060. Comparison of Several MV Means

Comparison of Several MV Means (wk5)

Paired Comparison

Recall:

for univariate, let $X_i-Y_i=D_i\sim N(\delta ,{\sigma }_d^2)$, $i=1,\cdots ,n$

then for $H_0:\delta =0$, test stat $t=\frac{\bar{D}}{\frac{S_d}{\sqrt{n}}}\stackrel{H_0}{\sim }{t}_{n-1}$.

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Assume independent rvec $D_1,\cdots ,D_n\sim N_p(\delta ,{\Sigma }_d)$.

then test stat $T^2=n(\bar{D}-\delta {)}^{\prime }{S}_d^{-1}(\bar{D}-\delta )\sim (n-1)\frac{p}{n-p}{F}_{p,n-p}$.

• Hypothesis Testing:

$$H_0:\delta =0$$ $$T^2=n(\bar{D}{)}^{\prime }{S}_d^{-1}(\bar{D})\sim \frac{(n-1)p}{n-p}{F}_{p,n-p}$$

reject $H_0$ if $T^2>\frac{(n-1)p}{n-p}{F}_{p,n-p}(\alpha )$.

1. $100(1-\alpha )$ CR for $\delta$:

$$(\bar{D}-\delta {)}^{\prime }{S}_d^{-1}(\bar{D}-\delta )\le \frac{1}{n}\frac{(n-1)p}{n-p}{F}_{p,n-p}(\alpha )$$

2. $100(1-\alpha )$ simultaneous CI for individual ${\delta }_i$:

3. Bonferroni’s $100(1-\alpha )$ simultaneous CI for individual ${\delta }_i$:

$$\begin{array}{rlr}{\bar{d}}_i\pm & \sqrt{\frac{(n-1)p}{n-p}{F}_{p,n-p}(\alpha )}& \sqrt{\frac{S_{d_i}^2}{n}}\\ {\bar{d}}_i\pm & t_{n-1}\left(\frac{\alpha }{2p}\right)& \sqrt{\frac{S_{d_i}^2}{n}}\end{array}\text{\hspace{1em}(2)(3)}$$

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Different Approach

let $X={\left[x_{11},\cdots ,x_{1p},x_{21},\cdots ,x_{2p}\right]}_{1\times 2p}^{\prime }\sim N_{2p}(\mu ,\Sigma )$.

then $D=CX$, where $C={\left(\begin{array}{ccccccc}1& & 0& ⋮& -1& & 0\\ & \ddots & & ⋮& & \ddots & \\ 0& & 1& ⋮& 0& & -1\end{array}\right)}_{p\times 2p}$.

at here,

$$\begin{array}{rl}E(D)& =E(CX)=C\mu \\ & =\delta \\ & \\ Cov(D)& =Cov(CX)=C\Sigma C^{\prime }\\ & ={\Sigma }_d\\ & \\ D& =CX\sim N_p(C\mu ,C\Sigma C^{\prime })\end{array}$$

therefore, given $H_0:C\mu =0$,

test stat $T^2=n(C\bar{X}{)}^{\prime }(CS{C}^{\prime }{)}^{-1}(C\bar{X})\stackrel{H_0}{\sim }\frac{(n-1)p}{n-p}{F}_{p,n-p}$

• graph, check normality:

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Comparing Mean Vectors from Two Populations

Recall: univariate, $t=\frac{{\bar{X}}_1-{\bar{X}}_2}{sqrt{S}_p^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}\stackrel{H_0}{\sim }{t}_{n_1+n_2-2}$

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for MV, assume below, where $(X_{11},\cdots ,X_{1n_1})$ and $(X_{21},\cdots ,X_{2n_2})$ are independent.

$$X_{11},\cdots ,X_{1n_1}\sim N_p({\mu }_1,{\Sigma }_1)$$ $$X_{21},\cdots ,X_{2n_2}\sim N_p({\mu }_2,{\Sigma }_2)$$

at here,

$$H_0:{\mu }_1-{\mu }_2=0$$

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case 1: ${\Sigma }_1={\Sigma }_2=\Sigma$

이하 대부분은 벡터에 관한 이야기이다.

${\bar{X}}_i$ estimates ${\mu }_i$, $i=1,2$.

