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\begin_body

\begin_layout Standard
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\begin_layout Standard
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\backslash
exerc1[00]
\end_layout

\end_inset

What line of the chi-square table should be used to check whether or not the value 
\begin_inset Formula $V=7\frac{7}{48}$
\end_inset

 of Eq.
 (5) is improbably high?
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\backslash
answer 
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\end_inset

Row 
\begin_inset Formula $\nu=k-1=11-1=10$
\end_inset

.
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\backslash
rexerc3[23]
\end_layout

\end_inset

Some dice that were loaded as described in the previous exercise were rolled 144 times,
 and the following values were observed:
\begin_inset Formula 
\[
\begin{array}{rrrrrrrrrrrr}
\text{value of }s= & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12\\
\text{observed number, }Y_{s}= & 2 & 6 & 10 & 16 & 18 & 32 & 20 & 13 & 16 & 9 & 2
\end{array}
\]

\end_inset

Apply the chi-square test to 
\emph on
these
\emph default
 values,
 using the probabilities in (1),
 pretending that the dice are not in fact known to be faulty.
 Does the chi-square test detect the bad dice?
 If not,
 explain why not.
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answer 
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\end_inset

We take the values 
\begin_inset Formula $np_{s}$
\end_inset

 from (2),
 to get:
\begin_inset Formula 
\begin{align*}
V & =\sum_{s=2}^{12}\frac{(Y_{s}-np_{s})^{2}}{np_{s}}=\frac{4}{4}+\frac{4}{8}+\frac{4}{12}+\frac{0}{16}+\frac{4}{20}+\frac{64}{24}+\frac{0}{20}+\frac{9}{16}+\frac{16}{12}+\frac{1}{8}+\frac{4}{4}\\
 & =1+\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{8}{3}+\frac{9}{16}+\frac{4}{3}+\frac{1}{8}+1=2+\frac{13}{3}+\frac{19}{16}+\frac{1}{5}\\
 & =7+\frac{80+45+48}{240}=7+\frac{173}{240}.
\end{align*}

\end_inset

Using 
\begin_inset Formula $n=10$
\end_inset

 we get a probability between 
\begin_inset Formula $.25$
\end_inset

 and 
\begin_inset Formula $.5$
\end_inset

,
 which is not suspect.
 This seems to be because the bias of one die compensates that of the other,
 smoothing out the probability differences.
 The difference could be discovered with a large enough value of 
\begin_inset Formula $n$
\end_inset

.
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\backslash
rexerc4[23]
\end_layout

\end_inset

The author actually obtained the data in experiment 1 of (9) by simulating dice in which one was normal,
 the other was loaded so that it always turned up 1 or 6.
 (The latter two possibilities were equally probable.) Compute the probabilities that replace (1) in this case,
 and by using a chi-square test decide if the results of that experiment are consistent with the dice being loaded in this way.
\end_layout

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answer 
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\end_inset

We compute the table with the sum of the two dice:
\begin_inset Formula 
\[
\begin{array}{r|rrrrrr}
 & 1 & 2 & 3 & 4 & 5 & 6\\
\hline 1 & 2 & 3 & 4 & 5 & 6 & 7\\
6 & 7 & 8 & 9 & 10 & 11 & 12
\end{array}
\]

\end_inset

This gives us the following table of probabilities:
\begin_inset Formula 
\[
\begin{array}{rrrrrrrrrrrr}
s= & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12\\
144p_{s}= & 12 & 12 & 12 & 12 & 12 & 24 & 12 & 12 & 12 & 12 & 12
\end{array}
\]

\end_inset

Thus,
\begin_inset Formula 
\begin{align*}
V & =\frac{1}{12}\left(8^{2}+2^{2}+2^{2}+1^{2}+8^{2}+\frac{6^{2}}{2}+6^{2}+1^{2}+1^{2}+2^{2}+1^{2}\right)\\
 & =\frac{1}{12}(64+4+4+1+64+18+26+1+1+2+1)=\frac{186}{12}=15+\frac{1}{2}.
\end{align*}

\end_inset

With 
\begin_inset Formula $n=10$
\end_inset

,
 this is somewhat in the middle of 
\begin_inset Formula $p=.75$
\end_inset

 and 
\begin_inset Formula $p=.95$
\end_inset

,
 which is consistent.
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\begin_layout Standard
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\backslash
exerc8[00]
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\end_inset

