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.

Sample size n = 400 stores
Sample mean \(\bar x\) = 153.7 bars
Sample SD s = 17.2 bars
Population mean m = 146.3 bars
Since population SD is unknown we can consider the sample SD s = \(\sigma\)
Null Hypothesis :
The advertising campaign is not successful i.e, H₀: \(\mu\) = 146.3 
(There is no significant difference between the mean weekly sales of soap bars in department stores before and after advertising campaign)
Alternative Hypothesis H₁:
\(\mu\) >143.3 (Right tail test). The advertising campaign was successful
Level of significance \(\sigma\) = 0.05
Test statistic :
\(Z=\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \sim N(0,1)\)
\(Z=\frac { 153.7-146.3 }{ \frac { 17.2 }{ \sqrt { 400 } } } \)
\(=\frac { 7.4 }{ 0.86 } =8.605\)
\(\therefore\) Z = 8.605
Comparing the calculated value Z=8.605 and the significant value or table value \({ Z }_{ \alpha }=1.645.\) We get 8.605 > 1.645 
Inference:
Since, the calculated value is much greater than table value i.e., Z > \({ Z }_{ \alpha }\), it is highly significant at 5% level of significance.
Hence we reject the null hypothesis H₀ and conclude that the advertising campaign was definitely successful in promoting sales.

Sample size n =100, Sample mean \(\bar x\) = 390 pages, Population mean \(\mu\) = 400 pages
Population SD \(\sigma\) = 20 pages
The sample is a large sample and so we apply Z -test
Null Hypothesis:
There is no significant difference between the sample mean and the population mean of writing life of pen he manufactures, i.e., H₀ : \(\mu\) = 400
Alternative Hypothesis:
There is significant difference between the sample mean and the population mean of writing life of pen he manufactures, i.e., H₁:\(\mu\neq\) 400 (two tailed test)
The level of significance \(\alpha\) = 1% = 0.01
Applying the test statistic
\(Z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} \sim N(0,1) ; \)
\(Z=\frac{390-400}{\frac{20}{\sqrt{100}}}=\frac{-10}{2}=-5, \therefore|Z|=5\)
Thus the calculated value |Z| = 5 and the significant value or table value \({ Z }_{ \frac { \sigma }{ 2 } }=2.58\)
Comparing the calculated and table values, we found Z > \({ Z }_{ \frac { \sigma }{ 2 } }\) i.e., 5 > 2.58
Inference: Since the calculated value is greater than table value i.e., \(Z>{ Z }_{ \frac { \sigma }{ 2 } }\) at 1% level of significance, the null hypothesis is rejected and Therefore we concluded that \(\mu \neq400\) and the manufacturer’s claim is rejected at 1% level of significance.

Given: n = 169, \(\bar x\) =1350 hours, \(\sigma\) =100 hours, since the level of significance is (100-90)% =10% thus \(\alpha\) is 0.1, hence the significant value at 10% is \({ Z }_{ \frac { \alpha }{ 2 } }\)=1.645
\(S.E.=\frac { \sigma }{ \sqrt { n } } =\frac { 100 }{ \sqrt { 169 } } =7.69\)
Hence 90% confidence limits for the population mean are
\(\bar { x } -{ Z }_{ \frac { \sigma }{ 2 } }<\mu <\bar { x } +{ Z }_{ \frac { \sigma }{ 2 } }SE\)
\(\begin{gathered} 1350-(1.645 \times 7.69) \leq \mu \leq 1350+(1.645 \times 7.69) \end{gathered}\)
\(1337.35 \leq \mu \leq 1362.65\)
Hence the mean life time of light bulbs is expected to lie between the interval (1337.35, 1362.65).

Here φ is the mean length of the components in the population.
The formula for the confidence interval is
\(\bar{x}-Z_{\alpha / 2} \frac{\sigma}{\sqrt{n}}<\mu<\bar{x}+Z_{\alpha / 2} \frac{\sigma}{\sqrt{n}}\)
\({ Here } \ \sigma=1.6, Z_{\alpha / 2}=1.96, \bar{x}=90 \text { and } \mathrm{n}=64\)
Then \(S.E=\frac { \sigma }{ \sqrt { n } } =\frac { 1.6 }{ \sqrt { 64 } } =0.2\)
Therefore, 90 - (1.96 x 0.2)\(\le φ \le\) 90 + (1.96 x 0.2)
\(\text { i.e. } \ (89.61 \leq \varphi \leq 90.39)\)
This implies that the probability that the true value of the population mean length of the components will fall in this interval (89.61,90.39) at 95% . Hence we concluded that 95% confidence interval ensures acceptance of the component by the consumer.