p = population proportion is 0.6
P = sample proportion \(=\frac{55}{150}=0.367\)
q = 1 - p = 1 - 0.367
= 0.633
n = 150
Null Hypothesis :
\(\mathrm{H}_{0}: \mathrm{P}=0.6\)
Alternative Hypothcsis :
\(\mathrm{H}_{1}: P \neq 0.6\)
Test statistic :
\( Z =\frac{p-p}{\sqrt{\frac{p q}{n}}}=\frac{0.367-0.6}{\sqrt{\frac{0.367}{\left.\frac{p}{150}\right)}}} \)
\(=\frac{-0.633}{0.03935}=-5.92 \)
\(|Z| =5.92\)
Level of significance:  \(\alpha=0.01 \text { (or) } 1 \%\)
\(|Z|=5.92>2.58\)
Hence null hypothesis is rejected

\(\mathrm{n}=1600, \bar{x}=99, \mu=100, \sigma=15\)
Null Hypothesis :
\(\mathrm{H}_{0}: \mu=100\)
Alternative Hypothesis :
\(\mathrm{H}_{1}: \mu \neq 100\)
Test statistic :
\( Z =\frac{\bar{x}-\mu}{\sigma / \sqrt{n}} \sim N(0,1) \)
\( =\frac{99-100}{\frac{15}{\sqrt{1600}}}=-2.67 \)
Level of significance = \(\alpha\) = 0.05
\(|Z|=2.67>1.96\)
Reject H_o

\(\mathrm{n}=400,\ \bar{x}=171.38, \ \sigma=3.3\)
Null Hypothesis :
\(\mathrm{H}_{0}: \mu=171.17\)
Alternative Hypothesis :
\(\mathrm{H}_{1}: \mu \neq 171.17\)
Test statistic :
\( \mathrm{Z} =\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} \sim N(0,1) \)
\(=\frac{171.38-171.17}{\frac{3.3}{\sqrt{400}}}=1.273 \)
Since |z| = 1.273 < 1.96 we accept null hypothesis at 5% level of significance. Thus the sample of 400 has come from the population with height of 171.17 cms

\( \mu=6000, \ \sigma^{2}=32400, \ \sigma=180 \bar{x}=5950, \mathrm{n}=64 \)
Null Hypothesis :
H_o : \(\mu\) = 6000
Alternative Hypothesis :
\(\mathrm{H}_{1}: \mu \neq 6000\)
Test statistic :
\( Z =\frac{\bar{x}-\mu}{\sigma / \sqrt{n}} \sim N(0,1) \)
\( =\frac{5950-6000}{\frac{180}{\sqrt{64}}} \)
\( =-2.2 \)
\(|Z| =2.2 \)
(a) Level of significance
\( \alpha=0.05 \text { (or) } 5 \% \)
\( |Z|=2.2>1.96 \)
Since calculated value of z is greater then the tabulated value, the null hypothesis is rejected.
(b) Level of significance
\( \alpha=0.01 \text { (or) } 1 \% \)
\( |Z|=2.2<2.58 \)
Since calculated value of Z is lesser then the tabulated value, the null hypothesis is  accepted.

\(\mathrm{n}=64, \bar{x}=51250, \sigma=2000\)
Null Hypothesis :
\(\mathrm{H}_{0}: \mu=50000\)
Alternative Hypothesis :
\(\mathrm{H}_{1}: \mu \neq 50000\)
Level of significance: \(\alpha\) = 0.05 (or) 5%
Test statistic :
\( Z =\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} \sim \mathrm{N}(0,1) \)
\( =\frac{51250-50000}{\frac{2000}{\sqrt{64}}}=5 \)
\(|Z| =5>1.96 \)
Since calculated value of Z is greater than 1.96, it is highly significant.
H_o is rejected
The sample mean differs significantly from the population mean.

Null Hypothesis : H_o : \(\mu=900\)
Alternative Hypothesis : \(\mathrm{H}_{1}: \mu \neq 900\)
\(\bar{x}=825, \mu=900, \sigma=110, \mathrm{n}=5\)
Test statistic :
\(Z=\frac{\bar{x}-\mu}{\sigma / \sqrt{n}}\)
where z is standard normal variable
\(Z=\frac{825-900}{\frac{110}{\sqrt{50}}}=-4.82\)
\(|Z|=4.82>1.96\)
Level of significance \(\alpha\) = 0.05 (or) 5%
Since calculated value of z is greater then the tabulated value we reject H_o. Hence we conclude that the mean life time of the population of electric bulbs cannot be taken as 900 hours.

