Given sample size n = 64
Sample mean \(\bar { x } \) = 5950
Population mean μ = 6000
Population variance σ² = 32400
Population Standard deviation σ =\(\sqrt { 32400 } \) =180
Null hypothesis: H₀: population mean μ = 6000 Alternative hypothese : H₁: μ ≠ 6000
The test statistic, Z = \(\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
=\(\frac { 5950-6000 }{ \frac { 180 }{ \sqrt { 64 } } } =\frac { -50 }{ \frac { 180 }{ 8 } } \)
= -50\(\left( \frac { 8 }{ 180 } \right) \) = -2.2
|z| = 2.2
As the level of significance is α =0.001, \(Z_{ \frac { \alpha }{ 2 } }\) = 2.58 Here |z| < z_α
Inference: Null hypotheses H₀ is accepted.
Hence, we can conclude that the sample of 64 persons with a mean income of Rs.5950 belong to the population.

Given sample size n = 400
Sample mean \(\bar { x } \) = 171.38
Population mean μ = 171.17
Population Standard deviation σ = 3.3
Null hypotheses: H₀ : μ = 171.17
Alternative hypotheses: H₁ : μ ≠ 171.17
The test statistic, z = \(\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } =\frac { 171.38-171.17 }{ \frac { 3.3 }{ \sqrt { 400 } } } \)
=\(\frac { 0.21 }{ 0.165 } \) = 1.273
As the level of significance is α = 0.005, \(Z_{ \frac { \alpha }{ 2 } }\) = 1.96
Here z < \(Z_{ \frac { \alpha }{ 2 } }\) as 1.273 < 1.96
Inference: since z < \(Z_{ \frac { \alpha }{ 2 } }\), we accept the null hypotheses at 5% level of significance.
Hence, we can conclude that the sample of 400 has taken from the population with mean height of 171.17 cm.

Given population size N = 1500
Sample size n = 150
Sample proportion p =\(\frac { 10 }{ 150 } \)=0.07
∴ q = 1 - P = 1 - 0.07 = 0.93
Standard error of sample proportion =\(\sqrt { \frac { pq }{ n } } \)
=\(\sqrt { \frac { (0.07)(0.93) }{ 150 } } \)
S.E_(p) = 0.02 

Sample size n = 1000
Sample proportion of T.V. viewers
p=\(\frac { 320 }{ 1000 } \)=0.32
∴ q = 1 - P = 1 - 0.32 =0.68
Standard error = \(\sqrt { \frac { pq }{ n } } =\sqrt { \frac { (.32)(.68) }{ 1000 } } \)
S.E = 0.0147

Give sample size = 50
sample mean \(\bar { X } \) =67.9
S.E. =\(\sqrt { 0.7 } \)
95% confidence interval for population mean μ are \(\bar { X } -Z_{ \frac { \alpha }{ 2 } }(S.E)\le \bar { X } +Z_{ \frac { \alpha }{ 2 } }(S.E)\)
As the level of significance is α = 0.05, \(Z_{ \frac { \alpha }{ 2 } }\) = 1.96
∴ 67.9 - (1.96) \(\sqrt { 0.7 } \) ≤ μ ≤ 67.9 + (1.96) \(\sqrt { 0.7 } \)
⇒ 67.9 - 1.64 ≤ μ ≤  67.9 + 1.64
⇒ 66.2 ≤ μ ≤ 69.54
Thus, the 95% confidence intervals for estimating μ is given by (66.2,69.54).

When the hypothesis about the population parameter is rejected only for the value of sample statistic falling into one of the tails of the sampling distribution, then it is known as one tailed test.

A region corresponding to a test statistic in the sample space which tends to rejection of Ho is called critical region.

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 statistical constants of the population like mean (μ), variance (σ²) are referred as population parameters.

A selection of a group of individuals from a population in such a way that it represents the population is called as sample.

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

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 = 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.

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.

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)