Practicality of life

## Set A

## 1) Model Sampling from Cauchy and Laplace distribution
> # a)
> # X follows cauchy disribution with parameter theta=10 and lambda=1
> #using inverse Transformation CDF
> n=11 #size of random sample to be drawn from Cauchy distribution
> lambda=1 #Scale Parameter
> theta=10 #Location parameter
> # As we Know that evry distribution cdf follows standard uniform distribution
> y=runif(n) # Generating random sample of size 11 from uniform distribution
> x=lambda*tan(pi* (v-0.5))+theta;x
> # b) Median of the cauchy sample obtain From obove
> median(x)
## Cauchy= L/2*exp(-L|x-u|)



## 2) Log normal Cancer
# (1/(x-a)sigrt2pi)exp(-1/2sigsq(logx-a - u)sq)
> # Fiting log normal distribution
> LL=seq(0,60,10);LL #lower Limit of given data
> UL=seq(10,70,10);UL #upper limit of the given data
> Freq=c(20,90,52,11,6,4,1);Freq
> N=sum(Freq)
> x=(UL+LL)/2; #mid point of the gien data
> d=data.frame("lower limit"=LL, "Upper limit"=UL,"mid point"=x, "frequency"=Freq)
> d
> M12=sum(Freq*x/sum(Freq); M12 # Frist sample raw Movement used for estimating mean and Sigma
> M13=sum(Freq*(x^2))/sum(Freq);M13 # Second sample raw moment used for estimating mean sigma
> sigma_est=log(M13,base=exp(1))-(2*log(M12,base =exp (1))); sigma_est ##estimating sigma for calculating Probabilities
> mu_est=(2*log(M12,base=exp(1)))-(log(M13,base=exp(1))/2); mu_est ##estimating mean for calculating probabilities
> a=plnorm(UL,mu_est,sqrt(sigma_est));a #calculating cumulative Probabilities
> px=с()
> px[1]=a[1]
> for (i in 1: (length(a)-1)){
+ px[i+1]=ali+1]-a[i]
+}
> ExFreq=round(N*px,2);ExFreq
> sum(ExFreq)
> D1= data.frame(d, "expected"= Exfreq)




# 3) literates 2 year Ratio and Regression mthd
› #Ratio and regression method of estimation Comparsion with SRSWOR
> x=с(109,101,125,254,559,359,427,481);X
> y=c(99,112,111,278,634,355,399,489); Y
> N=170
#No observation of given data X.
> n=length(y)
> Xt=21288000 #population total of given data X
> Х_bar_N=xt/N;X_bar_N #Population Mean of given data of X
> y_bar_n=mean(y)
> x_bar_n=mean(x)
> #Ratio method.
> Rn=y_bar_n/x_bar_n;Rn #ratio of sample mean of y and x
> Y_bar_N=Rn*X_bar_N;Y_bar_N #Estimate of Population mean of Y
> sy_sq=var(y);sy_sq #sample mean square of y
> sx_sq=var(x);sx_sq #sample mean square of x
> sxy_sq=var(x,y);sxy_sq #sample mean square of xy
> SE_Y_totalhat=N*sqrt(((1/n)-(1/N))*(sy_sq+Rn^2*sx_sq-(2*Rn*sxy_sq)));SE_Y_totalhat #Estimate of SE of population total of Y
> #Regression method:
> byx=sxy_sq/sx_sq;byx #regression coefficient of Y on X
> Y_hat=y_bar_n+(byx*(X_bar_N-x_bar_n));Y_hat #Estimate of Population mean of Y
> Y_hat_total=N*Y_hat;Y_hat_total #Estimate of Population total of Y
> SE_Y_total1=N*sqrt((N-n)/(N*n))*(sy_sq+(byx^2*sx_sq)-(2*byx*sxy_sq)));SE_Y_total1 #Estimate of SE of population total of Y
> #comparison between SRSwor
> var_y_bar=((1/n)-(1/N))*sy_sq;var_y_bar #variance of sample mean under SRSWOR
› SE_Y_hat=N*sqrt(var_y_bar);SE_Y_hat #S.E of sample mean under SRSWOR
> #conclusion- SE(Y_hat)SRSWOR>SE(Y_hat)RATIO>SE(Y_hat)REGRESSION
># This implies that variation of regression estimator is minimum and use supplementary information increases the precision.




