Usuari:Lixiaoxu

Xiaoxu LI szpku.lixiaoxu@gmail.com Shenzhen Graduate School of Peking University, Guangdong, China

Regression of Inputted Data

<Rform name="owndata"> Enter your own data for a scatterplot:
You can use <a href="https://spreadsheets.google.com/ccc?key=0Aic4pmEZm32xclhJZm9hNWFyZlZOV1RSV19xWXRlbmc&hl=en">the free online spreadsheet </a> of Google Docs to edit your data before pasting. Just click the link, need NO login.
<textarea name="mydata" rows="8"> 1.262954285 3.8739569 -0.326233361 1.0400041 1.329799263 2.0161824 1.272429321 2.8284819 0.414641434 2.1324980 -1.539950042 0.4565291 -0.928567035 1.6093698 -0.294720447 0.9723025 -0.005767173 2.5310696 2.404653389 2.7861843 </textarea> <input type="submit" value=" Submit "> </Rform>

<R output="display" name="owndata" iframe="height:500px;"> if (exists("mydata")) {

 main <- "Data from user"
 x <- readdataSK(mydata, format="txt") 

} else {

 main <- "Default data"
 set.seed(0);
 x<-matrix(rnorm(20),10,2);
 x[,2]=2.1+x[,1]*.8+x[,2];
 colnames(x)<-c('V1','V2');

}

pdf(rpdf, width=6, height=6) lm.m<-lm(x[,2]~x[,1]); main<-paste(main,'\nV2 =',round(lm.m$coefficients[1],3),'+',round(lm.m$coefficients[2],3),'*V1 + ',round(summary(lm.m)$sigma,3),'*e') plot(x, cex=2, main=main) abline(lm.m)

</R>

回归分析课件

输入参数

<Rform name="Tri"> 向量${\vec {Y}}$与${\vec {X}}_{1}$、${\vec {X}}_{2}$的夹角(90度为直角)分别是

<Input name="cy1" value="89" size="5"/>度

和

<Input name="cy2" value="89" size="5"/>度。

向量${\vec {X}}_{1}$与${\vec {X}}_{2}$的夹角是

<Input name="c12" value="177.9" size="5"/>度。

这三个角度应当满足两两之和大于第三者。 这些向量的$N$个分量代表$N$个标准化之后的样本。

请设定样本量为

<Input name="N" value="100" size="5"/>，输出模拟数据<Input name="rawdata" type="checkbox"/>

<input type="submit" /> </Rform>

练习

请观察${\vec {X}}_{2}$加入前后，回归方程

${\vec {Y}}=\beta _{1}{\vec {X}}_{1}+{\vec {\epsilon }}$

与

${\vec {Y}}=\beta _{1}{\vec {X}}_{1}+\beta _{2}{\vec {X}}_{2}+{\vec {\epsilon }}$

的$R^{2}$的变化。

两个与DV相关极小的IV却能极好地预测DV

三个角度分别为 89,89,177.9

两个与DV高正相关的IV却出现负回归系数

三个角度分别为 5,2.6,2.6

两个不相关的DV对IV的预测能力($R^{2}$)可以相加

第三个角度为90

($R_{1}^{2}+R_{2}^{2}-R_{12}^{2}$)从0变大再变小甚至变负的情形

零:三个角度分别为：60,45,90

正:三个角度分别为：60,45,45

负:三个角度分别为：60,45,15.1

与Redundancy的关系: Cohen & Cohen (2003, p. 76)

结果

<R output="html" name="Tri" iframe="height:400px;"> cy1 <- ifelse(exists("cy1"),as.numeric(cy1),89); cy2 <- ifelse(exists("cy2"),as.numeric(cy2),89); c12 <- ifelse(exists("c12"),as.numeric(c12),177.9); N <- ifelse(exists("N"),as.integer(N),100); rawdata <- ifelse(exists("rawdata"),as.logical(N),FALSE);

