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-#LyX 2.4 created this file. For more info see https://www.lyx.org/
-\lyxformat 620
-\begin_document
-\begin_header
-\save_transient_properties true
-\origin unavailable
-\textclass book
-\begin_preamble
-\input defs
-\end_preamble
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-\language english
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-\index Index
-\shortcut idx
-\color #008000
-\end_index
-\secnumdepth 3
-\tocdepth 3
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-\end_header
-
-\begin_body
-
-\begin_layout Standard
-\begin_inset Note Note
-status open
-
-\begin_layout Plain Layout
-TODO 1,
- 3,
- 4,
- 8,
- 9,
- 10,
- 11 (2 pp.,
- 1:11)
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\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?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Row 
-\begin_inset Formula $\nu=k-1=11-1=10$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\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.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\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
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\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
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\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.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc8[00]
-\end_layout
-
-\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
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\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.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc9[20]
-\end_layout
-
-\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.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\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.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\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?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\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
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc11[10]
-\end_layout
-
-\end_inset
-
-Solve exercise 10 substituting the KS test for the chi-square test.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\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 
-\[
-K_{2n}^{-}=\sqrt{2n}\max_{1\leq j\leq2n}\left(F(X'_{j})-\frac{j-1}{2n}\right)=\sqrt{2n}\max_{1\leq j\leq n}\left(F(X_{j})-\frac{2j-2}{2n}\right)=\sqrt{2}K_{n}^{-}.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\end_body
-\end_document