Question

To compare the means of two populations, independent random samples of 400 observations are selected from each population, with the following results:

Sample 1	Sample 2
$\begin{array}{l}{\overline{\mathbf{\text{x}}}}_{\mathbf{\text{1}}}\mathbf{\text{=5,275}}\\ {\mathbf{\text{σ}}}_{\mathbf{\text{1}}}\mathbf{\text{=150}}\end{array}$	$\begin{array}{l}{\overline{\mathbf{\text{x}}}}_{\mathbf{\text{2}}}\mathbf{\text{=5,240}}\\ {\mathbf{\text{σ}}}_{\mathbf{\text{2}}}\mathbf{\text{=200}}\end{array}$

a. Use a $95\%$confidence interval to estimate the difference between the population means $\mathbf{(}{\mathbf{\text{μ}}}_{\mathbf{\text{1}}}\mathbf{-}{\mathbf{\text{μ}}}_{\mathbf{\text{2}}}\mathbf{)}$. Interpret the confidence interval.

b. Test the null hypothesis ${\mathbf{\text{H}}}_{\mathbf{\text{0}}}\mathbf{\text{:}}\mathbf{(}{\mathbf{\text{μ}}}_{\mathbf{\text{1}}}\mathbf{-}{\mathbf{\text{μ}}}_{\mathbf{\text{2}}}\mathbf{)}\mathbf{\text{=0}}$versus the alternative hypothesis ${\mathbf{\text{H}}}_{\mathbf{\text{a}}}\mathbf{\text{:}}\mathbf{(}{\mathbf{\text{μ}}}_{\mathbf{\text{1}}}\mathbf{-}{\mathbf{\text{μ}}}_{\mathbf{\text{2}}}\mathbf{)}\mathbf{\ne }\mathbf{\text{0}}$ . Give the significance level of the test and interpret the result.

c. Suppose the test in part b was conducted with the alternative hypothesis ${\mathbf{\text{H}}}_{\mathbf{\text{a}}}\mathbf{\text{:}}\mathbf{(}{\mathbf{\text{μ}}}_{\mathbf{\text{1}}}\mathbf{-}{\mathbf{\text{μ}}}_{\mathbf{\text{2}}}\mathbf{)}\mathbf{\ne }\mathbf{\text{0}}$ . How would your answer to part b change?

d. Test the null hypothesis ${\mathbf{\text{H}}}_{\mathbf{\text{0}}}\mathbf{\text{:}}\mathbf{(}{\mathbf{\text{μ}}}_{\mathbf{\text{1}}}\mathbf{-}{\mathbf{\text{μ}}}_{\mathbf{\text{2}}}\mathbf{)}\mathbf{\text{=25}}$ versus ${\mathbf{\text{H}}}_{\mathbf{\text{a}}}\mathbf{\text{:}}\mathbf{(}{\mathbf{\text{μ}}}_{\mathbf{\text{1}}}\mathbf{-}{\mathbf{\text{μ}}}_{\mathbf{\text{2}}}\mathbf{)}\mathbf{\ne }\mathbf{\text{25}}$. Give the significance level and interpret the result. Compare your answer with the test conducted in part b.

e. What assumptions are necessary to ensure the validity of the inferential procedures applied in parts a–d?

Short answer

A hypothesis is a concept that has been proposed as a reasonable reason for a certain condition as well as situation, however, has never been proven right.

Step-by-step solution

Step-by-Step Solution Step 1: Explanation.

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.

(a) Find the confidence interval.

It is given that, ${\overline{x}}_1=5275$, ${\sigma }_1=150$, ${\overline{x}}_2=5240$, ${\sigma }_2=200$and $n=400$ .

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

A large sample $95\%$confidence interval $({\mu }_1-{\mu }_2)$is $({\overline{x}}_1-{\overline{x}}_2)\pm Z_{\frac{\alpha }{2}}\sqrt{\frac{{{\sigma }_1}^2}{n}+\frac{{{\sigma }_2}^2}{n}}$

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

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

(b) Test the null hypothesis H0:(μ1 - μ2)= 0 .

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

The level of significance is $5\%$.

$\begin{array}{l}Z=\frac{({\overline{x}}_1-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{\frac{{{\sigma }_1}^2}{n}+\frac{{{\sigma }_2}^2}{n}}}\\ =\frac{(5275-5240)-0}{\sqrt{\frac{{\left(150\right)}^2}{400}+\frac{{\left(200\right)}^2}{400}}}\\ =\frac{35}{12.5}\\ =2.80\end{array}$

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.

(c) Test the alternate hypothesis Ha:(μ1 − μ2)> 0

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 $5\%$

$\begin{array}{l}Z=\frac{({\overline{x}}_1-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{\frac{{{\sigma }_1}^2}{n}+\frac{{{\sigma }_2}^2}{n}}}\\ =\frac{(5275-5240)-0}{\sqrt{\frac{{\left(150\right)}^2}{400}+\frac{{\left(200\right)}^2}{400}}}\\ =\frac{35}{12.5}\\ =2.80\end{array}$

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, $H_a:{\mu }_1>{\mu }_2$.

(d) Test the null hypothesis H0:(μ1 - μ2)= 25

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

The level of significance is $5\%$.

$\begin{array}{l}Z=\frac{({\overline{x}}_1-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{\frac{{{\sigma }_1}^2}{n}+\frac{{{\sigma }_2}^2}{n}}}\\ =\frac{(5275-5240)-25}{\sqrt{\frac{{\left(150\right)}^2}{400}+\frac{{\left(200\right)}^2}{400}}}\\ =\frac{10}{12.5}\\ =0.8\end{array}$

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.

(e) State the necessary assumptions.

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.