91Ó°ÊÓ

A type of statistical analysis in which the assumptions about a population parameter are tested is called Hypothesis Testing. It estimates the relation between 2 statistical variables.

It is given that, ${\stackrel{¯}{x}}_1=5275$, ${\mathrm{σ}}_1=150$, ${\stackrel{¯}{x}}_2=5240$, ${\mathrm{σ}}_2=200$and $n=400$ .

The critical value Z at $5\%$the level of significance is$1.96$.

A large sample $95\%$confidence interval $({\mathrm{μ}}_1−{\mathrm{μ}}_2)$is $({\stackrel{¯}{x}}_1−{\stackrel{¯}{x}}_2)Â\pm Z_{\frac{\mathrm{Î\pm }}{2}}\sqrt{\frac{{{\mathrm{σ}}_1}^2}{n}+\frac{{{\mathrm{σ}}_2}^2}{n}}$

$\begin{array}{l}=(5275−5240)Â\pm 1.96\sqrt{\frac{{\left(150\right)}^2}{400}+\frac{{\left(200\right)}^2}{400}}\\ =35Â\pm 1.96\sqrt{56.25+100}\\ =35Â\pm 1.96\left(12.5\right)\\ =35Â\pm 24.5\\ =(10.5,59.5)\end{array}$

Therefore, the difference between the population means at the confidence interval is $(10.5,59.5)$.

It is given that,

Null Hypothesis,${\text{H}}_{\text{0}}\text{:}\left({\text{μ}}_{\text{1}}{\text{-μ}}_{\text{2}}\right)\text{=0}$and

Alternate Hypothesis,${\text{H}}_{\text{a}}\text{:}\left({\text{μ}}_{\text{1}}{\text{-μ}}_{\text{2}}\right)≠\text{0}$

The level of significance is $5\%$.

$\begin{array}{l}Z=\frac{({\stackrel{¯}{x}}_1−{\stackrel{¯}{x}}_2)−({\mathrm{μ}}_1−{\mathrm{μ}}_2)}{\sqrt{\frac{{{\mathrm{σ}}_1}^2}{n}+\frac{{{\mathrm{σ}}_2}^2}{n}}}\\ =\frac{(5275−5240)−0}{\sqrt{\frac{{\left(150\right)}^2}{400}+\frac{{\left(200\right)}^2}{400}}}\\ =\frac{35}{12.5}\\ =2.80\end{array}$

So, the p-value is $.00511$.

As the p-value is less than the significance level, so the null hypothesis is rejected. Therefore, it can be concluded that there is a significant difference between the two means.

It is given that,

Null Hypothesis,${\text{H}}_{\text{0}}\text{:}({\text{μ}}_{\text{1}}−{\text{μ}}_{\text{2}})\text{=0}$and

Alternate Hypothesis, ${\text{H}}_{\text{a}}\text{:}({\text{μ}}_{\text{1}}−{\text{μ}}_{\text{2}})>\text{0}$

The level of significance is $5\%$

$\begin{array}{l}Z=\frac{({\stackrel{¯}{x}}_1−{\stackrel{¯}{x}}_2)−({\mathrm{μ}}_1−{\mathrm{μ}}_2)}{\sqrt{\frac{{{\mathrm{σ}}_1}^2}{n}+\frac{{{\mathrm{σ}}_2}^2}{n}}}\\ =\frac{(5275−5240)−0}{\sqrt{\frac{{\left(150\right)}^2}{400}+\frac{{\left(200\right)}^2}{400}}}\\ =\frac{35}{12.5}\\ =2.80\end{array}$

So, the value of$p$ is $.0026$.

As the value $p$is less than the significance level, so the null hypothesis is rejected. Therefore, it can be concluded that the mean of the first population is greater than the mean of the second population, $H_a:{\mathrm{μ}}_1>{\mathrm{μ}}_2$.

It is given that,

Null Hypothesis,${\text{H}}_{\text{0}}\text{:}\left({\text{μ}}_{\text{1}}{\text{-μ}}_{\text{2}}\right)\text{=25}$and

Alternate Hypothesis,${\text{H}}_{\text{a}}\text{:}\left({\text{μ}}_{\text{1}}{\text{-μ}}_{\text{2}}\right)≠\text{25}$

The level of significance is $5\%$.

$\begin{array}{l}Z=\frac{({\stackrel{¯}{x}}_1−{\stackrel{¯}{x}}_2)−({\mathrm{μ}}_1−{\mathrm{μ}}_2)}{\sqrt{\frac{{{\mathrm{σ}}_1}^2}{n}+\frac{{{\mathrm{σ}}_2}^2}{n}}}\\ =\frac{(5275−5240)−25}{\sqrt{\frac{{\left(150\right)}^2}{400}+\frac{{\left(200\right)}^2}{400}}}\\ =\frac{10}{12.5}\\ =0.8\end{array}$

So, the value of $p$is $.423711$.

As the value$p$ is more than the significance level, so the null hypothesis is accepted. Therefore, it can be concluded that there is no significant difference between the two means.

The following requirements must be met to make accurate large-sample inferences:

1. The two samples are randomly selected independently from the two target populations.
2. The sample sizes are large (more than 30). So, as per the Central Limit Theorem, the sampling distribution of the sample means will approach a normal distribution, irrespective of the shape of the population distribution.
3. Since the sample size is large, so Z-test will be used for analysis.