$S_p$ estimates $\Sigma$, where $S_p=\frac{(n_1-1)S_1+(n_2-1)S_2}{(n_1-1)+(n_2-1)}$.

the test stats Hotelling’s $T^2=\frac{n_1{n}_2}{n_1+n_2}{\left(\overline{X_1}-\overline{X_2}\right)}^{\prime }{S}_p^{-1}\left(\overline{X_1}-\overline{X_2}\right)$

where $\frac{(n_1-1)+(n_2-1)-(p-1)}{p[(n_1-1)+(n_2-1)]}{T}^2=\frac{n_1+n_2-p-1}{p[(n_1+n_2-2)]}{T}^2\stackrel{H_0}{\sim }{F}_{p,n_1+n_2-p-1}$. (p.285 for pf)

• CR for ${\mu }_1-{\mu }_2$ will be

$$Pr\left[\left({\left(({\bar{X}}_1-{\bar{X}}_2)-({\mu }_1-{\mu }_2)\right)}^T{\left(\left(\frac{1}{n_1}+\frac{1}{n_2}\right)S_p\right)}^{-1}\left(({\bar{X}}_1-{\bar{X}}_2)-({\mu }_1-{\mu }_2)\right)\le c^2\right)\right]=1-\alpha$$

where $c^2=\frac{p[(n_1+n_2-2)]}{n_1+n_2-p-1}{T}^2\stackrel{H_0}{\sim }{F}_{p,n_1+n_2-p-1}(\alpha )$.

• 이때 constant가 역수가 되었음을 눈치.
• The equality will define the boundary of a region.
• The region is an ellipsoid centered at $({\bar{X}}_1-{\bar{X}}_2)$.

Example) Testing $H_0:{\mu }_1-{\mu }_2=0$ at $\alpha =0.05$ is equivalent to see whether falls within the confidence region

• Axes of the confidence region
  • let ${\lambda }_1,\cdots ,{\lambda }_p$ are ev of $S_p$.
  • let $e_1,\cdots ,e_p$ are evc of $S_p$.
    • then $e_i$‘s are the direction of CI
    • $\sqrt{{\lambda }_i}\sqrt{\left(\frac{1}{n_1}+\frac{1}{n_2}\right)c^2}$ are the half-length of the CR Link

let $c^2=\frac{p[(n_1+n_2-2)]}{n_1+n_2-p-1}{T}^2\stackrel{H_0}{\sim }{F}_{p,n_1+n_2-p-1}(\alpha )$.

• $100(1-\alpha )$ simultaneous CI for $a^{\prime }({\mu }_1-{\mu }_2)$, $\forall a$:
  $

$$a^{\prime }\left({\bar{X}}_1-{\bar{X}}_2\right)\pm c\sqrt{a^{\prime }\left(\frac{1}{n_1}+\frac{1}{n_2}\right)S_{p}a}$$

Example) simultaneous CI for $({\mu }_{1i}-{\mu }_{2i}),i=1,\cdots ,p$.

let $a^{\prime }=\left[0,\cdots ,0,1,0,\cdots ,0\right]$. 이때 $a^{\prime }$가 하나만 1이고 나머지 0이면, 어떤 특별한 한 axis로 proj하라는 의미. link

let ${\mu }_1-{\mu }_2={\left[{\mu }_{1i}-{\mu }_{2i}\right]}_{i=1,\cdots ,p}$.

$a^{\prime }({\bar{X}}_1-{\bar{X}}_2)={\bar{X}}_{1i}-{\bar{X}}_{2i}$, $a^{\prime }\left(\frac{1}{n_1}+\frac{1}{n_2}\right)S_{p}a=\left(\frac{1}{n_1}+\frac{1}{n_2}\right)S_{pii}$

• $S_{pii}$ : p번째 변수의 표본 cov. 이는 단변량에서 나왔던 공통 cov, 즉 샘플 se와 표기법이 동일해지며 유사하다. (ch1) link

the Bonferroni’s $100(1-\alpha )\%$ simultaneous CI for $({\mu }_{1i}-{\mu }_{2i})$ is $({\bar{X}}_1-{\bar{X}}_2)\pm t_{n_2+n_2-2,(\frac{\alpha }{2p})}\sqrt{\left(\frac{1}{n_1}+\frac{1}{n_2}\right)S_{pii}}$.

