京都大学 情報学研究科 知能情報学専攻 2021年8月実施 専門科目 S-2

Author

Isidore

Description

Kai

設問1

(1)

According to the rule of Power analysis, we have the Statistical Power:

\[ \begin{aligned} 1-\beta &= 1 - Pr[\textrm{accept } H_0|H_0 \textrm{ is false}] \\ &= 1-Pr[\frac{\bar{X}-\mu_0}{\sigma/\sqrt{n}} \leq Z_{\alpha}] \\ &= 1-Pr[\frac{\bar{X}-\mu}{\sigma/\sqrt{n}} \leq Z_{\alpha} - \frac{\mu-\mu_0}{\sigma/\sqrt{n}}] \\ &= Pr[\frac{\bar{X}-\mu}{\sigma/\sqrt{n}} \geq Z_{\alpha} - \frac{\mu-\mu_0}{\sigma/\sqrt{n}}] \end{aligned} \]

Insert the values and we immediately have So \(Z_{\alpha} - \frac{\mu-\mu_0}{\sigma/\sqrt{n}}\) is the asked \(u\). Insert the value and we have

\[ u = Z_{\alpha} - \frac{\mu-\mu_0}{\sigma/\sqrt{n}} = 1.645 - \frac{1.2}{4/\sqrt{16}} = 0.445 \]

(2)

By (1), we have

\[ \begin{aligned} 1-\beta &= Pr[\frac{\bar{X}-\mu}{\sigma/\sqrt{n}} \geq Z_{\alpha} - \frac{\mu-\mu_0}{\sigma/\sqrt{n}}] \end{aligned} \]

To make sure the power larger than \(95\%\), we have

\[ Z_{\alpha} - \frac{\mu-\mu_0}{\sigma/\sqrt{n}} \leq Z_{0.95} = -Z_{0.05} = -1.645 \]

Insert the values and we immediately have

\[ \sqrt{n} \geq \frac{Z_{\alpha} + Z_{0.05}}{\mu-\mu_0}\sigma \approx 10.967 \\ n \geq 10.967^2 = 120.27 \]

So, the minimum value of \(n\) is \(121\)

設問2

(1)

\[ Pr = \textrm{C}^0_8(p)^0(1-p)^8+\textrm{C}^1_8(p)^1(1-p)^7 = \frac{9}{256} \]

(2)

\[ Pr = \textrm{C}^0_{10000}(p)^0(1-p)^{10000} \]

The approximation PMF is \(Pr[X=r] = (\frac{5}{2})^r\frac{e^{-\frac{5}{2}}}{r!}\). So the answer is

\[ Pr[X=0] = e^{-\frac{5}{2}} \]

(3)

Let the B's defective rate be denoted as \(p_0\). We design a test where the null hypothesis \(H_0\) is \(p_0 = p\) and the alternative hypothesis \(H_1\) is \(p_0 \neq p\).

The statistic \(Z = \frac{\bar{p} - p}{\sqrt{p(1-p)/n}}\) follow the standard normal distribution. By using the significance level \(\alpha = 0.05\), the rejection region is

\[ Z \geq Z_{\frac{\alpha}{2}} \textrm{ or } Z \leq -Z_{\frac{\alpha}{2}} \]

Insert the values and we immediately have

\[ Z = 2 > Z_{0.025} = 1.96 \]

So we reject \(H_0\) and accept \(H_1\) that the defect rate of B is not as the same as A

設問3

(1)

\[ \begin{align} &E[X^2] = Var[X] + E^2[X] = 10 \nonumber \\ &E[Y^2] = Var[Y] + E^2[Y] = 17 \nonumber \\ &Cov(X,Y) = E[XY] - E[X]E[Y] = 0.3 \nonumber \end{align} \]

(2)

\[ Cov(U, V) =E[(5X-3)(-3Y+2)] - E[5X-3]E[-3Y+2] = -4.5 \]

\[ \rho(U,V) = \frac{Cov(U,V)}{\sqrt{Var[U]Var[V]}} = -0.3 \]