S <- matrix(rep(1,9),3); S[1,2]<-S[2,1]<-cos(cy1/180*pi); S[1,3]<-S[3,1]<-cos(cy2/180*pi); S[2,3]<-S[3,2]<-cos(c12/180*pi);

if ((det(S)<= 0 )|(N<1)) outHTML(rhtml,NA,title='Please check your input!\n Sum of any two angles should be larger than the third one.'); require(MASS);

x<-mvrnorm(n=N,mu=c(0,0,0),Sigma=S,empirical= TRUE); Y<-x[,1];X_1<-x[,2];X_2<-x[,3];

colnames(x)<-colnames(S)<-rownames(S)<-c('Y','X_1','X_2');

lm1 <- lm(Y~0+ X_1); lm2 <- lm(Y~0+ X_2); lm12 <- lm(Y~0+ X_1+X_2); R2<-matrix(rep(NA,3),nrow=3); rownames(R2)<-c('Y ~ 0+ X_1','Y ~ 0+ X_2','Y ~ 0+ X_1 + X_2'); R2[,1] <- c( summary(lm1)$r.squared, summary(lm2)$r.squared, summary(lm12)$r.squared); colnames(R2)[1]<-round( summary(lm1)$r.squared + summary(lm2)$r.squared - summary(lm12)$r.squared,4); outHTML(rhtml, t(R2), title="R^2_1+R^2_2-R^2_12", format="f", digits=4);

outHTML(rhtml, summary(lm1)$coefficients, title=rownames(R2)[1], format="f", digits=4);

outHTML(rhtml, summary(lm2)$coefficients, title=rownames(R2)[2], format="f", digits=4);

outHTML(rhtml, summary(lm12)$coefficients, title=rownames(R2)[3], format="f", digits=4);

outHTML(rhtml, S, title="correlation\n", format="f", digits=4); if (rawdata) outHTML(rhtml, x, title="Raw data\n", format="f", digits=4); </R>

R 代码

cy1 <- 89; ## \angle YX_1
cy2 <- 89; ## \angle YX_2
c12 <- 177.9; ## \angle X_1X_2
N <- 100;
rawdata=TRUE;

S <- matrix(rep(1,9),3);
S[1,2]<-S[2,1]<-cos(cy1/180*pi);
S[1,3]<-S[3,1]<-cos(cy2/180*pi);
S[2,3]<-S[3,2]<-cos(c12/180*pi);

require(MASS);## install.packages('MASS');

x<-mvrnorm(n=N,mu=c(0,0,0),Sigma=S,empirical= TRUE);
Y<-x[,1];X_1<-x[,2];X_2<-x[,3];

colnames(x)<-colnames(S)<-rownames(S)<-c('Y','X_1','X_2');

R2<-matrix(rep(NA,3),nrow=3);
colnames(R2)<-c('R^2');
rownames(R2)<-c('Y = b_1*X_1 + e','Y =b_2*X_2 + e','Y =b_1*X_1 + b_2*X_2 + e');
lm1 <- lm(Y~0+X_1);
lm2 <- lm(Y~0+X_2);
lm12 <- lm(Y~0+X_1+X_2);
R2[,1] <- c( summary(lm1)$r.squared, summary(lm2)$r.squared, summary(lm12)$r.squared);

R2
R2[1,1]+R2[2,1]-R2[3,1]
summary(lm1)
summary(lm2)
summary(lm12)
cat('\ncorr')
S
cat('\nraw data')
if (rawdata) (x);

Online calculator for critical values, cumulative probabilities, and critical noncentral parameters

Input

<Rform name="dncpx"> Choose a statistic: <input type="radio" name="name" value="t" checked/>$t$, <input type="radio" name="name" value="chisq"/>$\chi ^{2}$, or <input type="radio" name="name" value="F"/>$F$ (noncentral parameter of $\chi ^{2}$ or $F$ must be non-negative)
noncentral parameter (ncp)