\(\mathrm{p}=\frac{55}{100}, \mathrm{q}=\frac{45}{100}, \mathrm{n}=100\)
(a) 95% confidence limits for proportion is 2 given by
\( =p \pm z_{c} \sqrt{\frac{p q}{n}} \)
\(=0.55 \pm 1.96 \sqrt{\frac{0.55 \times 0.45}{100}} \)
\(=0.55 \pm 1.96(0.0497) \)
\(=0.55 \pm 0.97 \)
\(=(0.45,0.65) \)
The limits are between 45% and 65%
(b) 99% confidence limits for proportion is
\( =p \pm z_{c} \sqrt{\frac{p q}{n}} \)
\( =0.55 \pm 2.58 \sqrt{\frac{0.55 \times 0.45}{100}} \)
\(=0.55 \pm 2.58(0.0497) \)
\( =0.55 \pm 0.128 \)
\( =(0.42,0.68) \)
The limits lie between 42% and 68%

\(\mathrm{n}=200, \bar{x}=0.824, \mathrm{~s}=0.042\)
(a) 95% confidence limits for \(\mu\) are given by
\( =\bar{x} \pm z_{\epsilon} \frac{s}{\sqrt{n}} \)
\( =0.824 \pm(1.96) \frac{(0.042)}{\sqrt{200}} \)
\( =0.824 \pm 0.00582 \)
\( =(0.818,0.83) \)
(b) 99% confidence limits for \(\mu\) are given by
\( =\bar{x} \pm z_{c} \frac{s}{\sqrt{n}} \)
\( =0.824 \pm(2.58) \frac{(0.042)}{\sqrt{200}} \)
\( =(0.816,0.832) \)

n = 10000
Sample proportion = p =  \(\frac{560}{1000}=0.56\)
Null Hypothesis :
\(\mathrm{H}_{0}: \mathrm{P}=0.5\) (population proportion)
Alternative Hypothesis :  \(\mathrm{H}_{1}: P \neq 0.5\)
Level of significance : \(\alpha\) = 1%
Test statistic :
\( Z =\frac{p-P}{\sqrt{\frac{P Q}{n}}} \sim N(0,1) \)
\(Q =1-p=1-0.5=0.5 \)
\( =\frac{0.56-0.50}{\sqrt{\frac{(0.5)(0.5)}{1000}}}=3.79\)
Since H, is two sided alt othesis, the critical value at 1% level of significance is 2.58
\(|Z|=3.79>2.58\)
Hence H_o is rejected.
Therefore thus is a significant evidence to conclude that the preference of tea and coffee are different.

\(\mathrm{n}=100, \bar{x}=1950, \sigma=150\)
Null Hypothesis :
\(\mathrm{H}_{0}: \mu=2000\)
Alternative Hypothesis :
\(\mathrm{H}_{1}: \mu \neq 2000\)
Level of significance = \(\alpha\) = 5%
Test statistic :
\( Z =\frac{\bar{x}-\mu}{\sigma / \sqrt{n}} \sim N(0,1) \)
\(=\frac{1950-2000}{\frac{150}{\sqrt{100}}}=-3.33 \)
\(|Z| =3.33 \)
Since H, is a two sided alternative the critical values at \(\alpha\) = 0.05 is 1.96
\(|Z|=3.33>1.96\)
Thus the mean life spar of the LED bulbs is significantly different from 2000 hours.

\(\mathrm{n}=100, \bar{x}=30, \mathrm{~S}=3\)
Null Hypothesis:
\(\mathrm{H}_{0}: \mu=27\)
Alternative Hypothesis: 
\(\mathrm{H}_{1}: \mu>27\)
Level of significance :
\(\alpha\) = 5%
= 0.05
Test statistic :
\(Z=\frac{\bar{x}-\mu}{s / \sqrt{n}}\)
\( \frac{30-27}{3} \)
\(= \frac{3}{\sqrt{100}} \sim N(0,1) \)
= 10
Since H₁ is one sided (right) alternative hypothesis, the critical value at \(\alpha\) = 0.05 is 1.645
\(|Z|=10>1.645\)
Hence H_o is rejected.

