統計検定準1級例題集解答/解答例と解説

論述問題問3 [2]

設問の要約

観測データ

	発生	非発生	計
投与群	$a$	$b$	$m$
非投与群	$c$	$d$	$n$
計	$s$	$t$	$N$

オッズ比の推定値
$O R^{*} = \frac{a d}{b c}$
$\begin{equation} OR^{∗} = \frac{ad}{bc} \end{equation}$
対数オッズ比　 $\log OR^∗$ の標準誤差
$S E [\log O R^{*}] = \sqrt{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}}$
$\begin{equation} SE[\log OR^∗] = \sqrt{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}} \end{equation}$
$RR$ の推定値 $RR^∗$ を求めよ．
$\log RR^∗$ の標準誤差を求めよ．

解答例

$p_1$ の推定値 $\hat{p}_1$ と， $p_0$ の推定値 $\hat{p}_0$ は，

{\hat{p}}_{1} = \frac{a}{m}

$\begin{equation} \hat{p}_1 = \frac{a}{m} \end{equation}$

{\hat{p}}_{0} = \frac{c}{n}

$\begin{equation} \hat{p}_0 = \frac{c}{n} \end{equation}$

R R

$RR$ の推定値

R R^{*}

$RR^{*}$ は，

R R^{*} = \frac{{\hat{p}}_{1}}{{\hat{p}}_{0}} = \frac{a / m}{c / n} = \frac{a n}{c m}

$\begin{equation} RR^{*} = \frac{\hat{p}_1}{\hat{p}_0} = \frac{a/m}{c/n} = \frac{a\,n}{c\,m} \end{equation}$

\log R R^{*}

$\log RR^{*}$ の標準偏差

S E [\log R R^{*}]

$SE[\log RR^{*}]$ は，

S E [\log R R^{*}] = \sqrt{V a r [\log R R^{*}]}

$\begin{equation} SE[\log RR^{*}] = \sqrt{\mathrm{Var}[\log RR^{*}]} \end{equation}$

V a r [\log R R^{*}] = V a r [\log \frac{{\hat{p}}_{1}}{{\hat{p}}_{0}}] = V a r [\log {\hat{p}}_{1} - \log {\hat{p}}_{0}]

$\begin{equation} \mathrm{Var}[\log RR^{*}] = \mathrm{Var}\left[\log \frac{\hat{p}_1}{\hat{p}_0}\right] = \mathrm{Var}\left[\log\hat{p}_1 - \log\hat{p}_0\right] \end{equation}$

ここで，

a, b

$a,~b$ の確率変数を

A, C

$A,~C$ とおくと，

A \sim B (n, p_{1}), C \sim B (m, p_{0})

$A \sim B(n, p_1),~C \sim B(m, p_0)$ である．

E [A] = m p_{1}

$\begin{equation} \mathrm{E}[A] = m p_1 \end{equation}$

V a r [A] = m p_{1} (1 - p_{1})

$\begin{equation} \mathrm{Var}[A] = m p_1(1 - p_1) \end{equation}$

これらより，

E [\frac{A}{m}] = E [{\hat{p}}_{1}] = p_{1}

$\begin{equation} \mathrm{E}\left[\frac{A}{m}\right] = \mathrm{E}[\hat{p}_1] = p_1 \end{equation}$

V a r [\frac{A}{m}] = V a r [{\hat{p}}_{1}] = \frac{p_{1} (1 - p_{1})}{m}

$\begin{equation} \mathrm{Var}\left[\frac{A}{m}\right] = \mathrm{Var}[\hat{p}_1] = \frac{p_1(1 - p_1)}{m} \end{equation}$

デルタ法により，

\begin{aligned} V a r [\log {\hat{p}}_{1}] & = {({(\log p_{1})}^{'})}^{2} V a r [{\hat{p}}_{1}] \\ = {(\frac{1}{p_{1}})}^{2} \frac{p_{1} (1 - p_{1})}{m} \\ = \frac{1 - p_{1}}{m p_{1}} \\ = \frac{1}{m p_{1}} - \frac{1}{m} \end{aligned}

$\begin{align} \mathrm{Var}[\log\hat{p}_1] &= \left(\left(\log p_1\right)'\right)^2 \mathrm{Var}\left[\hat{p}_1\right] \\ &= \left(\frac{1}{p_1}\right)^2 \frac{p_1(1 - p_1)}{m} \\ &= \frac{1 - p_1}{m p_1} \\ &= \frac{1}{m p_1} - \frac{1}{m}\\ \end{align}$

m p_{1}

$m p_1$ の推定値は，

a

$a$ であるので，

V a r [\log {\hat{p}}_{1}] = \frac{1}{a} - \frac{1}{m}

$\begin{equation} \mathrm{Var}[\log\hat{p}_1] = \frac{1}{a} - \frac{1}{m} \\ \end{equation}$

同様にして，

V a r [\log {\hat{p}}_{0}] = \frac{1}{c} - \frac{1}{n}

$\begin{equation} \mathrm{Var}[\log\hat{p}_0] = \frac{1}{c} - \frac{1}{n} \\ \end{equation}$

ところで，

A

$A$ と

C

$C$ は互いに独立であるので，

{\hat{p}}_{0} = \frac{A}{m}

$\hat{p}_0 = \frac{A}{m}$ と

{\hat{p}}_{1} = \frac{C}{n}

$\hat{p}_1 = \frac{C}{n}$ も互いに独立である．

\begin{aligned} V a r [\log R R^{*}] & = V a r [\log {\hat{p}}_{1}] + V a r [- \log {\hat{p}}_{0}] \\ = V a r [\log {\hat{p}}_{1}] + V a r [\log {\hat{p}}_{0}] \\ = \frac{1}{a} - \frac{1}{m} + \frac{1}{c} - \frac{1}{n} \end{aligned}

$\begin{align} \mathrm{Var}[\log RR^{*}] &= \mathrm{Var}\left[\log\hat{p}_1\right] + \mathrm{Var}\left[- \log\hat{p}_0\right] \\ &= \mathrm{Var}\left[\log\hat{p}_1\right] + \mathrm{Var}\left[\log\hat{p}_0\right] \\ &= \frac{1}{a} - \frac{1}{m} + \frac{1}{c} - \frac{1}{n} \end{align}$

よって，

S E [\log R R^{*}] = \sqrt{V a r [\log R R^{*}]} = \sqrt{\frac{1}{a} - \frac{1}{m} + \frac{1}{c} - \frac{1}{n}}

$\begin{equation} SE[\log RR^{*}] = \sqrt{\mathrm{Var}[\log RR^{*}]} = \sqrt{\frac{1}{a} - \frac{1}{m} + \frac{1}{c} - \frac{1}{n}} \end{equation}$

統計検定 準1級 例題集 解答/解答例と解説

論述問題 問3 [2]

設問の要約

解答例

統計検定準1級例題集解答/解答例と解説

論述問題問3 [2]