= <input name="ncp" type="text" size="5" maxlength="10" value="4.5"> ( vs. <input name="ncp_c" type="text" size="5" maxlength="10" value="0"> )

degree freedom

= <input name="df1" type="text" size="5" maxlength="10" value="4">

degree freedom of denominator (only for $F$)

= <input name="df2" type="text" size="5" maxlength="10" value="3">

Mark the points with cumulative probability

= <input type="input" name="q0" maxlength="10" value=".025" size="5">

and the critical statistic

= <input type="input" name="x1" maxlength="10" value="5.1" size="5">

Click to update display with precision <input type="submit" name="submit" value=".001"><input type="submit" name="submit" value=".0001"><input type="submit" name="submit" value=".000001"><input type="submit" name="submit" value="fully">
</Rform>

Results of critical statistic, cumulative probability, and critical noncentral parameter

How to cite this result

<R output="display" name="dncpx" iframe="height:400px;"> digits <- ifelse(exists("submit"),ifelse(as.character(submit)=='fully',Inf,round(log(as.numeric(submit),.1))),3) if (exists("ncp")) ncp <- as.numeric(ncp) else ncp <- 4.5 ncp_c <- ifelse(exists("ncp_c"), as.numeric(ncp_c),0); if (exists("df1")) df1 <- as.numeric(df1) else df1 <- 4 if (exists("df2")) df2 <- as.numeric(df2) else df2 <- 3 if (exists("name")) name <- as.character(name) else name <- 't' if (exists("q0")) q0 <- as.numeric(q0) else q0 <- .025 if (exists("x1")) x1 <- as.numeric(x1) else x1 <- 5.1

ncpt<-function(x,q,df,confirm=FALSE){ .f<-function(ncp,x,df,q)abs(q-pt(x,df=df,ncp=ncp)) .n<-1; while ( ( (pt(x,df=df,ncp=-.n) < q+(1-q)/2 ) | (pt(x,df=df,ncp=.n) > q/2) ) & (.n < Inf) ) .n <- .n *2 ; if (confirm) optimize(f=.f,x=x,df=df,q=q,interval=c(-.n,.n)) else optimize(f=.f,x=x,df=df,q=q,interval=c(-.n,.n))$minimum }

ncpchisq<-function(x,q,df,confirm=FALSE){ .f<-function(ncp,x,df,q)abs(q - pchisq(x,df=df,ncp=ncp)) if (pchisq(x,df=df)<=q){ if (confirm) { minimum <-0; objective <- pchisq(x,df=df)-q; data.frame(minimum,objective) }else 0 }else { .n<- 1; while ( (pchisq(x,df=df,ncp=.n) > q/2) & (.n < Inf) ) .n <- .n + 1; if (confirm) optimize(f=.f,x= x,df=df,q=q,interval=c(0,.n)) else optimize(f=.f,x= x,df=df,q=q,interval=c(0,.n))$minimum } }; ncpf<-function(x,q,df1,df2,confirm=FALSE){ .f<-function(ncp,x,df1,df2,q)abs(q - pf(x,df1=df1,df2=df2,ncp=ncp)) if (pf(x,df1=df1,df2=df2)<=q){ if (confirm) { minimum <-0; objective <- pf(x,df1=df1,df2=df2)-q; data.frame(minimum,objective) }else 0 }else { .n<- 1; while ( (pf(x,df1=df1,df2=df2,ncp=.n) > q/2) & (.n < Inf) ) .n <- .n +1 ; if (confirm) optimize(f=.f,x= x,df1=df1,df2=df2,q=q,interval=c(0,.n)) else optimize(f=.f,x= x,df1=df1,df2=df2,q=q,interval=c(0,.n))$minimum } };

d.f <- function(x,df1,df2,ncp=0) { delta <- 10^-6; (pf(x+delta,df1,df2,ncp)-pf(x-delta,df1,df2,ncp))/(2*delta) } df <- df1

if (name=='chisq'){ x <- seq(.001,ncp+df*9,length.out=500); pr <- dchisq(x,df=df,ncp=ncp); pb <- dchisq(x,df=df,ncp=ncp_c); x0 <- qchisq(q0,df=df,ncp=ncp); q1 <- pchisq(x1,df=df,ncp=ncp); ncp2 <- ncpchisq(x=x1,q=q0,df=df); pg <- dchisq(x,df=df,ncp=ncp2); q2 <- pchisq(x1,df=df,ncp=ncp2);