## SET B
## 1. B
> #que no 4
> mu=5
> sigma=sqrt(4)
> set.seed (8)
> u=runif(8,0,1)
#where u and v follows U(0,1)
> U
> v=runif (8,0,1)
> V
> z=sqrt(-2*log(x=u)) *cos(2*pi*v)
#random sample from N (0,1)
> Z
1>x=mu+(2* sigma)
#random sample from N(5,4)
> X
> mean (X)
> #conclusion-Hence our sample mean is close to population mean but sample variance shows some deviation but as the sample size tends to infinity it will close to population variance


#1.C
> #Weibull distribution with alpha=15 and beta=10
> alpha3=3
#parameter values of weibull distribution
> beta3=1
> x=seq (-0.01,15,0.1);x
> px4=dweibull(x,shape=beta3,scale=alpha3);px
> plot(x,px4,col="pink",xlab="variable", ylab="probability", lwd=4,type="1", ylim=c(0,0.92))

> #а)
> alpha=1
> beta=2 #parameter values of weibull distribution
> x=seq(-0.01,15,0.1);x
> px=dweibull(x,shape=beta,scale=alpha);px
> lines(x=x,y=px,lty=3,col="yellow", lwd=3)
>#D）
> alpha1=5 #parameter values of weibull distribution
> beta1=1
> px1=dweibull(x,shape=beta1,scale=alpha1);px
> lines(x=x,Y=px1,col="blue", lwd=4,Ity=2)
> #C)
> alpha1=1 #parameter values of weibull distribution
> beta1=1
> px2=dweibull(x,shape=beta1,scale=alphal);px
> lines(x=x,y=px2,col="green", lwd=3,Ity=2)
>
>#d)
> alpha2=10 #parameter values of weibull distribution
> beta2=5
> px3=dweibull(x,shape=beta2,scale=alpha2); px
> lines(x=x,y=px3,col="violet", wd=4,Ity=2)
legend (locator(1), legend=c("W(alpha=3,beta=1)", "W(alpha=1,beta=2)", "W(alpha=5,beta=1)", "W(alpha=1,beta=1)", "W(alpha=10, beta=5)") fill=c("pink", "yellow","blue", "green", "violet"))
> #conclusion-As we we increase the values of both the shape parameter it slowly becomes symmetric from strong negatively skewed.




###proportional and neyman allocation
> N1=80 #population
> N2=60
> N3=40
> N4=100
> N5=100
> Ni=c(80,60,40,100,120); Ni #population of each strata
> N=sum(N1,N2,N3, N4, N5);N
> y_bar_i=c(82.3,161.5,139.2,239.1,200);Y_bar_i #Popn mean of each strata
> Si=c(21.3,17.7,14.7,15.3,18.9);Si
> n=40
> # Obtain the size of the sub sample from each stratum
> # for population allocation
> ni=(n/N)*Ni;ni
> n1=ni[1];n1
> n2=ni|2];n2
> n3=ni[3];n3
> n4=ni[4];n4
> n5=ni[5];n5

># for Nayman's allocation
> ni=(n*Ni*Si)/sum(Ni*Si)
> n1=round(ni(1],0);n1
> n2=round(ni[2],0);n2
> n3=round(ni[3],0);n3
> n4=round(ni[4],0);n4
> n5=round (ni[5],0);n5
> # To obtain S.E of the estimate of population mean and population total
> # under population allocation
> pi=c(N1/N,N2/N,N3/N, N4/N,N5/N);pi
> S_E_est_ Ybar1=sqrt(((1/n)-(1/N))*sum(pi*Si^2));S_E_est_Ybar1
> S.E_est_Yt1=sqrt(N*S_E_est_Ybar1);S.E_est_Yt1
› # Under Nayman's allocation
> S.E_est_ Ybar2=sqrt((1/n)*(sum(pi*Si))*2)-((1/N)*sum(pi*Si^2)));S.E_est_Ybar2
> S.E_est_Vt2=sqrt(N*S.E_est_Ybar2);S.E_est_Yt2
> #UNDER SRSWOR
> Y_bar_N=sum(Y_bar_i*Ni)/N;Y_bar_N
> S.E_est_Ybar3=sart((N-n)/(N*n))*(1/(N-1)) *(Isum(Ni-1) *Si^2)+(sum((Y_bar_i-Y_bar _N)*2*Ni));S.E_est_Ybar3
> S.E_est_Yt3=sqrt(N*S.E_est_Ybar3);S.E_est_Yt3
› # conclusion:since varience of ybarst of population mean under SRSWOR is more than the varience under population allocation and Neyman's allocation



SET C
Q1. A(i)
> x=sort(c(24,38,61,22,16,57,31,29,35));x
> length(x)
> LL=seq(1,9,1);LL
> UL=sort(seq(1,9,1), decreasing = TRUE); UL
> d=data.frame(LL,UL);a
> d1=subset(d,LL<UL); d1
> d2=data.frame(d1,LB=x[d1$LL],UB=[d1$UL]);d2
> con_Fin=pbinom(d2$UL-1,9,0.5)-pbinom(d2$LL-1,9,0.5)
> data.frame(d2,con_Fin)
> #Median would lie within 24,38 confidence interval with 82% confidence coefficent.

