Sample size n = 50,
Sample mean \(\bar { X } \) = 1850
Population mean μ = 1800
Population standard deviation σ = 100
Null Hypotheses H₀:
μ = 1800 (i.e., To claim that the breaking strength of the cables have increased)
Alternative Hypotheses H₁:
μ≠1800(To claim that the breaking strength of the cables have not increased)
The level of significance ∝ = 1% =.001
Applying the test statistic,
\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
\(\Rightarrow Z=\frac { 1850-1800 }{ \frac { 100 }{ \sqrt { 50 } } } =\frac { 50 }{ \frac { 100 }{ 7.07 } } =\frac { 50 }{ 14.144 } =3.535\)
\(\Rightarrow \therefore Z=3.535\)
The Significant value \({ Z }_{ \frac { \alpha }{ 2 } }=2.58\)
Here \(Z<{ Z }_{ \frac { \alpha }{ 2 } }i.e.,3.535<2.58\)
Inference: Since \(Z<{ Z }_{ \frac { \alpha }{ 2 } }\) at 1% level of significance, the null hypothesis Ho is rejected.
Hence, we conclude that μ≠ 1800 and we cannot support the claim that the breaking strength of the cables have increased.

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 = 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

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:

Sample size n = 50
Sample mean \(\bar { x } =9.3\) minutes
Sample S.D s = 1.6 minutes
Population mean  μ = 8.9 minutes
Null hypothesis H₀:  μ = 8.9
Alternative hypothesis H₁:  μ = 8.9 (two tail)
Level of significance  μ = 0.05 
Test statistic  \(Z=\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \sim N(0,1)\)
\(\\ Z=\frac { 9.3-8.9 }{ \frac { 1.6 }{ \sqrt { 50 } } } =\frac { 0.4 }{ 0.2263 } =1.7676\)
Calculated value Z = 1.7676
Critical value at 5% level of significance is \({ Z }_{ \frac { \alpha }{ 2 } }=1.96\)
Inference: Since the calculated value is less than table value i.e., \(Z<{ Z }_{ \frac { \alpha }{ 2 } }\) at 5% level of significance, the null hypothesis is accepted.
Therefore we conclude that an ambulance service claims on the average 8.9 minutes to reach its destination in emergency calls.