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\hfill December 16, 2008
\vskip 12 pt
\centerline{\bf Final Exam --- Math 102 / Core 143 CX}
\vskip 12 pt
\noindent
Points are in parentheses.  An unsimplified
answer like $12\sqrt{3.51}+6/7$ is usually worth more than 23.3,
because it is easier to understand where it came from.

\begin{itemize}

\qu{1}{25} An educational researcher wants to test whether students using
a new curriculum for teaching fractions learn the material better than those
using an older curriculum.  She teaches 100 students, chosen at random, by
the new method, and then administers a standardized exam.  Long experience
has shown that using the old method, students score an average of 50 on this
exam, but her students' average score is 54, with an SD of 20.
\begin{itemize}
\item[(a)] What kind of test ($z$ or $t$ or 2-sample $z$ or $\chi^2$) should
be used to decide whether the new curriculum is better than the old?
\item[(b)] Find the value of the variable ($z$ or $t$ or $\chi^2$) used in
this test.
\item[(c)] Find the $P$-value, i.e., the probability that, if the new
curriculum were no better than the old, her students would average 54
or more just by chance.
\item[(d)] Should we conclude that the new curriculum results in higher
scores than the old?
\item[(e)] Suppose another researcher reaches a conclusion that is
the opposite of ours.  Should we conclude that his data is ``tainted''
(i.e., obtained by faulty methods or even perhaps falsified)?
\end{itemize}

\qu{2}{20} The cathat plant has flowers that are either pink, blue or
lavender.  A genetic model holds that a single gene controls the color of
the flowers, with pure forms (genotypes p/p or b/b) having pink or blue
flowers respectively; but
neither is dominant, so that hybrids (p/b or b/p) have lavender flowers.
In an experiment, lavender-flowered cathats are crossed with each other,
and, out of 40 offspring selected at random, 15 have pink flowers, 22 have
lavender flowers and 3 have blue flowers.
\begin{itemize}
\item[(a)] What kind of test would be used to decide whether
these results are significant evidence against the genetic model?
\item[(b)] In computing the value of the variable ($z$ or $t$ or $\chi^2$)
used in this test, should we use the numbers of plants with flowers of
each color, or the percentages of each color?  Or doesn't it matter?
\item[(c)] Compute the value of the variable.
\item[(d)] Is this data significant evidence against the model?
\end{itemize}

\qu{3}{15} An ecologist projects the level $N$ of a certain nutrient
in a stream by using linear regression on the amount $F$ of fertilizer
used in a few fields near the stream. Suppose the
averages of $F$ and $N$ are 3 (tons) and 3.5 (parts per thousand)
respectively, with standard deviations of 2.5 and 2 respectively,
and a correlation of 0.3.
\begin{itemize}
\item[(a)] What is his regression equation for projecting $N$ from $F$?.
\item[(b)] Roughly how far off should the ecologist expect the
projections to be as he makes them using his equation in (a)?
\item[(c)] Suppose his regression equation was $N = .4F+5$ (which it isn't).
If the nutrient level turns out to be 7 when 8 tons of fertilizer were used,
what is the corresponding residue (or residual) relative to his projection?
\end{itemize}

\qu{4}{15} A market survey of 400 randomly chosen households in a large
city finds that 280 of them have an internet connection and the average
amount of money spent per meal is \$8, with a standard deviation of \$4.
\begin{itemize}
\item[(a)] Find a 95\% confidence interval for the percentage of
households in the city that have internet connections.
\item[(b)] How many households must be surveyed so that the confidence
interval requested in (a) will turn out only a third as large as the
one you found in (a)?
\item[(c)] Find a 85\% confidence interval for the average amount of
money spent per meal in the city.
\end{itemize}

\qu{5}{20} On the modified boxplot, name the values (a), (b) and (c) (the
dots) relative to the distribution.  For the scatterplot (d), estimate the
correlation. For (e), three of the histograms were generated from a very large
survey of individuals' numbers of years of schooling; one is the actual data,
another is the averages of samples of 50 taken from the data, a third
is averages of samples of 500 taken from the data, and the fourth is not
related.  Arrange the histograms in the order just described. \newline
\centerline{\scalebox{.6}{\includegraphics{feF07n5.png}}}