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case 2: ${\Sigma }_1\ne {\Sigma }_2$

assume $n_1-p,n_2-p$ are large.

for $H_0:{\mu }_1-{\mu }_2=0$, test stat becomes $T^2=({\bar{X}}_1-{\bar{X}}_2{)}^{\prime }{\left[\frac{1}{n_1}{S}_1+\frac{1}{n_2}{S}_2\right]}^{-1}({\bar{X}}_1-{\bar{X}}_2)\stackrel{H_0}{\sim }{\chi }_p^2$.

note

$$E({\bar{X}}_1-{\bar{X}}_2)={\mu }_1-{\mu }_2$$ $$Cov({\bar{X}}_1-{\bar{X}}_2)=Cov({\bar{X}}_1)+Cov({\bar{X}}_2)-2Cov({\bar{X}}_1,{\bar{X}}_2)=\frac{1}{n_1}{\Sigma }_1+\frac{1}{n_2}{\Sigma }_2-0$$ $${\bar{X}}_1-{\bar{X}}_2\stackrel{\cdot }{\sim }{N}_p\left({\mu }_1-{\mu }_2,\frac{1}{n_1}{\Sigma }_1+\frac{1}{n_2}{\Sigma }_2\right)\text{\hspace{1em}(∵ CLT)}$$

\[3ex]

$$\text{under }{H}_0,$$ $$S_1\stackrel{p}{\to }{\Sigma }_1,S_2\stackrel{p}{\to }{\Sigma }_2\text{\hspace{1em}(∵ WLLN)}$$ $$({\bar{X}}_1-{\bar{X}}_2{)}^{\prime }{\left[\frac{1}{n_1}{S}_1+\frac{1}{n_2}{S}_2\right]}^{-}1({\bar{X}}_1-{\bar{X}}_2)\stackrel{app}{\sim }{\chi }_p^2\text{\hspace{1em}(∵ Slutsky’s thm)}$$

why Cov become 0???

i.e. reject $H_0$ if $T^2>{\chi }_p^2(\alpha )$.

CI becomes

$$Pr\left[\left({\left(({\bar{X}}_1-{\bar{X}}_2)-({\mu }_1-{\mu }_2)\right)}^T{\left(\left(\frac{1}{n_1}+\frac{1}{n_2}\right)S_2\right)}^{-1}\left(({\bar{X}}_1-{\bar{X}}_2)-({\mu }_1-{\mu }_2)\right)\le {\chi }_p^2\right)\right]=1-\alpha$$

차이는~~

Remark: if $n_1=n_2=2$,

$$\begin{array}{rl}\frac{1}{n_1}{S}_1+\frac{1}{n_2}{S}_2& =\frac{1}{n}(S_1+S_2)\\ & =\frac{1}{n}\left[\frac{1}{n-1}\sum _{n=1}^n(X_{1i}-\overline{X_1})(X_{1i}-\overline{X_1}{)}^{\prime }+\frac{1}{n-1}\sum _{n=1}^n(X_{2i}-\overline{X_2})(X_{2i}-\overline{X_2}{)}^{\prime }\right]\\ & =\frac{1}{n}\frac{1}{n-1}{S}_p\ast 2(n-1)=\frac{2}{n}{S}_p\end{array}$$

i.e. case 1 and case 2 are the same procedure when the sample sizes are the same for large sample sizes.

• $100(1-\alpha )$ simultaneous CI for $a^{\prime }({\mu }_1-{\mu }_2)$, $\forall a$:

$

$$a^{\prime }\left(\overline{X_1}-\overline{X_2}\right)\pm \sqrt{{\chi }_p^2(\alpha )}\sqrt{a^{\prime }\left(\frac{1}{n_1}{S}_1+\frac{1}{n_2}{S}_2\right)a}$$

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Other Statistics for Testing two Mean Vectors

• let $W=(n_1-1)S_1+(n_2-1)S_2$: within SS, $B=n_1(\overline{X_1}-\bar{X})(\overline{X_1}-\bar{X}{)}^{\prime }+n_2(\overline{X_2}-\bar{X})(\overline{X_2}-\bar{X}{)}^{\prime }$

• Wilk’s Lambda:

  • when two-sample procedure, Hotelling’s $T^2$

$${\Lambda }^{\ast }=\frac{|W|}{|B+W|}$$

• Lawley-Hotelling’s Trace:

$$tr(B{W}^{-1})$$

• Pillai Trace:

$$tr\left[B(B+W{)}^{-1}\right]$$

• Roy’s Largest Root:
  • maximum ev of $B(B+W{)}^{-1}$.