}else if (name=='F'){ x <- seq(.001,ncp/df1+9,length.out=500); pr <- d.f(x,df1=df,df2=df2,ncp=ncp); pb <- d.f(x,df1=df,df2=df2,ncp=ncp_c); x0 <- qf(q0,df1=df1,df2=df2,ncp=ncp); q1 <- pf(x1,df1=df1,df2=df2,ncp=ncp); ncp2 <- ncpf(x=x1,q=q0,df1=df1,df2=df2); pg <- d.f(x,df1=df1,df2=df2,ncp=ncp2); q2 <- pf(x1,df1=df1,df2=df2,ncp=ncp2);

}else{ x <- seq(min(ncp,ncp_c)-8,max(ncp,ncp_c)+8,length.out=500) pr <- dt(x,df=df,ncp=ncp); pb <- dt(x,df=df,ncp=ncp_c); x0 <- qt(q0,df=df,ncp=ncp); q1 <- pt(x1,df=df,ncp=ncp); ncp2 <- ncpt(x=x1,q=q0,df=df); pg <- dt(x,df=df,ncp=ncp2); q2 <- pt(x1,df=df,ncp=ncp2); } if (df <= 2) { pr[pr<=0] <- NA; pb[pb<=0] <- NA; pg[pg<=0] <- NA; pr[pr>10] <- NA; pb[pb>10] <- NA; pg[pg>10] <- NA; }

x0 <- round(x0,digits); q1 <- round(q1,digits); ncp2 <- round(ncp2,digits); q2 <- round(q2,digits);

pdf(rpdf, width=5, height=5) main <- paste('Pr(',name,'<',x0,';df=',df,',ncp=',ncp,')=',q0,'\nPr(',name,'<',x1,';df=',df,',ncp=',ncp,')=',q1,'\nPr(',name,'<',x1,';df=',df,',ncp=',ncp2,')=',q2, sep=) sub <-paste('\nnoncentral parameter = ',ncp,'(Red), ',ncp_c,'(Blue),\nand ',round(ncp2,digits),'(Black) which fits \nPr(',name,'<',x1,';ncp=',round(ncp2,digits),')=',round(q2,digits), sep=) plot(c(x,x),c(pr,pb),type='n',main=main,sub=sub,xlab=,ylab=paste(name,'probability density')) points(x[x<=max(x0,x1)],pr[x<=max(x0,x1)],col='green',type='h') points(x[x<=min(x0,x1)],pr[x<=min(x0,x1)],col='yellow',type='h') lines(x,pb,col='blue') lines(x,pr,col='red') lines(x,pg) </R>

z and noncentral distributions (chi-square, t, and F)

Noncentral chi-square

Let $Z_{i}$,i=0,1,2,... denote a series of independent random variables of standard normal distribution.

$V=\sum _{i=1}^{df}{Z_{i}}^{2}$

will be a random variable of $\chi ^{2}$ distribution with df degrees of freedom. For any given series of constants $\mu _{i}$,i=1,2,...,df,

$\sum _{i=1}^{df}(Z_{i}+\mu _{i})^{2}$

will be a random variable of the respective noncentral $\chi ^{2}$ distribution with the same df and the distinct noncetral parameter

$ncp=\sum _{i=1}^{df}{\mu _{i}}^{2}$

It is different from the random variable $V+ncp=\sum _{i=1}^{df}{Z_{i}}^{2}+\sum _{i=1}^{df}{\mu _{i}}^{2}$ of the respective central $\chi ^{2}$ distribution with a central drift.