\qu{6}{20} A pinochle deck has 48 cards, in the usual four suits but only
six ranks (9-10-J-Q-K-A), two of each card.  (Thus, for example, there are
8 aces and 12 spades.) In each case below, a card is
selected only after the deck is shuffled.
\begin{itemize}
\item[(a)] If two cards are selected \underbar{without} replacement,
what is the probability that both are kings?
\item[(b)] If one card is selected, what is the probability that it
is either a king or a club, or both?
\item[(c)] If five cards are selected \underbar{with} replacement,
what is the probability that at least one is a club?
\item[(d)] If five cards are selected \underbar{with} replacement,
what is the probability that exactly three are clubs?
\end{itemize}

\qu{7}{10} Relative to the article ``Who'll Stop the Rain?''
by Sharon Begley: An article in the Irish Times shortly after 
the Olympics began reported that there was no rain on Beijing 
during the Friday opening ceremonies, but there was rain on 
Sunday. Does that mean Begley is wrong?

\end{itemize}

\newpage
{\parindent=0pt
\parskip=12pt
\def\ind{\strut\hskip .3 true in}

\centerline{\bf Formulas and Terms in Math 102 / Core 143}

\vskip 12 pt

In a histogram,\newline
\ind\# of scores (or total probability of outcomes) in
an interval\newline
\ind\ind$=$ area under the histogram above
that interval

\vskip 12 pt

For a list of $n$ scores ($x$-values):\newline
\ind Average $\overline{x} = \mu = \displaystyle\frac{\sum x}{n}$,\ind
Standard deviation $\sigma  = SD
   = \sqrt{\displaystyle{\frac{\sum(x-\overline{x})^2}{n}}}$
\newline
\ind $N$-th percentile $=$ $x$-value for which $N\%$ of scores
are $\leq x$\newline
\ind Median $=$ 50-th percentile,\ind $IQR =$ 75-th $-$ 25-th
percentile

\vskip 12 pt

Converting a score $x$ to standard units $(z$-value or $t$-value):
$$
z {\rm\ or\ }t = \displaystyle{\frac{x-\overline{x}}{\sigma}}
\qquad{\rm or}\qquad \displaystyle{\frac{x-EV}{SE}}\ {\rm (see\ below)}
$$

Correlation: $r = $ average of products $(z_x z_y)$, where $z_q$ means
$q$ in standard units

Regression line of $y$ on $x$ (for predicting $y$ from $x$,
or for estimating average $y$ within a vertical strip at $x$):  Denote
predicted $y$-value by $\hat{y}$. Then
$$
\hat{y} - \overline{y}
   = r\left(\displaystyle\frac{\sigma_y}{\sigma_x}\right)(x - \overline{x})
$$
Residual corresponding to data point $(x,y)$: $y-\hat{y}$ or $(x,y-\hat{y})$.

RMS error for regression of $y$ on $x$
$(=$ approximate standard deviation of data in any vertical strip, if
scatter diagram is homoscedastic):
$$
\sigma_y\sqrt{1 - r^2}
$$

Multiplication rule: Prob($A{\rm\ and\ }B)
   =$ Prob($A)\cdot$Prob($B{\rm\ given\ }A)$

Independent events: Prob($B{\rm\ given\ }A) =$ Prob($B)$

Addition rule:
Prob($A{\rm\ or\ }B) =$ Prob$(A)$ $+$ Prob($B) -$ Prob($A$ and $B$)

Mutually exclusive events: Prob($A{\rm\ and\ }B) = 0$

Binomial probabilities: If an event has probability~$p$ on each
trial, the probability of its occurring exactly $k$~times
in $n$~independent trials:
$$
C(n,k)\cdot p^k\cdot(1-p)^{n-k}\ , \qquad {\rm where}
\qquad C(n,k) = \displaystyle\frac{n!}{k!(n-k)!}
$$

\newpage

For a sample of size~$n$ from a population with average~$\mu$
and standard deviation $\sigma$:\newline
\ind $EV$ of sum of scores in sample $= n\mu$\newline
\ind $SE$ of sum  $= \sigma\cdot \sqrt{n}$\newline
\ind $EV$ of average of sample = $\mu$\newline
\ind $SE$ of average = $\sigma/\sqrt{n}$\newline
For significance tests (especially with small samples), approximate 
(bootstrap) population standard
deviation $\sigma$ with \underbar{sample} standard deviation $s =
SD^+ = \sqrt{\displaystyle{\frac{\sum(x-\overline{x})^2}{n-1}}} =
(SD{\rm\ of\ sample})\sqrt{\frac{n}{n-1}}$.  (The null hypothesis
will give a value to use for $\mu$.)  For large samples $(n\geq 30)$,
$s$ is close to $\sigma$.\newline
For confidence intervals, also approximate population average~$\mu$
with sample average~$\overline{x}$.