Testing Equality of Covariance Matrices

$$H_0:{\Sigma }_1={\Sigma }_2$$

let $S_p=\frac{1}{n_1+n_2-2}\left[(n_1-1)S_1+(n_2-1)S_2\right]$.

$$\begin{array}{rl}M& =(n_1+n_2-2)\ln |S_p|-(n_1-1)\ln |S_1|-(n_2-1)\ln |S_2|\\ C^{-1}& =1-\frac{2p^2+3p-1}{6(p+1)}\left(\frac{n_1+n_2-2}{(n_1-1)(n_2-1)}-\frac{1}{n_1+n_2-2}\right)\\ M{C}^{-1}& \sim {\chi }_v^2,v=\frac{p(p+1)}{2}\end{array}\text{\hspace{1em}(Test statistic)(Scale factor)}$$

reject $H_0$ if $M{C}^{-1}>{\chi }_v^2(\alpha )$

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Profile Analysis (for $g=2$)

Recall:

$H_0:{\mu }_1={\mu }_2$, when ${\Sigma }_1={\Sigma }_2=\Sigma$

$$\begin{array}{rl}{T}^2& =(\overline{X_1}-\overline{X_2}{)}^{\prime }{\left[\left(\frac{1}{n_1}+\frac{1}{n_2}\right)S_p\right]}^{-1}(\overline{X_1}-\overline{X_2})\\ & \stackrel{H_0}{\sim }\frac{(n_1+n_2-2)p}{n_1+n_2-p-1}{F}_{p,n_1+n_2-p-1}\end{array}$$

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let’s $H_0:C{\mu }_1=C{\mu }_2$, when ${\Sigma }_1={\Sigma }_2=\Sigma$, where $C_{q\times p}$, $q\le p$ and $rank(C)=q$.

$$\begin{array}{rl}{T}^2& =(\overline{X_1}-\overline{X_2}{)}^{\prime }{C}^{\prime }{\left[\left(\frac{1}{n_1}+\frac{1}{n_2}\right)C{S}_p{C}^{\prime }\right]}^{-1}C(\overline{X_1}-\overline{X_2})\\ & \stackrel{H_0}{\sim }\frac{(n_1+n_2-2)q}{n_1+n_2-q-1}{F}_{p,n_1+n_2-p-1}\end{array}$$

Profiles are constructed for each group.

Consider two groups. Questions:

1. Are the profiles parallel?

$$\begin{array}{rl}& H_0:{\mu }_{11}-{\mu }_{12}={\mu }_{21}-{\mu }_{22},{\mu }_{12}-{\mu }_{13}={\mu }_{22}-{\mu }_{23},{\mu }_{13}-{\mu }_{14}={\mu }_{23}-{\mu }_{24},\cdots ,{\mu }_{1,p-1}-{\mu }_{1,p}={\mu }_{2,p-1}-{\mu }_{2,p}\\ ⟺& H_0:{\mu }_{11}-{\mu }_{21}={\mu }_{12}-{\mu }_{22}=\cdots ={\mu }_{1p}-{\mu }_{2p}\\ ⟺& C_{(p-1)\times p}\\ ⟺& H_0:C{\mu }_1=C{\mu }_2\end{array}$$

This is equivalent to test the equal mean vector of the transformed data $C{X}_1$ and $C{X}_2$.

Populations 1: $C{X}_{11},\cdots ,C{X}_{1n_1}\sim N_{p-1}(C{\mu }_1,C\Sigma C^{\prime })$ Populations 2: $C{X}_{21},\cdots ,C{X}_{2n_2}\sim N_{p-1}(C{\mu }_2,C\Sigma C^{\prime })$

reject $H_0:C{\mu }_1=C{\mu }_2$ (i.e. paralle profiles), if

$$T^2=(\overline{X_1}-\overline{X_2}{)}^{\prime }{C}^{\prime }{\left[\left(\frac{1}{n_1}+\frac{1}{n_2}\right)C{S}_p{C}^{\prime }\right]}^{-1}C(\overline{X_1}-\overline{X_2})>d^2=(n_1+n_2-2)\frac{p-1}{n_1+n_2-p}{F}_{p-1,n_1+n_2-p}(\alpha )$$

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2. Coincident Profiles

2. Assuming that the profiles are parallel, are the profiles coincident?

$$\begin{array}{rl}& H_0:{\mu }_{1i}={\mu }_{2i},i=1,\cdots ,p\\ ⟺& H_0:1^{\prime }{\mu }_1=1^{\prime }{\mu }_2\end{array}$$

is the case where $C$ is replaced by $1^{\prime }$.