Noncentral t

For any given constant $\mu _{0}$,

${\frac {Z_{0}+\mu _{0}}{\sqrt {V/df}}}=$${\frac {Z_{0}+\mu _{0}}{\sqrt {\sum _{i=1}^{df}{Z_{i}}^{2}/df}}}$

is a random variable of noncentral t-distribution with noncentrality parameter

$ncp=\mu _{0}$,

which is different from ${\frac {Z_{0}}{\sqrt {V/df\ }}}+\mu _{0}$, the central t-distributed random variable drifted with the same mean.

If df on this display is set to $\infty$ (Inf in R) and noncentral parameter set to 0, a standard normal distribution will be plotted and critical z score calculated.

Noncentral $F$

The noncentral parameter of F is only defined on its numerator. The noncentral F distributed

${\frac {\sum _{i=1}^{df_{1}}(Z_{i}+\mu _{i})^{2}/df_{1}}{\sum _{i=df_{1}+1}^{df_{1}+df_{2}}Z_{i}^{2}/df_{2}}}$

with noncentral parameter

ncp=$\sum _{i=1}^{df_{1}}\mu _{i}^{2}$

is different from the central F distributed random variable plus the respective constant ${\frac {\sum _{i=1}^{df_{1}}Z_{i}^{2}/df_{1}}{\sum _{i=df_{1}+1}^{df_{1}+df_{2}}Z_{i}^{2}/df_{2}}}+{\frac {\sum _{i=1}^{df_{1}}\mu _{i}^{2}}{df_{1}}}$ .

Confidence interval of standardized effect size by noncentral parameters

Confidence interval of unstandardized effect size like difference of means $(\mu _{1}-\mu _{2})$ can be found in common statistics textbooks and software, while confidence intervals of standardized effect size, especially Cohen's ${\tilde {d}}:={\frac {\mu _{1}-\mu _{2}}{\sigma }}$ and ${\tilde {f}}^{2}:={\frac {SS(\mu _{1},\mu _{2},...,\mu _{K})}{K\cdot \sigma ^{2}}}$, rely on the calculation of confidence intervals of noncentral parameters (ncp).

A common method to find confident interval limits of ncp is to solve the critical ncp value for marginal extreme quantile. The ncp parameter of the black curve in the above diagram could be directly adopted. For example, $\left(-\infty ,8.968\right)$ can be 97.5% one-way confidence interval of ncp if observed $t_{df=4}=5.1$, while change quantile from .025 to .975, we shall find that the two-way interval (1.139, 8.968) can be of 95% confidence level.

T test for mean difference of single group or two related groups

In case of single group, M ($\mu$) denotes the sample (population) mean of single group , and SD ($\sigma$) denotes the sample (population) standard deviation. N is the sample size of the group. T test is used for the hypothesis on the difference between mean and a baseline $\mu _{baseline}$. Usually, $\mu _{baseline}$ is zero, while not necessary. In case of two related groups, the single group is constructed by difference in each pair of samples, while SD ($\sigma$) denotes the sample (population) standard deviation of differences rather than within original two groups.

$t:={\frac {M}{SD/{\sqrt {N}}}}={\frac {{\sqrt {N}}{\frac {M-\mu }{\sigma }}+{\sqrt {N}}{\frac {\mu -\mu _{baseline}}{\sigma }}}{\frac {SD}{\sigma }}}$

$ncp={\sqrt {N}}{\frac {\mu -\mu _{baseline}}{\sigma }}$ and Cohen's $d:={\frac {M-\mu _{baseline}}{SD}}$ is the point estimate of ${\frac {\mu -\mu _{baseline}}{\sigma }}$.

So,

${\tilde {d}}={\frac {ncp}{\sqrt {N}}}$.

T test for mean difference between two independent groups

$n_{1}$ or $n_{2}$ is sample size within the respective group.

