7 7th Tutorial

Exercise 7.1 An assignment was given to four students attending a sample survey course. The problem was to estimate the average time per week devoted to study in Punjab Agricultural University (PAU) library by the students of this university. The university is running undergraduate, master’s degree and doctoral programs. Number of students registered for the three programs is \(1300,450\), and \(250\) respectively. Since the value of the study variable is likely to differ considerably with the program, the investigator divided the population of students into 3 strata: undergraduate program (stratum I), master’s program (stratum II), and doctoral program (stratum III). First of the four students selected WOR simple random samples of sizes \(20,10\) , and \(12\) students from strata I, II, and ill respectively, so that, the total sample is of size \(42\). The information about weekly time devoted in library is given in the csv file here

LBhour = read.csv("LBhour.csv", header = TRUE)
stratum1 = LBhour$HOUR[LBhour$PROGRAM==1]
stratum1

##  [1]  0  4  3  5  2  0  3  1  4  3  6  8 10  2  9  4  6  1  2  3

stratum2 = LBhour$HOUR[LBhour$PROGRAM==2]
stratum2

##  [1] 12  9 11 13  8  6 10  9 11  7

stratum3 = LBhour$HOUR[LBhour$PROGRAM==3]
stratum3

##  [1] 10 14 20 11 16 13 24 15 14 18 19 20

(n1 = length(stratum1))

## [1] 20

(n2 = length(stratum2))

## [1] 10

(n3 = length(stratum3))

## [1] 12

(ybar1 = mean(stratum1))

## [1] 3.8

(ybar2 = mean(stratum2))

## [1] 9.6

(ybar3 = mean(stratum3))

## [1] 16.16667

(yvar1 = var(stratum1))

## [1] 7.957895

(yvar2 = var(stratum2))

## [1] 4.933333

(yvar3 = var(stratum3))

## [1] 17.06061

N1 = 1300
N2 = 450
N3 = 250
N = N1 +N2 + N3
N

## [1] 2000

#determine an estimate of the population mean by stratified sample
W1=N1/N; W2=N2/N ; W3=N3/N 
W1; W2; W3

## [1] 0.65

## [1] 0.225

## [1] 0.125

str_mean=(W1*ybar1+W2*ybar2+W3*ybar3)
str_mean

## [1] 6.650833

#Variance of sample mean = sum(((1-fh)/nh)*(Wh^2*Vh^2), fh=nh/Nh

var_str=((W1)^2)*((N1-n1)/N1)*(yvar1/n1)+((W2)^2)*((N2-n2)/N2)*(yvar2/n2)+((W3)^2)*((N3-n3)/N3)*(yvar3/n3)

var_str

## [1] 0.2110923

CIU=str_mean+2*(sqrt(var_str))
CIU

## [1] 7.569729

CIL=str_mean-2*(sqrt(var_str))
CIL

## [1] 5.731938

#estimation for total and its variance
t_hat=N*str_mean
t_hat

## [1] 13301.67

Var_t=(N^2)*var_str
Var_t

## [1] 844369

#95% CI for  total estimation
CI_totU=t_hat+2*sqrt(Var_t)
CI_totU

## [1] 15139.46

CI_totL=t_hat-2*sqrt(Var_t)
CI_totL

## [1] 11463.88

Exercise 7.2 An insurance company’s records show that out of the total of \(500\) claims, 280 are major claims (from Rs \(1000\) to Rs \(2500\)) and \(220\) are minor (below Rs \(1000\)). A WOR simple random sample of \(10\) claims was drawn from each category (stratum), and claim amounts were recorded as :

Stratum1	1200	1600	1800	1400	1980	2110	2440	1660	1790	1910
Stratum2	720	880	760	660	790	840	550	960	640	800

Estimate the total amount of all the \(500\) claims, and construct the confidence interval for it.

\[{\bar{y}}_h=\frac{\sum_{i=1}^{n}y_{hi}}{n_h}; ~~~~\ \ s_h^2=\frac{\sum_{i=1}^{n}\left(y_{hi}-{\bar{y}}_{hi}\right)^2}{n_h-1}\ ; ~~~~\ \ W_h=\frac{N_h}{N}\ ;~~~~\ N=N_1+N_2=500\ \]

The estimate of average amount of claims is \(\ \ {\bar{y}}_{str}=W_1{\bar{y}}_1+W_2{\bar{y}}_2=1336.24\)
Also, the variance of the mean is \[v\left({\bar{y}}_{str}\right)=\sum_{h=1}^{L}W_h^2\ \frac{s_h^2}{n_h}\frac{N_h-n_h}{N_h}=4066.313\] we obtain the limits of confidence interval as

\[{\bar{y}}_{str}\pm Z_{1-\propto/2}\ \sqrt{v\left({\bar{y}}_{str}\right)} ; Z_{1-\propto/2}\approx2\] \[1336.24\pm2\ \sqrt{4066.313} \] \[ [1208.705,\ 1463.775\ ]\] We are \(95\%\) confidence that the average amount of claims is between \([1208.705, 1463.775]\)
Note: Multiply these limits by \(N\) to obtain the limits for the population total (T). OR \[{y\prime}_{str}=N \times {\bar{y}}_{str}=500\times 1336.24=668120\] Estimator of the variance of population total are obtained by \[ v(y^\prime_{str})=N^2\times v({\bar{y}_{str}})=\left(500\right)^2\left(4066.313\right)=1016578267\]
\(95\%\) CI for total estimation \[{y\prime}_{str}\ \ \pm Z_{1-\propto/2}\ \sqrt{v\left({y\prime}_{str}\right)}\]

\[668120\pm2\ \sqrt{1016578267\ }\] \[= [604352.4\ \ ,\ 731887.6]\] Code:

y1=c(1200,1600,1800,1400,1980,2110,2440,1660,1790,1910)
n1=length(y1)
n1

## [1] 10

y1_bar=mean(y1)
y1_bar

## [1] 1789

v1=var(y1)
v1

## [1] 125410

y2=c(720,880,760,660,790,840,550,960,640,800)
n2=length(y2)
n2

## [1] 10

y2_bar=mean(y2)
y2_bar

## [1] 760

v2=var(y2)
v2

## [1] 14822.22

N1=280
N2=220
N=N1+N2
N

## [1] 500

#determine an estimate of the population mean by stratified sample
W1=N1/N; W2=N2/N 
W1; W2

## [1] 0.56

## [1] 0.44

str_mean=(W1*y1_bar+W2*y2_bar)
str_mean

## [1] 1336.24

#Variance of sample mean =sum(((1-fh)/nh)*(Wh^2*Vh^2), fh=nh/Nh
var_str=((W1)^2)*((N1-n1)/N1)*(v1/n1)+((W2)^2)*((N2-n2)/N2)*(v2/n2)
var_str

## [1] 4066.313

CIU=str_mean+2*(sqrt(var_str))
CIU

## [1] 1463.775

CIL=str_mean-2*(sqrt(var_str))
CIL

## [1] 1208.705

#estimation for total and its variance
t_hat=N*str_mean
t_hat

## [1] 668120

Var_t=(N^2)*var_str
Var_t

## [1] 1016578267

#95%CI for  total estimation
CI_totU=t_hat+2*sqrt(Var_t)
CI_totU

## [1] 731887.6

CI_totL=t_hat-2*sqrt(Var_t)
CI_totL

## [1] 604352.4