There many ways to select 15 random samples from the given Tippet’s random number table. Since the population size is 83(two-digit number). Here the door numbers are assigned from 1 to 83. Assume that at random we first choose 2^nd column. So the first sample is 66 and other 14 samples are 74, 52, 39, 15, 34, 11, 14, 13, 27, 61, 79, 72, 35, and 60. If the numbers are above 83, choose the next number ranging from 1 to 83.

The errors that arise due to human factors which always vary from one investigator to another in selecting, estimating or using measuring instruments are called Non-Sampling errors.
It may arise in the following ways:
a) Due to negligence and carelessness of the part of either investigator or respondents.
b) Due to lack of trained and qualified investigators.
c) Due to framing of a wrong questionnaire.
d) Due to applying wrong statistical measure.
e) Due to incomplete investigation and sample survey.

Errors, which arise in the normal course of investigation or enumeration on account of chance, are called sampling errors. Sampling Errors arise due to the following reasons:
a) Faulty selection of the sample instead of correct sample by defective sampling technique.
b) The investigator substitutes a convenient sample if the original sample is not available while investigation.
c) In area surveys, while dealing with border lines it depends upon the investigator whether to include them in the sample or not. This is known as Faulty demarcation of sampling units.

Given \(\sigma\) = 10, S.E. \(\bar { X } \) = 3
We know that S.E = \(\frac { \sigma }{ \sqrt { n } } \)
Therefore, \(3=\frac { 10 }{ \sqrt { n } } \Rightarrow \sqrt { n } =\frac { 10 }{ 3 } \)
Taking Squaring on both sides we get
\(n=\left(\frac{10}{3}\right)^{2}=\frac{100}{9}=11.11 \cong 11\), 
The required sample size is 11.

Here, we have to select 10 random numbers ranging from 1550 to 8000 but the given random number table has only 2 digit numbers. To solve this, two - 2 digit numbers can be combined together to make a four- figure number. Let us select the 5^th and 6^th column and combine them to form a random number, then select the random number with given range. This gives 5 random numbers, similarly, 8^th and 9^th is selected and combined to form a random numbers, then select the random number with given range. This gives 5 random numbers, totally 10 four- figure numbers have been selected. The following table shows the 10 random numbers which are combined and selected.

Therefore the selected 10 random numbers are

When the population is heterogeneous with respect to the variable, then Stratified Random Sampling method is used.
Following steps are involved:
a) The population is divided into different classes so that each stratum will consist of more or less homogeneous elements. The strata are so designed that they do not overlap each other.
b) After the population is stratified, a sample is drawn at random from each stratum using Lottery Method or Table of Random Number Method.
Example:
From the following data, select 68 random samples from the population of heterogeneous group with size of 500 through stratified random sampling, considering the following categories as strata.
Category 1 : Lower income class - 39%
Category 2 : Middle income class - 38%
Category 3 : Upper income class- 23%
solution:

Sample size n = 400 stores
Sample mean \(\bar x\) = 153.7 bars
Sample SD s = 17.2 bars
Population mean m = 146.3 bars
Since population SD is unknown we can consider the sample SD s = \(\sigma\)
Null Hypothesis :
The advertising campaign is not successful i.e, H₀: \(\mu\) = 146.3 
(There is no significant difference between the mean weekly sales of soap bars in department stores before and after advertising campaign)
Alternative Hypothesis H₁:
\(\mu\) >143.3 (Right tail test). The advertising campaign was successful
Level of significance \(\sigma\) = 0.05
Test statistic :
\(Z=\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \sim N(0,1)\)
\(Z=\frac { 153.7-146.3 }{ \frac { 17.2 }{ \sqrt { 400 } } } \)
\(=\frac { 7.4 }{ 0.86 } =8.605\)
\(\therefore\) Z = 8.605
Comparing the calculated value Z=8.605 and the significant value or table value \({ Z }_{ \alpha }=1.645.\) We get 8.605 > 1.645 
Inference:
Since, the calculated value is much greater than table value i.e., Z > \({ Z }_{ \alpha }\), it is highly significant at 5% level of significance.
Hence we reject the null hypothesis H₀ and conclude that the advertising campaign was definitely successful in promoting sales.

Sample size n =100, Sample mean \(\bar x\) = 390 pages, Population mean \(\mu\) = 400 pages
Population SD \(\sigma\) = 20 pages
The sample is a large sample and so we apply Z -test
Null Hypothesis:
There is no significant difference between the sample mean and the population mean of writing life of pen he manufactures, i.e., H₀ : \(\mu\) = 400
Alternative Hypothesis:
There is significant difference between the sample mean and the population mean of writing life of pen he manufactures, i.e., H₁:\(\mu\neq\) 400 (two tailed test)
The level of significance \(\alpha\) = 1% = 0.01
Applying the test statistic
\(Z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} \sim N(0,1) ; \)
\(Z=\frac{390-400}{\frac{20}{\sqrt{100}}}=\frac{-10}{2}=-5, \therefore|Z|=5\)
Thus the calculated value |Z| = 5 and the significant value or table value \({ Z }_{ \frac { \sigma }{ 2 } }=2.58\)
Comparing the calculated and table values, we found Z > \({ Z }_{ \frac { \sigma }{ 2 } }\) i.e., 5 > 2.58
Inference: Since the calculated value is greater than table value i.e., \(Z>{ Z }_{ \frac { \sigma }{ 2 } }\) at 1% level of significance, the null hypothesis is rejected and Therefore we concluded that \(\mu \neq400\) and the manufacturer’s claim is rejected at 1% level of significance.