Special case: Population is 0's and 1's (or yeses and nos, or
ins and outs, or $\ldots$), fraction of 1's is
$p$, for a sample of size~$n$:\newline
\ind $EV$ of count $= np$\newline
\ind $SE$ of count $= \sqrt{p(1-p)}\cdot\sqrt{n}$\newline
\ind $EV$ of \% (or proportion) $= p$\newline
\ind $SE$ of \% (or proportion) = $\sqrt{p(1-p)}/\sqrt{n}$ \newline
For CIs, approximate (bootstrap) population
proportion~$p$ with sample proportion $\hat{p}$.

For use with confidence interval or $t$-test for significance on
small ($n<30$) samples: degrees of freedom $= n-1$

$k\%$ confidence interval for the average of a population: Let $z_k$
denote the $z$-value for which $k$ percent of the data is between $-z_k$ and
$z_k$.  Then the CI is
$$
\overline{x} \pm z_k\cdot(SE{\rm\ for\ average})
$$
(Similar for ``proportion'' in place of ``average''.)

For significance test for difference of $\mu$'s in two
populations:\newline
\ind $SE$ for difference of averages of 2 samples
$= \sqrt{(SE{\rm\ of\ first})^2
   + (SE{\rm\ of\ second})^2}$\newline
$EV$ of difference $=0$ by $H_0$. (For more than 2 samples, 
use one-way ANOVA.)

For deciding significance of differences in frequency
distributions among categories:\newline
\ind $\chi^2 = \sum$[(observed $-$ expected)$^2/$expected]\newline
\ind degrees of freedom: in ``list'' distributions, \# in list $-1$;\newline
\ind\hphantom{degrees of freedom:} in ``table'' distributions,
(\# of rows $-1)\cdot($\# of columns $-1$)