reject $H_0$ if

$$\begin{array}{rlrl}{T}^2& =1^{\prime }(\overline{X_1}-\overline{X_2}){\left[\left(\frac{1}{n_1}+\frac{1}{n_2}\right)1^{\prime }{S}_{p}1\right]}^{-1}(\overline{X_1}-\overline{X_2})& & \\ & ={\left(\frac{1^{\prime }(\overline{X_1}-\overline{X_2})}{\sqrt{\left(\frac{1}{n_1}+\frac{1}{n_2}\right)1^{\prime }{S}_{p}1}}\right)}^2& & >F_{1,n_1+n_2-2}(\alpha )(n_1+n_2-2)\frac{p-1}{n_1+n_2-p}{F}_{p-1,n_1+n_2-p}(\alpha )\end{array}$$

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3. Flat Profiles

3.Assuming that the profiles are coincident, are the profiles level?

$$H_0:{\mu }_{11}={\mu }_{12}=\cdots ={\mu }_{1p}={\mu }_{21}={\mu }_{22}=\cdots ={\mu }_{2p}$$

by 1 and 2, we can collapse two groups into one.

$$X_{11},\cdots ,X_{1n_1},X_{21},\cdots ,X_{2n_2}\sim N_p(\mu ,\Sigma )$$

this is one population problem

$$\exists C_{(p-1)\times p},H_0:C\mu =0$$

reject $H_0$, iff

$$T^2=(n_1+n_2){\bar{X}}^{\prime }{C}^{\prime }[CS{C}^{\prime }{]}^{-1}C\bar{X}>d^2=(n_1+n_2-1)\frac{p-1}{n_1+n_2-p+1}{F}_{p-1,n_1+n_2-p+1}(\alpha )$$

이는 1번에서의 그것과는 $F$분포의 df가 변화했다는 점에 주목.

• $\bar{X}=\frac{1}{n_1+n_2}\left({\sum }_{j=1}^{n_1}{X}_{1j}+{\sum }_{j=1}^{n_2}{X}_{2j}\right)$.
• $S=n_1+n_2$ sample covariance matrix, using data.

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Comparing Several Multivariate Population Means

Recall:

In univariate, two-sample t-test is extended to Analysis of Variance(ANOVA).

$$H_0:{\mu }_1=\cdots ={\mu }_g$$ $$F^{\ast }=\frac{SSR/d{f}_1}{SSE/d{f}_2}\stackrel{H_0}{\sim }{F}_{d{f}_1,d{f}_2}$$

• where
  • SSR: sum of squared regression,
  • SSE: sum of squared error,
  • SST: sum of squared total
  • $d{f}_1=g-1,d{f}_2=N-g,N={\sum }_{i=1}^g{n}_i$.

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Assume $g$ population or treatment groups, and each groups are independent. 각 population은 같은 Cov를 갖고 같은 숫자의 패러미터를 갖되 총 observation 숫자랑 각각의 population mean은 다름.

Population 1~g: $X_{i1},\cdots ,X_{i{n}_i}\sim N_p({\mu }_i,\Sigma )$.

• Model

$$X_{ij}={\mu }_i+{ϵ}_{ij},i=1,\cdots ,g,j=1,\cdots ,n_i$$ $$H_0:{\mu }_1=\cdots {\mu }_g$$ $$\text{where }{X}_{ij}={\left[\begin{array}{c}{X}_{ij1}\\ X_{ij2}\\ ⋮\\ X_{ijp}\end{array}\right]}_{p\times 1},{\mu }_{ij}={\left[\begin{array}{c}{\mu }_{i1}\\ {\mu }_{i2}\\ ⋮\\ {\mu }_{ip}\end{array}\right]}_{p\times 1},{ϵ}_{ij}={\left[\begin{array}{c}{ϵ}_{ij1}\\ {ϵ}_{ij2}\\ ⋮\\ {ϵ}_{ijp}\end{array}\right]}_{p\times 1}$$

• Assumptions
  1. The random samples from different populations are independent.
  2. All populations have a common covariance matrix $\Sigma$.
  3. Each population is Multivariate Normal. This assumption can be relaxed by C.L.T., when the sample sizes $n_1,\cdots ,n_g$ are large.

