2 2nd Tutorial

Exercise 2.1 What is a parameter? Work out mean and variance for the following given population values

\[44, 56, 60, 48, 55, 50, 58, 62, 60, 40\]
The parameter is any real valued function of variable values for all the population is known as a population parameter or simply a parameter.
The population mean is \[\text{Mean}:\bar{Y} = \frac{44+56+60+48+55+50+58+62+60+40}{10}=53.3\]
To calculate the population variance \[\text{Population Variance:}~~ \sigma^2=\frac{1}{N}\sum_{i=1}^{N}Y_i^2-\bar{Y}^2 = \ 50.01\] In R one can use the function mean() to calculate the mean and var() to calculate the variance.

Y <- c(44,56,60,48,55,50,58,62,60,40)
(Ybar <- mean(Y))

## [1] 53.3

(Sigma <- var(Y)*(9/10))

## [1] 50.01

Exercise 2.2 Assuming that \(20, 12, 15, 16, 18, 14,22,28,24,\) and \(26\) are the observations for a sample of \(10\) units, calculate sample mean and the variance.

\[\text{Mean}:\bar{y} = \frac{20+12+15+16+18+14+22+28+24+26}{10}=19.5\] \[\text{Population Variance:}~~ \sigma^2=\frac{1}{n-1}\sum_{i=1}^{n}y_i^2-\bar{y}^2 = 29.16667\]

y <- c(20,12,15,16,18,14,22,28,24,26)
(ybar <- mean(y))

## [1] 19.5

(Ssquare <- var(y))

## [1] 29.16667

Exercise 2.3 Five babies were born in a particular year in village Beonhin of Mathura district. The age (in years) of mothers at the time of child birth were \(29, 32, 26, 28,\) and \(36\). Enumerate all possible WR equal probability samples of size 2, and show numerically that the sample mean age is an unbiased estimator of population mean age of the mothers.

Let’s name the mothers (A,B,C,D,E) with weights (29,32,26,28,36) respectively. \[N=5 ~~~~~~~~~~~~ n=2\] \[\Rightarrow ~~~ \mu = \frac{1}{5}~\sum_{i=1}^{5}X_i = 30.2\] The number of all possible samples is: \(5^2=25\)

Sample	Mothers in the sample	Sample mean
1	AA	29.0
2	AB	30.5
3	BA	30.5
4	AC	27.5
5	CA	27.5
6	AD	28.5
7	DA	28.5
8	AE	32.5
9	EA	32.5
10	BB	32.0
11	BC	29.0
12	CB	29.0
13	BD	30.0
14	DB	30.0
15	BE	34.0
16	EB	34.0
17	CC	26.0
18	CD	27.0
19	DC	27.0
20	CE	31.0
21	EC	31.0
22	DD	28.0
23	DE	32.0
24	ED	32.0
25	EE	36.0

Distribution of the sample mean:

Sample mean	Frequency	Probabilty
26.0	1	0.04
27.0	2	0.08
27.5	2	0.08
28.0	1	0.04
28.5	2	0.08
29.0	3	0.12
30.0	2	0.08
30.5	2	0.08
31.0	2	0.08
32.0	3	0.12
32.5	2	0.08
34.0	2	0.08
36.0	1	0.04

Now \(E(\bar{y})=\sum_{i=1}^{13} \bar{y_i}~P(\bar{y_i}) = 30.2\)
As \(E(\bar{y})=\mu\) so \(\bar{y}\) is unbias estimator of \(\mu\).
By R

#install.packages("tidyverse")
library(tidyverse)
Y <- c(29,32,26,28,36)
(Ybar <- mean(Y))

## [1] 30.2

(sample <- crossing(Var1=Y, Var2=Y)) # A tibble: 25 x 2

## # A tibble: 25 x 2
##     Var1  Var2
##    <dbl> <dbl>
##  1    26    26
##  2    26    28
##  3    26    29
##  4    26    32
##  5    26    36
##  6    28    26
##  7    28    28
##  8    28    29
##  9    28    32
## 10    28    36
## # ... with 15 more rows

