Sampling distribution of a statistic is the frequency distribution which is formed with various values of a statistic computed from different samples of the same size drawn from the same population.
Standard Error:
The standard deviation of the sampling distribution of a statistic is known as its Standard Error.

Sample size n = 100,
Sample mean \(\\ \bar { X } =3.5\)
Population mean μ = 4
Population standard deviation σ = 3
Null Hypotheses: There is no significant difference in the mean. i.e., H_o : μ = 4
Alternative Hypotheses : There is Significant difference in the mean.
i.e., H₁ : μ ≠ H
The level of significance ∝ = 5% = 0.05
Applying the test statistic,\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
\(\Rightarrow Z=\frac { 3.5-4 }{ \frac { 3 }{ \sqrt { 100 } } } =\frac { -.5 }{ .3 } =-1.667\)
\(\Rightarrow |Z|=1.667\)
\({ Z }_{ \frac { \alpha }{ 2 } }=1.96\)
Here Z < \({ Z }_{ \frac { \alpha }{ 2 } }\)i.e., 1.667<1.96
Inference: Since Z<\({ Z }_{ \frac { \alpha }{ 2 } }\)at 5% level of significance, the null hypothesis H₀ is accepted. Hence there is no Significant difference in the mean.

1. It can be kept small in size without losing its accuracy.
2. It is easy to administer, if the population under study is sub-divided.
3. It reduces the time and expenses in dividing the strata into geographical divisions, since the government itself had divided the geographical areas.

Sample size n = 50
Sample S.D s = 6.3
Population S.D \(\sigma\) = 6
The standard error for sample S.D is given by 
\(S.E=\sqrt { \frac { { \sigma }^{ 2 } }{ 2n } } =\frac { 6 }{ \sqrt { 2(50) } } =\frac { 6 }{ \sqrt { 100 } } =0.6\)
Thus standard error for sample S.D = 0.6.

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

Any sample statistic which is used to estimate an unknown population parameter is called an estimator (i.e.,). an estimator is a sample statistic used to estimate a population parameter.

The standard deviation of the sampling distribution of a statistic is known as its Standard Error.

Any statistical measure computed from sample is known as statistic.

The group of individuals considered under study is called as population. It refers not only to people but to all items that have been chosen for the study.

Give n = 100, \(\bar x\) =162 cm, s = 8 cm is known in this problem
since σ is unknown , so we consider \(\hat{\sigma}\) = s and \(\varphi\) = 160 cm
\(S.E = \frac { \hat{\sigma} }{ \sqrt { n } } =\frac { s }{ \sqrt { n } } =\frac { 8 }{ \sqrt { 100 } } =0.8\)
Therefore the standard error for the average height of large group of students of 160 cm is 0.8.

Givens \(\sigma^2\) = 4 which implies \(\sigma\) = 2, n = 1 hour = 60 min, \(\bar { X } \) = 20/min
Standard Error \(=\frac { \sigma }{ \sqrt { n } } =\frac { 2 }{ \sqrt { 60 } } =0.2582\)

Sample size n = 1600
Sample mean \(\bar { X } \) = 99
Population mean μ = 100
Population standard deviation σ = 15
Null Hypotheses H_o : μ = 100
Alternative Hypotheses H₁ : μ ≠ 100
Level of significance α = 0.05
Applying the test statistic,
\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
\(\Rightarrow Z=\frac { 99-100 }{ \frac { 15 }{ \sqrt { 1600 } } } =\frac { -1 }{ \frac { 15 }{ 40 } } =-\frac { 40 }{ 15 } =-2.667\)
∴ |Z| = 2.667
Critical value at 5% level of significance is \({ Z }_{ \frac { \alpha }{ 2 } }=1.96\)
Here |Z|>\({ Z }_{ \frac { \alpha }{ 2 } }\) as 2.667>1.96
Inference: since |Z|>\({ Z }_{ \frac { \alpha }{ 2 } }\) at 5% level of sinificance, the null hypothesis is rejected. 
Hence, we can conclude that the sample has not been taken from the population with mean μ = 100.

Sample size n = 50 workers
Total wages  \(\sum { x } =2550\)
Sample mean \(\bar { x } =\frac { \text{total wages} }{ n } -\frac { \sum { x } }{ n } =\frac { 2550 }{ 50 } \) = 51units
Population mean \(\mu\) = 52 
Population variance \(\sigma ^{ 2 }=25\)
Population SD  \( \sigma =5\)
Under the null hypothesis H₀:\(\mu\) = 52
Against the alternative hypothesis H₁:\(\mu\) \(\neq\) 52 (Two tail)
Level of significance  \(\mu\) = 0.01
Test statistic \(Z=\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \sim N(0,1)\)
\(Z=\frac { 51-52 }{ \frac { 5 }{ \sqrt { 50 } } } =\frac { -1 }{ 0.7071 } =-1.4142\)
Since alternative hypothesis is of two tailed test we can take |Z| = 1.4142
Critical value at 1% level of significance is \(\\ { Z }_{ \frac { \alpha }{ 2 } }=2.58\)
Inference: Since the calculated value is less than table value i.e., Z<\({ Z }_{ \frac { \alpha }{ 2 } }\) at 1% level  of significance, the null hypothesis H₀ is accepted.
Therefore, we conclude that there is no significant difference between the sample mean and population mean \(\mu\) = 52 and SD \(\sigma\) = 5.Therefore μ = 49 is rejected.

Sample size n = 50 Sample mean \(\bar x\) = 10 km sample standard deviation s = 3.5 km 
Population mean \(\mu\) = 9.5km
Since population SD is unknown we consider \(\sigma\) = s
The sample is a large sample and so we apply Z-test
Null Hypothesis :
There is no significant difference between the sample average and the company’s claim, i.e., \(H_0 : \mu=9.5\)
Alternative Hypothesis :
There is significant difference between the sample average and the company’s claim, i.e., H₁ : \(\mu\)\(\neq \) 9.5 (two tailed test)
The level of significance \(\alpha\) = 5% = 0.05
Applying the test statistic
\(Z=\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \sim N(0,1);\ \)
\(Z=\frac { 10-9.5 }{ \frac { 3.5 }{ \sqrt { 50 } } } \sim N(0,1)=\frac { 0.5 }{ 0.495 } =1.01\)
Thus the calculated value 1.01 and the significant value or table value \({ Z }_{ \frac { \sigma }{ 2 } }=1.96\)
Comparing the calculated and table value ,Here Z<\({ Z }_{ \frac { \sigma }{ 2 } }\)i.e., 1.01<1.96.
Inference : Since the calculated value is less than table value i.e.,  Z < \({ Z }_{ \frac { \sigma }{ 2 } }\) at 5% level of sinificance, the null hypothesis H₀ is accepted. Hence we conclude that the company’s claim that the new car petrol consumption is 9.5 km per litre is acceptable.