Q1. C
> #Q 2: 6 students mark out of 50
> L=c(42,18,25,35,20,30);L
> N=length(L);N #L is total no of sampling unit IN population
> n=2;n
> 0b1=1:6
> 002=1:6
> m=merge(ob1,0b2);m
> sub=subset(m,m$x>m$y);sub #position of observation in SRSWOR sample
> d=data.frame(ob1=L|sub|,11],ob2=L[sub|,2]]);d
> d1=data.frame(d,m=apply(d,1,mean),sv=(apply(d,1,var))*((n-1)/n));d1
> n1= length(d1$sm)
> #verify(E(y_bar)=Y_bar)
> est_y_bar=mean(d15sm);est_y_bar
> #conclusion:population mean is unbiased estimator of sample mean
> ybar=mean(L); ybar
> #veryfy (var(ybar)=((N-n)/Nn)s^2)
> n=length(d1$sv);n
> var_y_bar-var(d15sm)*(n1-1)/(n1);var_y_bar
> RHS=var(L)*((N-n)/(N*n)); RHS

> #confidence interval for population mean
> alpha=0.05
> z_alpha_2=gnorm(1-(alpha/2));z_alpha_2
> sigma=sqrt(d1$sv);sigma
> d2-data.frame(d2, L_cl-d15m-(z_alpha_2*sigma), U_CL=15sm+(z_alpha 2*sigma));d2
> d3=data.frame(d2,T_L_cl=N*(d1$sm-(z_alpha_2*sigma)), T_U_CL=N*(d1$sm+(z_alpha_2*sigma)));d3

## list of 3100 , 41 wrong .. Find SE and CI
> N=3100;N # where N is Population size.
> n=200;n #where n is random sample size from population.
> a=41 #No. of names and address which are wrong and need to be corrected in list in sample of 200.
> р=а/п;р #where p is proportion of sample which is unbaised estimator of population proportiona
> q=1-p;9

> #To find estimate for standard error for population Proportion
> est_S.E_est_P=sqrt((N-n)/(n*N))*(n/(n-1))*p*q);est_S.E_est_P
> #To find 95% confidence interval for population proprotion
> #for n>30
> alpha=0.05
> Z_alpha=qnorm (1-(alpha/2),0,1);Z_alpha
> L_CI=p-(Z_alpha*est_S.E_est_P);L_CI
> U_CI=p+(Z_alpha*est _S.E_est_P);U_Cl

> #To find 95% confidence interval for population Total
> L_CI=N*(p-(Z_alpha*est_S.E_est_P));L_Cl
> U_CI=N*(p+(Z_alpha*est_S.E_est_P));U_Cl



Sale_last_year=c(50,35,12,10, 15,30,9,25, 100,250,50,50,150,100,40);Sal e_last_year
111 50 35 12 10 15 30 9 25 100 250 50 50 150 100 40
> Sale_this_year=c(56,48,22,14,18,26,11,30,165,409,73,70,95,55,83);Sale _this_year
> N=300
#No observation of given data X
> n=length(y)
> Xt=21300
#population total of given data X
> X_bar_N=Xt/N;_bar_N
[1] 71
#Population Mean of given data of x
> y_bar
_n=mean (y)
> x_
bar_n=mean(x)
#sample mean of y #sample mean of y
> #Ratio method.
> Rn=y_bar_n/x_bar_n;Rn
#ratio of sample mean of y and
sample meanof x
[1] 0.9622642
> Y_bar_N=Rn *X_bar_N;Y_bar_N
#Estimate of Population mean of
Y
[1] 68.32075
> sy_sq=var(y);sy_sq
#sample mean square of y
[1] 88.1
> sx_sq=var(x);sx_sq
#sample mean square of x
[1] 100.6778
> sxy_sq=var(x,y);sxy_sq
#sample mean square of xy
[1] 79.63333
_Y_totalhat=N*sqrt(((1/n)-(1/N))*(sy_sq+Rn^2*sx_sq-
(2*Rn*sxy_sq)));SE_Y.
_totalhat #Estimate of SE of population total of Y
[1] 494.1418
> #Regression method:
> byx=sxy_sq/sx_sq;byx
#regression coefficient of Y on X
[1] 0.7909723
> Y_hat=y_bar_n+(byx*(X_bar_N-x_bar_n));Y__hat
Population mean of Y
[1] 66.14535
> Y_hat_total=N*Y_hat;Y_hat_total
mean of X
[1] 19843.6
#Estimate of Population
> SE_Ytotal1=N*sqrt((N-n)/(N*n))*(sy_sa+(byx^2*sx_sa)-(2*byx*sxy_sq)))
> #The Standard error of estimator from Ratio and regression method is494.1418 and 467.4147