$t:={\frac {M_{1}-M_{2}}{SD_{within}/{\sqrt {\frac {n_{1}n_{2}}{n_{1}+n_{2}}}}}}$, wherein $SD_{within}:={\sqrt {\frac {SS_{within}}{df_{within}}}}={\sqrt {\frac {(n_{1}-1)SD_{1}^{2}+(n_{2}-1)SD_{2}^{2}}{n_{1}+n_{2}-2}}}$.

$ncp={\sqrt {\frac {n_{1}n_{2}}{n_{1}+n_{2}}}}{\frac {\mu _{1}-\mu _{2}}{\sigma }}$ and Cohen's $d:={\frac {M_{1}-M_{2}}{SD_{within}}}$ is the point estimate of ${\frac {\mu _{1}-\mu _{2}}{\sigma }}$.

So,

${\tilde {d}}={\frac {ncp}{\sqrt {\frac {n_{1}n_{2}}{n_{1}+n_{2}}}}}$.

One-way ANOVA test for mean difference across multiple independent groups

One-way ANOVA test applies noncentral F distribution. While with a given population standard deviation $\sigma$, the same test question applies noncentral chi-square distribution.

$F:={\frac {{\frac {SS_{between}}{\sigma ^{2}}}/df_{between}}{{\frac {SS_{within}}{\sigma ^{2}}}/df_{within}}}$

For each j-th sample within i-th group $X_{i,j}$, denote $M_{i}\left(X_{i,j}\right):={\frac {\sum _{w=1}^{n_{i}}X_{i,w}}{n_{i}}};\;\mu _{i}\left(X_{i,j}\right):=\mu _{i};$.

While,

$SS_{between} \over \sigma ^{2}$

$={\frac {SS\left(M_{i}\left(X_{i,j}\right);i=1,2,\cdots ,K,\;j=1,2,\cdots ,n_{i}\right)}{\sigma ^{2}}}$
$=SS\left({\frac {M_{i}\left(X_{i,j}-\mu _{i}\right)}{\sigma }}+{\frac {\mu _{i}}{\sigma }};i=1,2,\cdots ,K,\;j=1,2,\cdots ,n_{i}\right)$
$\sim \chi ^{2}\left(df=K-1,\;ncp=SS\left({\frac {\mu _{i}\left(X_{i,j}\right)}{\sigma }};i=1,2,\cdots ,K,\;j=1,2,\cdots ,n_{i}\right)\right)$

So, both ncp(s) of F and $\chi ^{2}$ equate

$SS\left(\mu _{i}(X_{i,j})/\sigma ;i=1,2,\cdots ,K,\;j=1,2,\cdots ,n_{i}\right)$.

In case of $n:=n_{1}=n_{2}=\cdots =n_{K}$ for K independent groups of same size, the total sample size is $N:=n\cdot K$.

$Cohen's\;{\tilde {f}}{}^{2}:={\frac {SS(\mu _{1},\mu _{2},...,\mu _{K})}{K\cdot \sigma ^{2}}}={\frac {SS\left(\mu _{i}\left(X_{i,j}\right)/\sigma ;i=1,2,\cdots ,K,\;j=1,2,\cdots ,n_{i}\right)}{n\cdot K}}={\frac {ncp}{n\cdot K}}={\frac {ncp}{N}}$.

T-test of pair of independent groups is a special case of one-way ANOVA. Note that noncentral parameter $ncp_{F}$ of F is not comparable to the noncentral parameter $ncp_{t}$ of the corresponding t. Actually, $ncp_{F}=ncp_{t}^{2}$, and ${\tilde {f}}=\left|{\frac {\tilde {d}}{2}}\right|$ in the case.

RMSEA of Structural Equation Model

ncp of $\chi ^{2}$ reported by Structural Equation Model softwares is proportional to the population value of $RMSEA^{2}$, or the squared distance per df from population var-cov matrix to the model space.