The text describes an experiment in which 20 values of the statistic 
\begin_inset Formula $K_{10}^{+}$
\end_inset

 were obtained in the study of a random sequence.
 These values were plotted,
 to obtain Fig.
 4,
 and a KS statistic was computed from the resulting graph.
 Why were the table entries for 
\begin_inset Formula $n=20$
\end_inset

 used to study the resulting statistic,
 instead of the table entries for 
\begin_inset Formula $n=10$
\end_inset

?
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answer 
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\end_inset

Because the value of 
\begin_inset Formula $n$
\end_inset

 to use is not about the underlying probability distribution (which can be an arbitrary real-valued one,
 not just 
\begin_inset Formula $K_{n}^{+}$
\end_inset

 or 
\begin_inset Formula $K_{n}^{-}$
\end_inset

),
 but rather it is the number of observations we make for this distribution,
 which is 20.
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\begin_layout Standard
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\backslash
rexerc9[20]
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\end_inset

The experiment described in the text consisted of plotting 20 values of 
\begin_inset Formula $K_{10}^{+}$
\end_inset

,
 computed from the maximum-of-5 test applied to different parts of a random sequence.
 We could have computed also the corresponding 20 values of 
\begin_inset Formula $K_{10}^{-}$
\end_inset

;
 since 
\begin_inset Formula $K_{10}^{-}$
\end_inset

 has the same distribution as 
\begin_inset Formula $K_{10}^{+}$
\end_inset

,
 we could lump together the 40 values thus obtained (that is,
 20 of the 
\begin_inset Formula $K_{10}^{+}$
\end_inset

's and 20 of the 
\begin_inset Formula $K_{10}^{-}$
\end_inset

's),
 and a KS test could be applied so that we would get new values 
\begin_inset Formula $K_{40}^{+}$
\end_inset

,
 
\begin_inset Formula $K_{40}^{-}$
\end_inset

.
 Discuss the merits of this idea.
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answer 
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\end_inset

The issue here is that the 40 points would not be independent;
 if the maximum of 5 is low,
 the minimum of 5 must be necessarily lower,
 the probability of it being higher is 0.
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\begin_layout Standard
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\backslash
rexerc10[20]
\end_layout

\end_inset

Suppose a chi-square test is done by making 
\begin_inset Formula $n$
\end_inset

 observations,
 and the value 
\begin_inset Formula $V$
\end_inset

 is obtained.
 Now we repeat the test on these same 
\begin_inset Formula $n$
\end_inset

 observations over again (getting,
 of course,
 the same results),
 and we put together the data from both tests,
 regarding it as a single chi-square test with 
\begin_inset Formula $2n$
\end_inset

 observations.
 (This procedure violates the text's stipulation that all of the observations must be independent of one another.) How is the second value of 
\begin_inset Formula $V$
\end_inset

 related to the first one?
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\backslash
answer 
\end_layout

\end_inset

Let 
\begin_inset Formula $Y'_{s}=2Y_{s}$
\end_inset

 be the number of observations of category 
\begin_inset Formula $s$
\end_inset

 in the second test,
 the second value of 
\begin_inset Formula $V$
\end_inset

 is
\begin_inset Formula 
\[
V'=\sum_{s=1}^{k}\frac{(Y'_{s}-2np_{s})^{2}}{2np_{s}}=\sum_{s=1}^{k}\frac{(2Y_{s}-2np_{s})^{2}}{2np_{s}}=2V.
\]

\end_inset


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\backslash
exerc11[10]
\end_layout

\end_inset

Solve exercise 10 substituting the KS test for the chi-square test.
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\backslash
answer 
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\end_inset

This time,
 after sorting the 
\begin_inset Formula $2n$
\end_inset

 observations 
\begin_inset Formula $X'_{1},\dots,X'_{2n}$
\end_inset

,
 we have 
\begin_inset Formula $X_{j}=X'_{2j-1}=X'_{2j}$
\end_inset

,
 so
\begin_inset Formula 
\[
K_{2n}^{+}=\sqrt{2n}\max_{1\leq j\leq2n}\left(\frac{j}{2n}-F(X'_{j})\right)=\sqrt{2n}\max_{1\leq j\leq n}\left(\frac{2j}{2n}-F(X_{j})\right)=\sqrt{2}K_{n}^{+},
\]

\end_inset

and similarly,
\begin_inset Formula 
\[
K_{2n}^{-}=\sqrt{2n}\max_{1\leq j\leq2n}\left(F(X'_{j})-\frac{j-1}{2n}\right)=\sqrt{2n}\max_{1\leq j\leq n}\left(F(X_{j})-\frac{2j-2}{2n}\right)=\sqrt{2}K_{n}^{-}.
\]

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\end_document