n = 50, \(\bar x\) = 67.9
95% confidence limits for population mean
\(\bar{x} \pm Z_{c}(\mathrm{~S} . \mathrm{E} \bar{x})\)
\( =67.9 \pm(1.96) \sqrt{0.7} \)
\(=67.9 \pm 1.64\)
95% confidence intervals for estimating is given by (66.2, 69.54)

\(S . \mathrm{E}=\frac{s}{\sqrt{n}}\)
\(=\frac{24}{\sqrt{64}}\)
\(=\frac{24}{8}=3\)

\(\mathrm{p}=0.5,\ \theta=1-\mathrm{p}=1-0.5=0.5 \)
n = 100
\(S . E=\sqrt{\frac{p q}{n}}=\sqrt{\frac{(0.5)(0.5)}{100}}\)
\(=\frac{0.5}{10}=0.05\)

A good estimator must possess the following characteristic
(i) unbiasedness
(ii) Consistency
(ii) Efficiency
(iv) Sufficiency

The method of obtaining the most likely value of the population parameter using statistic is called estimation.

The variance random number tables available are
(a) L.H.C Tippett random number series 
(b) Fisher and Yates random number series
(c) Kendall and smith random number series
(d) Rand corporation random number series.

(i) This requires a complete list of the population but such upto date lists are not available in many enquiries.
(ii) If the size of the sample is small, then it will not be a representative of the population.

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).