${\tilde {RMSEA}}={\sqrt {\frac {ncp}{(N-1)df}}}$

Power vs. Standardized Effect Size or ncp

Power of t test for a given Cohen's $\delta$

Example of one-group mean test

Input

<Rform name="tEx">

A normally distributed population, for example, IQ distribution of students, is sampled

$N$=<Input name="N" size="3" value="16"/>

times independently. The mean and standard deviation estimates from all $M$ samples are respectively denoted $M$ and $SD$ in the current replication.

The statistical interest is usually on the mean of population, named $\mu$; sometimes also on the standard deviation of population, named $\sigma$. The statistic $t$ is defined as following --

$t:={\frac {M-\mu _{null}}{SD/{\sqrt {N}}}}$

It measures whether or not $M$ is significantly

<select name="direction"><option value="gt" selected>greater than</option><option value="ne" >different from</option><option value="lt">less than</option></select> a baseline :$\mu _{null}$=<input name="mu0" size="5" value="100.00"/>,

relative to the scale of standard error estimate of $M$. If $\mu$ is really $\mu _{null}$, the $t$ statistic distribution is known with noncentral parameter 0 and degrees freedom $(N-1)$.

Type I error, denoted

$\alpha$=<Input name="alpha" size="3" value=".05"/>,

defines the probability domain of the extreme $t$ values.

However, the real $\mu$ may be $\mu _{alternative}$ rather than $\mu _{null}$. Then, the noncentral parameter of the $t$ statistic distribution will change to be

${\sqrt {N}}\times {\frac {\mu _{alternative}-\mu _{null}}{\sigma }}$

wherein $\delta :=(\mu _{alternative}-\mu _{null})/\sigma$ is estimated by Cohen's $d$$:=(M-\mu _{null})/SD$. A known/hypothesized eg.

$\delta =$<input name="delta" size="4" value="0.75">,

together with the sample size $N$, will give a known/hypothesized noncentral $t$ distribution, while a $\mu _{alternative}$ alone without a given $\sigma$ is helpless.

Change

$M$=<Input name="M" size="5" value="105.22"/> and $SD$=<Input name="s" size="5" value="16.72" />,

then verify whether they affect the statistical power.

<input type="submit" value="Click to Update"/> </Rform>

Results

How to cite this result

<R output="display" name="tEx" iframe="height:320px;"> N <- ifelse(exists("N"), as.numeric(N),16) mu0 <- ifelse(exists("mu0"), as.numeric(mu0),100) alpha <- ifelse(exists("alpha"), as.numeric(alpha),.05) delta <- ifelse(exists("delta"), as.numeric(delta),0.75) M <- ifelse(exists("M"), as.numeric(M),105.22) s <- ifelse(exists("s"), as.numeric(s),16.72) direction <- ifelse(exists("direction"), as.character(direction),"gt")

ncpt<-function(x,q,df,confirm=FALSE){ if (q<=0) (+Inf) else if (q>=1) (-Inf) else if ((q>0)&(q<1)) { .f<-function(ncp,x,df,q)abs(q-pt(x,df=df,ncp=ncp)) .n<-1; while ( ( (pt(x,df=df,ncp=-.n) < q+(1-q)/2 ) | (pt(x,df=df,ncp=.n) > q/2) ) & (.n < Inf) ) .n <- .n *2 ; if (confirm) optimize(f=.f,x=x,df=df,q=q,interval=c(-.n,.n)) else optimize(f=.f,x=x,df=df,q=q,interval=c(-.n,.n))$minimum } }

pdf(rpdf, horizontal=FALSE, width=8, height=4)