}

\newpage

\small

\centerline{\begin{tabular}{|cc|cc|cc|cc|cc|}
\hline\multicolumn{10}{|c|}{Normal table (Area between $-z$ and $z$)}\\\hline
$z$ & Area(\%) & $z$ & Area(\%) & $z$ & Area(\%) & $z$ & Area(\%) & $z$ & Area(\%)\\
\hline
0.0 & 0.0 & 0.9 & 63.19 & 1.8 & 92.81 & 2.7 & 99.31 & 3.6 & 99.968 \\
0.05 & 3.99 & 0.95 & 65.79 & 1.85 & 93.57 & 2.75 & 99.4 & 3.65 & 99.974 \\
0.1 & 7.97 & 1 & 68.27 & 1.9 & 94.26 & 2.8 & 99.49 & 3.7 & 99.978 \\
0.15 & 11.92 & 1.05 & 70.63 & 1.95 & 94.88 & 2.85 & 99.56 & 3.75 & 99.982 \\
0.2 & 15.85 & 1.1 & 72.87 & 2 & 95.45 & 2.9 & 99.63 & 3.8 & 99.986 \\
0.25 & 19.74 & 1.15 & 74.99 & 2.05 & 95.96 & 2.95 & 99.68 & 3.85 & 99.988 \\
0.3 & 23.58 & 1.2 & 76.99 & 2.1 & 96.43 & 3 & 99.73 & 3.9 & 99.99 \\
0.35 & 27.37 & 1.25 & 78.87 & 2.15 & 96.84 & 3.05 & 99.771 & 3.95 & 99.992 \\
0.4 & 31.08 & 1.3 & 80.64 & 2.2 & 97.22 & 3.1 & 99.806 & 4 & 99.9937 \\
0.45 & 34.73 & 1.35 & 82.3 & 2.25 & 97.56 & 3.15 & 99.837 & 4.05 & 99.9949 \\
0.5 & 38.29 & 1.4 & 83.85 & 2.3 & 97.86 & 3.2 & 99.863 & 4.1 & 99.9959 \\
0.55 & 41.77 & 1.45 & 85.29 & 2.35 & 98.12 & 3.25 & 99.885 & 4.15 & 99.9967 \\
0.6 & 45.15 & 1.5 & 86.64 & 2.4 & 98.36 & 3.3 & 99.903 & 4.2 & 99.9973 \\
0.65 & 48.43 & 1.55 & 87.89 & 2.45 & 98.57 & 3.35 & 99.919 & 4.25 & 99.9979 \\
0.7 & 51.61 & 1.6 & 89.04 & 2.5 & 98.76 & 3.4 & 99.933 & 4.3 & 99.9983 \\
0.75 & 54.67 & 1.65 & 90.11 & 2.55 & 98.92 & 3.45 & 99.944 & 4.35 & 99.9986 \\
0.8 & 57.63 & 1.7 & 91.09 & 2.6 & 99.07 & 3.5 & 99.953 & 4.4 & 99.9989 \\
0.85 & 60.47 & 1.75 & 91.99 & 2.65 & 99.2 & 3.55 & 99.961 & 4.45 & 99.9991 \\
\hline
\end{tabular}}
\vskip 12 pt
\parbox{3truein}{\begin{tabular}{|c|rrrrrr|}
\hline\multicolumn{7}{|c|}{$t$-table: column head is $P(t\geq$ entry)}\\\hline
$df$& 25\%& 10\%& 5\%& 2.5\%& 1\%&  0.5\%\\\hline
 1&    1.00&   3.08&   6.31&  12.71&  31.82&  63.66\\
 2&    0.82&   1.89&   2.92&   4.30&   6.96&   9.92\\
 3&    0.76&   1.64&   2.35&   3.18&   4.54&   5.84\\
 4&    0.74&   1.53&   2.13&   2.78&   3.75&   4.60\\
 5&    0.73&   1.48&   2.02&   2.57&   3.36&   4.03\\
 6&    0.72&   1.44&   1.94&   2.45&   3.14&   3.71\\
 7&    0.71&   1.41&   1.89&   2.36&   3.00&   3.50\\
 8&    0.71&   1.40&   1.86&   2.31&   2.90&   3.36\\
 9&    0.70&   1.38&   1.83&   2.26&   2.82&   3.25\\
10&    0.70&   1.37&   1.81&   2.23&   2.76&   3.17\\
11&    0.70&   1.36&   1.80&   2.20&   2.72&   3.11\\
12&    0.70&   1.36&   1.78&   2.18&   2.68&   3.05\\
13&    0.69&   1.35&   1.77&   2.16&   2.65&   3.01\\
14&    0.69&   1.35&   1.76&   2.14&   2.62&   2.98\\
15&    0.69&   1.34&   1.75&   2.13&   2.60&   2.95\\
16&    0.69&   1.34&   1.75&   2.12&   2.58&   2.92\\
17&    0.69&   1.33&   1.74&   2.11&   2.57&   2.90\\
18&    0.69&   1.33&   1.73&   2.10&   2.55&   2.88\\
19&    0.69&   1.33&   1.73&   2.09&   2.54&   2.86\\
20&    0.69&   1.33&   1.72&   2.09&   2.53&   2.85\\
21&    0.69&   1.32&   1.72&   2.08&   2.52&   2.83\\
22&    0.69&   1.32&   1.72&   2.07&   2.51&   2.82\\
23&    0.69&   1.32&   1.71&   2.07&   2.50&   2.80\\
24&    0.68&   1.32&   1.71&   2.06&   2.49&   2.80\\
25&    0.68&   1.32&   1.71&   2.06&   2.49&   2.79\\
\hline
\end{tabular}}
\hfill
\parbox{3truein}{\begin{tabular}{|c|rrrrr|}
\hline\multicolumn{6}{|c|}{$\chi^2$-table: column head is
   $P(\chi^2\geq$ entry)}\\\hline
$df$&    50\%&    30\%&    10\%&    5\%&     1\%\\\hline
 1&      0.46&    1.07&    2.71&    3.84&    6.64\\
 2&      1.39&    2.41&    4.60&    5.99&    9.21\\
 3&      2.37&    3.67&    6.25&    7.82&   11.34\\
 4&      3.36&    4.88&    7.78&    9.49&   13.28\\
 5&      4.35&    6.06&    9.24&   11.07&   15.09\\
 6&      5.35&    7.23&   10.65&   12.59&   16.81\\
 7&      6.35&    8.38&   12.02&   14.07&   18.48\\
 8&      7.34&    9.52&   13.36&   15.51&   20.09\\
 9&      8.34&   10.66&   14.68&   16.92&   21.67\\
10&      9.34&   11.78&   15.99&   18.31&   23.21\\
11&     10.34&   12.90&   17.28&   19.68&   24.73\\
12&     11.34&   14.01&   18.55&   21.03&   26.22\\
13&     12.34&   15.12&   19.81&   22,36&   27.69\\
14&     13.34&   16.22&   21.06&   23.69&   29.14\\
15&     14.34&   17.32&   22.31&   25.00&   30.58\\
16&     15.34&   18.42&   23.54&   26.30&   32.00\\
17&     16.34&   19.51&   24.77&   27.59&   33.41\\
18&     17.34&   20.60&   25.99&   28.87&   34.81\\
19&     18.34&   21.69&   27.20&   30.14&   36.19\\
20&   19.34&   22.78&   28.41&   31.41&   37.57\\
\hline
\end{tabular}}