#to view as a table run
#view(sample)

(ybars= apply(sample,1 , mean))

##  [1] 26.0 27.0 27.5 29.0 31.0 27.0 28.0 28.5 30.0 32.0 27.5 28.5 29.0 30.5 32.5
## [16] 29.0 30.0 30.5 32.0 34.0 31.0 32.0 32.5 34.0 36.0

(unbiased=mean(ybars))

## [1] 30.2

Let’s take an example of size three:
By R:

Y <- c(29,32,26,28,36)
Ybar <- mean(Y)
sample <- crossing (Var1=Y, Var2=Y, Var3=Y)
samples<- sample(Y, 3, replace= TRUE)
sample     #No. of observation = 5^3 = 125

## # A tibble: 125 x 3
##     Var1  Var2  Var3
##    <dbl> <dbl> <dbl>
##  1    26    26    26
##  2    26    26    28
##  3    26    26    29
##  4    26    26    32
##  5    26    26    36
##  6    26    28    26
##  7    26    28    28
##  8    26    28    29
##  9    26    28    32
## 10    26    28    36
## # ... with 115 more rows

# A tibble: 125 x 3

Calculate all possible means

(ybars= apply(sample,1 , mean))

##   [1] 26.00000 26.66667 27.00000 28.00000 29.33333 26.66667 27.33333 27.66667
##   [9] 28.66667 30.00000 27.00000 27.66667 28.00000 29.00000 30.33333 28.00000
##  [17] 28.66667 29.00000 30.00000 31.33333 29.33333 30.00000 30.33333 31.33333
##  [25] 32.66667 26.66667 27.33333 27.66667 28.66667 30.00000 27.33333 28.00000
##  [33] 28.33333 29.33333 30.66667 27.66667 28.33333 28.66667 29.66667 31.00000
##  [41] 28.66667 29.33333 29.66667 30.66667 32.00000 30.00000 30.66667 31.00000
##  [49] 32.00000 33.33333 27.00000 27.66667 28.00000 29.00000 30.33333 27.66667
##  [57] 28.33333 28.66667 29.66667 31.00000 28.00000 28.66667 29.00000 30.00000
##  [65] 31.33333 29.00000 29.66667 30.00000 31.00000 32.33333 30.33333 31.00000
##  [73] 31.33333 32.33333 33.66667 28.00000 28.66667 29.00000 30.00000 31.33333
##  [81] 28.66667 29.33333 29.66667 30.66667 32.00000 29.00000 29.66667 30.00000
##  [89] 31.00000 32.33333 30.00000 30.66667 31.00000 32.00000 33.33333 31.33333
##  [97] 32.00000 32.33333 33.33333 34.66667 29.33333 30.00000 30.33333 31.33333
## [105] 32.66667 30.00000 30.66667 31.00000 32.00000 33.33333 30.33333 31.00000
## [113] 31.33333 32.33333 33.66667 31.33333 32.00000 32.33333 33.33333 34.66667
## [121] 32.66667 33.33333 33.66667 34.66667 36.00000

(unbiased=mean(ybars))

## [1] 30.2

Exercise 2.4 Distinguish between sampling and nonsampling errors. Which of these errors are more likely to be present in a census or a sample survey?

• Sampling error is the resultant discrepancy between the sample estimate and the population parameter value is the error of the estimate. Conversely, non-sampling error is arising due to defective sampling procedures, ambiguity in definitions, faulty measurement techniques, mistakes in recording, errors in coding-decoding, tabulation and analysis, etc…

• The sampling errors usually decreases with increase in sample size. In contrary, the nonsampling errors are likely to increase with increase in sample size. It is quite possible that nonsampling errors in a complete enumeration survey are greater than both the sampling and nonsampling errors taken together in a sample survey.