\(\mathrm{n}=60, \bar{x}=1960, \sigma=120\)
Null Hypothesis :
\(\mathrm{H}_{0}: \mu=1900^{\circ}\)
Alternate Hypothesis :
\(\mathrm{H}_{1}: \mu>1900\)
Level of significance : 
\(\alpha=1 \%\)
Test statistic :
\( \mathrm{Z} =\frac{\bar{x}-\mu}{\sigma / \sqrt{n}} \sim N(0,1) \)
\(=\frac{1960-1900}{\frac{120}{\sqrt{60}}}=3.87 \)
\(|Z| =3.87 \)
Since H₁ is one sided (right) alternative hypothesis the critical value at \(\alpha \) = 0.01 level of significance is 2.33
\(|Z|=3.87>2.33\)
Hence Null hypothesis is rejected. Therefore, the mean breaking strength of the cables is significantly more than 1900\(\frac{n}{m^{2}}\). Hence the manufacture's claim that the breaking strength of the cables has increased by the new technique is found valid.

Given p = \(\frac { 55 }{ 100 } \)
∴ q = \(\frac { 45 }{ 100 } \) and n = 100
Standard error = \(\sqrt { \frac { pq }{ n } } =\sqrt { \frac { \frac { 55 }{ 100 } \times \frac { 45 }{ 100 } }{ 100 } } \)
= 0.0497
(a) As the level of significance α = 0.05 \(Z_{ \frac { \alpha }{ 2 } }\) = 1.96
∴ 95% confidence limits for proportion is given by \(p-Z_{ \frac { \alpha }{ 2 } }(S.E)\le p\ge p+Z_{ \frac { \alpha }{ 2 } }(S.E)\)
⇒ 0.55 - (1.96) (0.0497) ≤ p ≤ 0.55 + (1.96) (0.0497)
⇒ 0.453 ≤ p ≤ 0.647
∴ 95% confidence interval for proportion is (0.45, 0.65)
(b) As the level of significance is α = 0.01, \(Z_{ \frac { \alpha }{ 2 } }\) = 2.58
∴ 99% confidence limits for proportion is given by  \(p-Z_{ \frac { \alpha }{ 2 } }(S.E)\le p\ge p+Z_{ \frac { \alpha }{ 2 } }(S.E)\)
⇒ 0.55 - (2.58) (0.0497) ≤ p ≤ 0.55 + (2.58) (0.0497)
⇒ 0.422 ≤ p ≤ 0.678
Hence, 99% confidence interval for proportion is (0.42, 0.68).

Given sample size n = 200
Sample mean \(\bar { x } \) = 0.824
Sample S.D. s = 0.042
Standard error = \(\frac { s }{ \sqrt { n } } =\frac { 0.042 }{ \sqrt { 200 } } \)
= \(\frac { 0.042 }{ 14.14 } \) = 0.00270
(a) As the level of significance is α = 0.05, \(Z_{ \frac { \alpha }{ 2 } }\) =  1.96
∴ 95% confidence limits for μ are given by \(\bar { x } -Z_{ \frac { \alpha }{ 2 } }(S.E)\le \mu \le \bar { x } +Z_{ \frac { \alpha }{ 2 } }(S.E)\)
⇒ 0.824 - (1.96) (0.00270) ≤ μ ≤ 0.824 + (1.96) (0.00270)
⇒ 0.824 - 0.00582 ≤ μ ≤ 0.824 + 0.00582
⇒ 0.818 ≤ μ ≤ 0.832
Hence, the 95% confidence limits for μ is (0.818,0.832)
(b) As the level of significance is α =0.001, \(Z_{ \frac { \alpha }{ 2 } }\) = 2.58
∴ 99% confidence limits for μ are given by \(\bar { x } -Z_{ \frac { \alpha }{ 2 } }(S.E)\le \mu \le \bar { x } +Z_{ \frac { \alpha }{ 2 } }(S.E)\)
⇒ 0.824 - (2.58) (0.00270) ≤ μ ≤ 0.824 + (2.58) (0.00270)
⇒ 0.824 - 0.00582 ≤ μ ≤ 0.824 + 0.005825
⇒ 0.816 ≤ μ ≤ 0.832
Hence, the 99% confidence limits for μ is (0.816, 0.832)