Let x_i, (i = 1,2,3....n) is a ran dom sample of size n from a normal population with mean μ and variance σ², then the sample mean is distributed
normally with mean μ and variance \(\frac { { \sigma }^{ 2 } }{ n } \)
Thus for large samples, the standard normal variate corresponding to \(\bar { X } \) is:
\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \sim N(0,1)\)
Under the null hypothesis, that the sample has been drawn from a population with mean μ and variance σ², the test statistic is \(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)

Sample size n = 100
Sample mean \(\bar { X } \)= 72
Population mean μ = 76
Population standard deviation σ = 8
Null Hypotheses H₀:
μ = 76(i.e., There is no Significant difference between the state scores and the national scores)
\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
\(\Rightarrow Z=\frac { 72-76 }{ \frac { 8 }{ \sqrt { 100 } } } =\frac { -4 }{ \frac { 8 }{ 10 } } =\frac { -4 }{ 8 } =-5\)
\(\Rightarrow |Z|=5\)
\({ Z }_{ \frac { \alpha }{ 2 } }=1.96\)
\(Z>{ Z }_{ \frac { \alpha }{ 2 } }i.e.,\ 5>1.96\)
Inference : Since \(Z>{ Z }_{ \frac { \alpha }{ 2 } }\)  at 5% level of significance, the null hypothesis H₀ is rejected. 
Hence, we conclude that there is significant difference between the state scores and the national scores.

Given Sample size n = 400
Sample mean \(\\ \bar { X } =67.47\)inches
Population mean μ = 67.39 & σ = 1.30 inches)
Null Hypotheses H_o:
μ = 67.39 inches (i.e., the sample has been drawn from the population with μ = 67.39 & σ = 1.30 inches)
Alternative Hypotheses H₁:
μ ≠ 67.39 inches (two tail test)
(i.e., the sample has not been drawn from the population with μ = 67.39 & σ = 1.30 inches) The level of significance a = 5% = 0.05
Applying the test statistic,
\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
\(\Rightarrow Z=\frac { 67.47-67.39 }{ \frac { 1.30 }{ \sqrt { 400 } } } =\frac { 0.08 }{ 0.065 } =1.2308\)
\(\therefore |Z|=1.2308\)
The significant value \({ Z }_{ \frac { \alpha }{ 2 } }=1.96\)
Here \(Z={ Z }_{ \frac { \alpha }{ 2 } }i.e.,1.2308<1.96\)
Inference: Since \({ Z }_{ \frac { \alpha }{ 2 } }\)at 5% level of significance the null hypothesis H_o is accepted.
Hence, we conclude that the sample has been drawn from the population with mean height 67.39 inches and standard deviation 1.30 inches.

Sample size n = 400
Probability of vegetarian = \(\frac{230}{400}=0.575\)
∴ P = 0.575
Probability of non-vegetarian q = 1 - P = 1 - 0.575
⇒ q = 0.425
Standard error for sample proportion =\(\\ \\ \sqrt { \frac { pq }{ n } } \)
\(\sqrt { \frac { (0.575)(0.425) }{ 400 } } \)
\(=\sqrt { 0.000610 } \)
= 0.0246
S.E = 0.025(app)

Sample size n = 60
Population standard deviation σ = 3
The standard error for sample standard deviation
\(=\sqrt { \frac { { \sigma }^{ 2 } }{ 2n } } \)
\(=\sqrt { \frac { { 3 }^{ 2 } }{ 2(60) } } =\sqrt { \frac { 9 }{ 120 } } =\sqrt { 0.075 } \)
S.E.of sample standard deviation = 0.2739