\normalsize

\newpage

\noindent
{\bf Solutions to Final Exam}

\begin{itemize}

\item[1.]\begin{itemize}
\item[(a)] A $z$-test.
\item[(b)] $EV = 50$ by $H_0$, $SD = 20$ by bootstrapping, so
$SE = 20/\sqrt{100} = 2$: $z = (54-50)/2 = 2$.
\item[(c)] $P({\rm avg}\geq 54) = P(z\geq 2) = (100\%-95\%)/2 = 2.5\%$
\item[(d)] Yes, the probability of a score at least this high by
chance is less than 5\%, so we reject the null hypothesis that the
average if all students used the new curriculum is only 50.
\item[(e)] No, we know from our computer projects that significance
tests can be wrong, so either he or we are making a legitimate mistake
on the basis of validly obtained data.
\end{itemize}

\item[2.] \begin{itemize}
\item[(a)] A $\chi^2$-test.
\item[(b)] The numbers must be used; percentages would give the wrong value.
\item[(c)] Under the null hypothesis that the model is correct, we would
expect 10 pink, 20 lavender and 10 blue.  So $\chi^2 = (15-10)^2/10 +
(22-20)^2/20 + (3-10)^2/10 = 7.6$.
\item[(d)] From the $\chi^2$-table with $df=3-1=2$, we see that $\chi^2 =
7.6$ is significant, so we reject the null hypothesis.
\end{itemize}

\item[3.] \begin{itemize}
\item[(a)] $N - 3.5 = .3(2/2.5)(F - 3)$
\item[(b)] $2\sqrt{1-(.3)^2}$
\item[(c)] $7 - (.4(8)+5) = -1.2$.  Technically this is correct, but I won't
take off points if the sign is positive.
\end{itemize}

\item[4.] \begin{itemize}
\item[(a)] $280/400 \pm 2\sqrt{(280/400)(120/400)}/\sqrt{400} \approx 70\%
\pm 4.6\%$
\item[(b)] 9 times as many, or 3600
\item[(c)] Using the normal table: $8 \pm 1.45(4/\sqrt{400}) = 8 \pm .29$ dollars
\end{itemize}

\item[5.] \begin{itemize}
\item[(a)] Median.
\item[(b)] Third quartile, or 75-th percentile.
\item[(c)] Outliers.
\item[(d)] Anything from $-.5$ to $-.9$ will be accepted.
\item[(e)] III, IV, I, II (The distinction between III and IV is a bit
thin, so I will give credit if they are reversed.)
\end{itemize}

\item[6.] \begin{itemize}
\item[(a)] $(8/48)(7/47)$
\item[(b)] $8/48 + 12/48 - 2/48$
\item[(c)] $1 - (3/4)^5$
\item[(d)] $(5!/(3!2!))(1/4)^3(3/4)^2$
\end{itemize}

\item[7.] Begley claimed that the ``science'' of weather control 
was not capable of assuring that there would be no rain on the 
Olympic opening ceremonies, despite the Chinese government's 
claim that it had ways of doing that.  Sure enough, there was no 
rain on the ceremonies.  But of course the weather is very fickle;
the lack of rain that day probably had nothing to with the 
government's efforts.  In fact, their efforts were designed to 
bring rain \underbar{before} the ceremonies, in order to ``wring
out'' the clouds so that they would not rain on the Chinese
parade.  But the rain came after the ceremonies, not before.  
So in my opinion Begley was probably right.  But you may feel 
differently; as usual, I'm not looking for a ``right'' answer, but
rather for evidence that you had read and thought a bit about the 
article. 

\end{itemize}

\end{document}


\centerline{\includegraphics{.png}}
