6 6th Tutorial

Exercise 6.1 Assume that the sample of \(24\) hotels selected in Exercise 5.3 is a preliminary sample. Examine whether this sample is sufficient to estimate the average per day rent with a permissible error of Rs \(5\) ? If not, how many additional units need to be selected.

Solution

\[N=192,~~~~ n=24,,~~~~ B=5\] \[n_s=\frac{Ns^2}{ND+s^2} ,~~~~~ D=\ \frac{B^2}{4}=6.25, ~~~~ s^2=\frac{\sum_{i=1}^{n}\left(y_i-\bar{y}\right)^2}{n-1}=\ 393.4783 \]

y=c(100,120,125,115,110,80,130,120,90,110,125,80,70,125,130,105,125,85, 90,105,130,95,135,140)
n=length(y)
N = 192
K = N/n
# Compute the sample mean
(ybar = mean(y))

## [1] 110

# Compute the variance of the sample  
X = var(y)
X

## [1] 393.4783

Therefore,

\[n=\ \frac{Ns^2}{ND+s^2}=\frac{\ 192\ast393.4783\ }{(192\ast6.25)+393.4783}=47.41064\approx~47\] Since the sample size required to estimate the average per day rent with a permissible error of \(5\) is \(47\), the investigator will, therefore, need to select \(47-24= 23\) more hotels to get the estimate with specified magnitude of tolerable error.

Exercise 6.2 Some of the school buildings in a district collapsed during last few years, and caused damage to life and property. The district administration decided to have a quick estimate of the proportion of unsafe school buildings in the district. For this purpose, a systematic sample of 84 buildings, out of a total of \(1260\) school buildings, was selected. The selected school buildings were examined by experts. The number of unsafe buildings was found to be \(16\). Estimate the proportion of unsafe buildings in the district, and work out the confidence interval for it.

Solution The giving information are \(N=1260, n=84, a =16\) \(p_{sy}=\dfrac{16}{84}=0.1905\)
\(v\left(p_{sy}\right)=\frac{(N-n)p_{sy}(1-p_{sy})}{N\ast(n-1)}=\frac{\left(1260-84\right)\ast0.1905\ast(1-0.1905)}{1260\ast(84-1)}=\ 0.0017\)

\(se\left(p_{sy}\right)=\ \sqrt{\ 0.0017}=0.0416\)

The confidence limit for \(P\) is \(p_{sy}\pm1.96\ast se(p_{sy})\)

\[0.1905\pm2\ast0.0416\] The C.I is \([0.10730, 0.2737]\)

Exercise 6.3 Solve Example 6.4 by R and compare the answers to textbook:

y=c(5.614,8.202,6.115,9.765,8.550,9.225,6.640,7.350,5.843, 6.875,8.460,10.850,6.970,5.524,7.847)
n=length(y)
N=162
sum(y)

## [1] 113.83

ybar=mean(y)
ybar

## [1] 7.588667

ytotal=N*ybar
ytotal

## [1] 1229.364

#  compute the variance of the sample mean
X=  sum(diff(y)^2)
X

## [1] 67.59603

var_sysmean= ((N-n)/(2*N*n*(n-1)))*X
var_sysmean

## [1] 0.1460408

#  compute the variance of sample total 
ytotal_var=N^2*var_sysmean
ytotal_var

## [1] 3832.695

sd= sqrt(var_sysmean)
sd

## [1] 0.3821529

#CI for sample total
CIL=N*(ybar-(2*sd))
CIL

## [1] 1105.546

CIU=N*(ybar+(2*sd))
CIU

## [1] 1353.182