In systematic sampling, randomly select the first sample from the first k units. Then every k^th member, starting with the first selected sample, is included in the sample.
Procedure for selection of samples by systematic sampling method.
(I) If we want to select a sample of 10 students from a class of 100 students, then \(k=\frac{N}{n}=\frac{100}{10}=10\)
Thus, sampling interval = 10 denotes that for every 10 samples one sample has to be selected.
(II) If the selected first random sample is 5, then the rest of the samples are automatically selected as 5, 15, 25, 35, 45, 55, 65, 75, 85, 95.[∵ k = 10]

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:

Stratum	Homogenous group	Percentage from population	Number of people in each strata	Random Samples
Category 1	Lower income class	39	\(\frac{39}{100}\times 500=195\)	\(195\times \frac { 68 }{ 500 } =26.5\sim 26\)
Category 2	Middle income class	38	\(\frac{38}{100}\times 500=190\)	\(190\times \frac { 68 }{ 500 } =26.5\sim 26\)
Category 3	Upper income class	23	\(\frac{23}{100}\times 500=115\)	\(115\times \frac { 68 }{ 500 } =15.6\sim 16\)
Total		100	500	68

In simple random sampling the samples are selected in such a way that each and every unit in the population has an equal and independent chance of being selected as a sample. It can be done with or without replacement of the samples selected. If the sampling is With replacement, there is a possibility of selecting the same sample any number of times. So, in simple random sampling without replacement is followed. Thus in simple random sampling from a population of N units, the probability is 1/ N the probability of drawing any unit in the second draw from among the available (N - 1) units is 1/(N - 1) and so on.
Simple random sampling without replacement is followed. The following two methods are generally used.
(A) Lottery method:
This is the most popular and simplest method when the population is finite. In this method, all the items of the population are numbered on separate slips of paper of same size, shape and colour. They are folded and placed in a container and shuffled thoroughly. Then the required numbers of slips are selected.
(B) Table of Random number:
The random number table has been so constructed that each of the digits 0,1,2, ... ,9 will appear approximately with the same frequency and independently of each other.
The various random number tables available are
a. L.H.C. Tippet random number series
b. Fisher and Yates random number series
c. Kendall and Smith random number series
d. Rand Corporation random number series.
Example: Tippett's table of random numbers is 20 items out of 6000.
Here we consider row wise selection of random numbers.

2952	6641	3992	9792	7969	5911	3170	5624
4167	9524	1545	1396	7203	5356	1300	2693
2670	7483	3408	2762	3563	1089	6913	7991
0560	5246	1112	6107	6008	8125	4233	8776
2754	9143	1405	9025	7002	6111	8816	6446

Given sample size n = 500
Probability of bad apples in the sample =\(\frac{65}{500}\)
∴ P=0.13
∴ Probability of good apples in the sample
⇒ q = 1 - P = 1 - 0.13 = 0.87
Standard error of proportion \(=\sqrt { \frac { pq }{ n } } \)
\(=\sqrt { \frac { (0.13)(0.87) }{ 500 } } =\sqrt { \frac { 0.1131 }{ 500 } } =\sqrt { 0.000262 } \)
∴ S.E=0.015

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.

Null hypothesis is the hypothesis which is tested for possible rejection under the assumption that it is true and it is denoted by H₀.
Example: If we want to find the population mean has a specific value μ_o, then the null hypothesis H_o is H_o : μ = μ₀

The interval within which the unknown value of parameter is, expected to lie is called confidence interval. It indicates the probability that the population parameter lies within a specilied range. If o is the population paramcter, then we choose a small value a, known as level of significance (1% or 5%) and determine 2 constants c, and c,such that p(c₁ < 0 < c₂/t) = 1- \(\alpha\). When t is the value of statistic. The quantities c₁ and c₂ are determined as confidence limits and the interval [c₁, c₂]  within which the unknown value of the population parameter is expected to lieis known as confidence interval.

The two branches of statistical inference are
(i) Estimation and
(ii) Testing of hypotheses.

1. Systematic samples are not random samples.
2. If N is not a multiple of n, then the sampling interval (k) cannot be an integer, thus sample selection becomes difficult.