alpha_r <- ifelse(direction == "ne",alpha/2, ifelse(direction == "gt",alpha,0)); alpha_l <- alpha - alpha_r; se <- s/sqrt(N); df <- N-1; ncp <- sqrt(N)*delta; t <-(M-mu0)/se; tc_r <- qt(1-alpha_r,df=df); tc_l <- qt(alpha_l,df=df); d_l <- ncpt(x=t,q=1-alpha_r,df=df)/sqrt(N); d_r <- ncpt(x=t,q=alpha_l,df=df)/sqrt(N); sub <- paste("H0:blue central t; H1:red noncentral t\n",1-alpha," confidence interval of Cohen's delta\n",round(d_l,4)," ~ ",round(d_r,4),sep=""); if (direction == "ne") main1=paste("two-tail p value", round(1-pt(abs(t),df=df)+pt(-abs(t),df=df),4)); if (direction == "gt") main1=paste("right-tail p value", round(1-pt(t,df=df),4)); if (direction == "lt") main1=paste("left-tail p value", round(pt(t,df=df),4));

main1=paste("t value",round(t,4),",",main1,"\n",1-alpha,"confidence interval of mean\n",round(M+se*qt(alpha_r,df=df),4),"~",round(M-se*qt(alpha_l,df=df),4));

x <- seq(min(-4,ncp-5),max(4,ncp+5),length.out=200); op <-par(mfrow=c(1,2)); plot(x,dt(x,df=df),main=main1,sub=sub,xlab=,ylab=,type='l',col='blue'); points(x,dt(x,df=df,ncp=ncp),type='l',col='red'); x_reject <- c(x[(x > tc_r) | (x < tc_l)],tc_r,tc_l); points(x_reject,dt(x_reject,df=df,ncp=ncp),type='h',col='grey90'); points(x_reject,dt(x_reject,df=df,ncp=0),type='h',col='grey70'); points(t,dt(t,df=df,ncp=0));points(t,dt(t,df=df,ncp=0),type='h');

main2=paste("statistical power is", round(1-pt(tc_r,df=df,ncp=ncp)+pt(tc_l,df=df,ncp=ncp),4),"."); xN <- round(N/3):(3*N); plot(xN,1-pt(tc_r,df=df,ncp=sqrt(xN)*delta)+pt(tc_l,df=df,ncp=sqrt(xN)*delta),main=main2,xlab='sample size',ylab='statistical power',col='gray'); points(N,1-pt(tc_r,df=df,ncp=sqrt(N)*delta)+pt(tc_l,df=df,ncp=sqrt(N)*delta),col='red',type='h'); par(op); </R>

Two-related-group mean test

For two-related-group case, the difference scores between each pair of samples can apply one-group mean test interface. Usually $\mu _{null}$ is set to zero.

Power of $F$ test for a given Cohen's ${\tilde {f}}^{2}$

Let's use ${\tilde {f}}^{2}$ denote the population of Cohen's $f^{2}$, specially

$SS(\mu _{1},\mu _{2},\cdots ,\mu _{K}) \over {K\times \sigma ^{2}}$

in one-way ANOVA of $K$ groups setup with within-group sample size n and within-group population mean $\mu _{1},\mu _{2},\cdots ,\mu _{K}$ respectively. The noncentral parameter of the corresponding F or $\chi ^{2}$ distribution is $ncp=n\times K\times {\tilde {f}}^{2}$.

Power of SEM close-fit test for a given RMSEA

$(N-1)\cdot df\cdot RMSEA^{2}+df\sim \chi _{df,ncp=(N-1)\cdot df\cdot Distance^{2}perDf({\tilde {\Sigma }},Model\,Space)}^{2}$

$RMSEA={\sqrt {\frac {{\hat {\chi }}^{2}-df}{df\cdot (N-1)}}}$

How to cite this page in APA style

In APA style this page can be cited in reference lists like --

Comparison of noncentral and central distributions. (yyyy, Month dd). In SlideWiki. Retrieved MM:SS, Month dd, yyyy, from http://mars.wiwi.hu-berlin.de/mediawiki/slides/index.php/Comparison_of_noncentral_and_central_distributions

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External links

• Noncentral t-distribution on Wikipedia
• Noncentral $\chi ^{2}$ distribution on Wikipedia
• Noncentral F-distribution on Wikipedia
• Confidence interval of Effect Size on Wikipedia