\magnification=\magstep1
\centerline{MONT 107N -- Understanding Randomness}
\centerline{Selected Solutions for Problem Set 3}
\centerline{March 14, 2010}
\bigskip
\noindent
{\it Chapter 20 Review Problems}
\bigskip
\noindent
3)  a)  There should be $50,000$ tickets in the box -- one for each of the
forms in the whole ``population.''
\medskip
\noindent
b)  Since we are counting/classifying, this will be a 0/1 box;
tickets with 1's will represent forms with gross income over
$\$50,000$, and tickets with 0's will represent forms
with gross incomes under that level.
\medskip
\noindent
c)  The SD of the box is (using the ``shortcut formula'') 
$${\rm SD\ of\ box} = \sqrt{(.2) \times (.8)} = \sqrt{.16} = .4,$$
so this is False.
\medskip
\noindent
d)  The 900 draws from the box will represent the sample, so 
this is True.
\medskip
\noindent
e)  The number of forms in the sample with gross income over $\$50,000$ is
the sum of the tickets drawn from the box, so ${\rm EV} = 900 \times .2 = 180$,
and the SE for the sum is $\sqrt{900} \times .4 = 12$.  The number of
forms in the sample with gross incomes over $\$50,000$ will be about
180, plus or minus 12 or so.  
\medskip
Now $12/900 \times 100\% \doteq 1.33\%$ is the SE for the percent. 
The percent of forms in the sample with gross incomes over $\$50,000$ will be about
$20\%$ give or take $1.33\%$ or so.  
\medskip
Finally, we want to estimate the chance
that a particular sample will have between $19\%$ and $21\%$ of
forms with gross income over $\$50,000$.  Note that $1\% = .75\times 1.33$.
So $1\%$ corresponds to $.75$ in standard units (multiples of the SE for the percent).
So this chance will be estimated by the area under the normal curve
from $-.75$ to $.75$.  This is about $55\%$.  
\medskip
\noindent
f)  This cannot be done with the given information.  There is no information
about the percentage of forms with gross incomes over $\$75,000$.  
\bigskip
\noindent
5) The total weight for a random group of 50 people is like the 
sum of a random draw of 50 tickets from a box representing
the whole population of people.  The ticket for each person
has his or her weight in pounds.  The given information shows
that the average of the box is 150, and the SD of the box is 35.
Hence the EV for the sum of the draws is $50 \times 150 = 7500$ pounds
and the SE for the sum is $\sqrt{50} \times 35 \doteq 247.5$ pounds.
The chance that a fully-loaded elevator will be carrying more
than 4 tons $= 8000$ pounds is estimated as the area under the
normal curve with 
$$z > {8000 - 7500\over 247.5} \doteq 2.02,$$
which is a bit over $2\%$.  (Since overloading is not all that
unlikely, maybe a sturdier elevator is indicated.  You don't
want the elevator to fail roughly every 50 times it used
with a full load of people!!)
\bigskip
\noindent
6)  Only option (ii) is correct here.  It is important to 
realize that the size of the sample relative to the size of
the population {\it is essentially irrelevant} for questions
like this(!)  The {\it absolute size of the sample} is what 
matters because of the formula for the SE for the percentage:
$$\sqrt{{pq\over n}} \times 100\%,$$
where $p = $ observed proportion, $q = 1 - p$, and $n$ is the
size of the sample.  The larger the sample is, the smaller
the SE is (no matter what the size of the population the
sample is drawn from is).
\bigskip
\noindent
{\it Chapter 21 Review Exercises}
\bigskip
\noindent
5) a) Think of a (big!) 0/1 box with one 1 ticket for
each teenager who knows that Chaucer wrote the {\it Canterbury
Tales}, and a 0 ticket for each teenager who does not know
that fact.  Then the data is like the result of drawing
a random sample of size $6000$ from the box, and we
observe a percentage of $36.1\%$ of 1s in the sample.
Using the bootstrap, the SE is estimated by
$$\sqrt{(.361)(.639)\over 6000} \times 100\% = .6\%$$
(that's $.6$ of $1\%$).  Hence the $95\%$ is estimated
as 
$$36.1\% \pm 2 \times (.6\%) = 36.1\% \pm 1.2\%.$$ 
\medskip
\noindent
b)  This part is similar.  (There is no reason why one
cannot repeat the calculations for the percentage of
the population of teenagers who know who Thomas Edison
was.)  The answer is $95.2\% \pm 0.6\%$.  
\bye