One-Way MANOVA

The quantities SSR, SSE and SST become matrices in MANOVA.

$$\begin{array}{rl}B& =\sum _{i=1}^g{n}_i(X_i-X)(X_i-X{)}^{\prime }\\ W& =\sum _{i=1}^g\sum _{j=1}^{n_i}(X_{ij}-X_i)(X_{ij}-X_i{)}^{\prime }\\ & =(n_1-1)S_1+\cdots +(n_g-1)S_g\end{array}\text{\hspace{1em}(SSR)(SSE)}$$

• Note:

$$\begin{array}{rlrl}(X_{ij}-\bar{X})& =(\overline{X_i}-\bar{X})+(X_{ij}-\overline{X_i})& & \\ (X_{ij}-\bar{X})(X_{ij}-\bar{X}{)}^{\prime }& =(\overline{X_i}-\bar{X})(\overline{X_i}-\bar{X}{)}^{\prime }+& & (\overline{X_i}-\bar{X})(X_{ij}-\overline{X_i}{)}^{\prime }+(X_{ij}-\overline{X_i})(\overline{X_i}-\bar{X}{)}^{\prime }+(X_{ij}-\overline{X_i})(X_{ij}-\overline{X_i}{)}^{\prime }\\ \sum _{i=1}^g\sum _{j=1}^{n_i}(X_{ij}-\bar{X})(X_{ij}-\bar{X}{)}^{\prime }& =\sum _{i=1}^g{n}_i(\overline{X_i}-\bar{X})(\overline{X_i}-\bar{X}{)}^{\prime }& & +\sum _{i=1}^g\sum _{j=1}^{n_i}(X_{ij}-\overline{X_i})(X_{ij}-\overline{X_i}{)}^{\prime }\\ T& =B& & +W\end{array}$$

• B: Between Sum of Squares
• W: Within Sum of Squares

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Any test statistic will be a function of B and W. Popular test statistics use eigenvalues of $B{W}^{-1}$.

let ${\lambda }_1,\cdots ,{\lambda }_r$ be ev of $B{W}^{-1}$, where $r=$ of non-zero ev’s.

1. Wilk’s Lambda (LRT)

$$\Lambda =\frac{|W|}{|B+W|}=\frac{1}{|I+B{W}^{-1}|}=\prod _{i=1}^r(1+{\lambda }_1{)}^{-1}$$

2. Pillai’s Trace

$$\begin{array}{rl}V& =tr[B(B+W{)}^{-1}]=tr[B(B(I+B^{-1}W){)}^{-1}]=tr[B(I+B^{-1}W{)}^{-1}{B}^{-1}]\\ & =tr[B^{-1}B(I+B^{-1}W{)}^{-1}]=tr[(I+B^{-1}W{)}^{-1}]=tr[I+(B^{-1}W{)}^{-1}]\\ & =\sum _{i=1}^r\left(\frac{{\lambda }_i}{1+{\lambda }_i}\right)\end{array}$$

3. Lawley-Hotelling’s Trace

$$T=tr\left(B{W}^{-1}\right)=\sum _{i=1}^r{\lambda }_i$$

4. Roy’s Largest Root

$$U=\underset{i=1,\cdots ,r}{\max }\left\{{\lambda }_i\right\}$$

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• Sampling Distribution of Wilk’s Lambda

$$\begin{array}{rlrl}p=1,g\ge 2:& \left(\frac{{\sum }_{i=1}^g{n}_i-g}{g-1}\right)\left(\frac{1-{\Lambda }^{\ast }}{{\Lambda }^{\ast }}\right)& & \stackrel{H_0}{\sim }{F}_{g,{\sum }_{i=1}^g{n}_i-g}\\ p=2,g\ge 2:& \left(\frac{{\sum }_{i=1}^g{n}_i-g-1}{g-1}\right)\left(\frac{1-\sqrt{{\Lambda }^{\ast }}}{\sqrt{{\Lambda }^{\ast }}}\right)& & \stackrel{H_0}{\sim }{F}_{2(g-1),2({\sum }_{i=1}^g{n}_i-g-1)}\\ p\ge 1,g=2:& \left(\frac{n_1+n_2-p-1}{p}\right)\left(\frac{1-{\Lambda }^{\ast }}{{\Lambda }^{\ast }}\right)& & \stackrel{H_0}{\sim }{F}_{p,n_1+n_2-p-1}\\ p\ge 1,g\ge 3:& \left(\frac{{\sum }_{i=1}^3{n}_i-p-2}{p}\right)\left(\frac{1-\sqrt{{\Lambda }^{\ast }}}{\sqrt{{\Lambda }^{\ast }}}\right)& & \stackrel{H_0}{\sim }{F}_{2p,2({\sum }_{i=1}^g{n}_i-p-2)}\\ \text{large sample sizes}:& -\left(\sum _{i=1}^g{n}_i-1-\frac{p+q}{2}\right)\ln {\Lambda }^{\ast }& & \stackrel{H_0}{\sim }{\chi }_{p(g-1)}^2\end{array}\text{\hspace{1em}(Why?)}$$