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).