you understand this!})
\bigskip
\noindent
9. (a)  A score at the 5th percentile means that the
corresponding standard unit score $-z$ was located so
that the area in the lower tail below $-z$ was $5\%$.
By the symmetry of the normal curve, the upper tail
above $z$ was also $5\%$.  So the area in the central
region between $-z$ and $z$ was $90\%$.  From the 
normal curve area table, $z \doteq 1.65$.  So
the student's score was $1.65$ SDs below average.
On the final, the same student would be predicted
to score $r \times 1.65 = .5\times 1.65 \doteq .825$
SD's below average.  Using the approximation $.8$
in our table gives $A(.8) = 57.63$, so area in 
the lower tail below $-.8$ is about $(100 - 57.63)/2 \doteq 42.3/2 \doteq 21.0$.
The predicted percentile is approximately the $21$st percentile.
\bigskip
\noindent
(b)  Similar to (a):  A standard unit score $z$ at the 80th percentile
means the area in the upper tail to the right of $z$
is $20\%$, and the central area from $-z$ to $z$ is
$60\%$.  From the normal curve area table, this occurs
at about $z = .85$.  On the final, the predicted score
would be $r \times .85 = .5 \times .85 = .425$ SD's
above average.  Using the approximation, $.4$ in the
area table gives $A(.4) = 31.08\%$, so the area 
below $z = .4$ is about 
$31.08 + (100 - 31.08)/2 = 31.08 + 68.92/2 \doteq 65.54\%$. 
\bigskip
\noindent
(c) and (d) Both of these would be the 50th percentile.
\bigskip
\noindent
Chapter 11 Review Problems (p. 198 -- 201)
\bigskip
\noindent
4.  The general idea for this problem is that when 
the scatter diagram is football-shaped, the rms error
for regression functions as an SD for the predicted 
values within a vertical strip.  
\bigskip
\noindent
(a)  Recall that about 68 percent of a normally distributed
data set is within 1 SD of the average.  Hence we expect about
32 or roughly 1/3 of the data to be more than one SD from 
the average.  This means that in this problem roughly
1/3 of the predicted final scores would be off by
more than 1 rms error, or more than $\sqrt{1 - r^2} \times SD_y = 
\sqrt{1 - (.6)^2} \times 15 = .8 \times 15 = 12$ points.
\bigskip
\noindent
(b)  This is like the problems in the previous section, except 
that you must keep track of {\it which average and which SD to use}
by whether you are working with the midterm scores or
the final scores.
A midterm score of 80 is ${80 - 50\over 25} = 1.2$ $SD_x$'s
above the average.  The predicted final score is $r \times 1.2$ $SD_y$'s
above the average of the $y$'s, so $55 + (.6)(1.2)(15) = 55 + 10.8 = 65.8$.
\bigskip
\noindent
(c)  12 points = rms error (since SD is the expected size of 
the ``experimental error'').
\bigskip
\noindent
12.  To answer this question, of course we need to compute
the predicted blood pressure for the man with 20 years of 
education.  $x = 20$ is $7/3 = 2.33$ SD's above the average
for education.  The predicted blood pressure is 
$${\rm ave}_y + (r \times SD_y)(2.33) = 119 + (-0.1 \times 11)(2.33) \doteq 116.4$$
(Note that we {\it do not} divide by $SD_x = 3$ again here, since that is already
done in the computation of the $2.33$ SD's above the average for $x$.  The 
${r\times SD_y\over SD_x}$ is the slope of the regression line.  If you use
that equation, then ``raw values'' of $x$ and $y$ are used, not the corresponding
standard unit values.)
Thus, the blood pressure value of 118 for the man {\it is ``a bit high''} (though of
course a difference this small might not actually have any practical significance).  
 
\bye   