Question

Independent random samples from two populations with standard deviations ${\mathbf{\text{σ}}}_{\mathbf{\text{1}}}{\mathbf{\text{=2andσ}}}_{\mathbf{\text{2}}}\mathbf{\text{=8}}$, respectively, are selected. The sample sizes and the sample means are recorded in the following table:

Sample 1	Sample 2
$\begin{array}{l}{\mathbf{\text{n}}}_{\mathbf{\text{1}}}\mathbf{\text{=58}}\\ {\overline{\mathbf{\text{x}}}}_{\mathbf{\text{1}}}\mathbf{\text{=17.5}}\end{array}$	$\begin{array}{l}{\mathbf{\text{n}}}_{\mathbf{\text{2}}}\mathbf{\text{=62}}\\ {\overline{\mathbf{\text{x}}}}_{\mathbf{\text{2}}}\mathbf{\text{=16.23}}\end{array}$

a. Calculate the standard error of the sampling distribution for Sample 1.

b. Calculate the standard error of the sampling distribution for Sample 2.

c. Suppose you were to calculate the difference between the sample means $\mathbf{(}{\mathbf{\text{x}}}_{\mathbf{\text{1}}}\mathbf{-}{\mathbf{\text{x}}}_{\mathbf{\text{2}}}\mathbf{)}$. Find the mean and standard error of the sampling distribution $\mathbf{(}{\mathbf{\text{x}}}_{\mathbf{\text{1}}}\mathbf{-}{\mathbf{\text{x}}}_{\mathbf{\text{2}}}\mathbf{)}$.

d. Will the statistic $\mathbf{(}{\mathbf{\text{x}}}_{\mathbf{\text{1}}}\mathbf{-}{\mathbf{\text{x}}}_{\mathbf{\text{2}}}\mathbf{)}$ be normally distributed?

Short answer

The sample means distribution is described as the collection of averages from all feasible randomly selected of a certain size (n) drawn from a particular population.

Step-by-step solution

Step-by-Step Solution Step 1: Central Limit Theorem.

According to theCentral Limit Theorem,the sampling distribution of the sample means approaches a normal distribution, irrespective of the shape of population distribution if the sample size is over 30.

A statistic's standard error is the estimated standard deviation of a statistics population sample.

(a) Find the standard error for Sample 1.

It is given that,${\overline{x}}_1=17.5$, ${\sigma }_1=2$, and $n_1=58$.

The standard error is localid="1652760610266" $\frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}$and localid="1652760613813" ${\mu }_1={\overline{x}}_1$

So,

${\mu }_1=17.5$

Standard error for Sample 1 =localid="1652760584440" $\frac{{\sigma }_1}{\sqrt{n_1}}$

$$\begin{gathered} =\frac{2}{\sqrt{58}} \\ =0.26 \end{gathered}$$

Therefore, the standard error of the sampling distribution for Sample 1 is $0.26$.

(b) Find the interval extending standard deviations x¯2 on each side μ2. 

It is given that, ${\overline{x}}_1=16.23$,${\sigma }_1=8$, and $n_1=62$.

The standard error is $\frac{\sigma }{\sqrt{n}}$and${\mu }_2={\overline{x}}_2$

So,

${\mu }_2=16.23$

The standard error for Sample 2 =role="math" localid="1652760921438" $\frac{{\sigma }_2}{\sqrt{n_2}}$

$\begin{array}{l}=\frac{8}{\sqrt{62}}\\ =1.02\end{array}$

Therefore, the standard error of the sampling distribution for Sample 2 is $1.02$.

(c) Find the mean as well as the standard error of the sampling distribution (x¯1− x¯2).

It is given that, ${\mu }_1=17.5$, ${\mu }_2=16.23$, ${\sigma }_1=2$, ${\sigma }_2=8$,$n_1=58$, and $n_2=62$

$\begin{array}{l}{\overline{x}}_{{}_1}-{\overline{x}}_{{}_2}={\mu }_1-{\mu }_2\\ =17.5-16.23\\ =1.27\end{array}$

role="math" localid="1652761176006" $\begin{array}{l}{\sigma }_{{}_{{\overline{x}}_{{}_1}-{\overline{x}}_{{}_2}}}=\sqrt{\frac{{{\sigma }_1}^2}{n_1}+\frac{{{\sigma }_2}^2}{n_2}}\\ =\sqrt{\frac{4}{58}+\frac{64}{62}}\\ =\sqrt{1.1012}\\ =1.05\end{array}$

Therefore, the mean and standard deviation of the sampling distribution$({\overline{x}}_{{}_1}-{\overline{x}}_{{}_2})$ are$1.27$ and $1.05$respectively.

(d) State the distribution of (x¯1− x¯2)

The difference in sample means will be normally distributed according to the Central Limit Theorem as $n>30$.