Question

A variable of two population has a mean of $40$and standard deviation of $12$for one of the population and a mean of $40$and a standard deviation of $6$ for the other population.

a. For independent samples of sizes $9and4$respectively find the mean and standard deviation of $\overline{x_1}-\overline{x_2}$

b. Must the variable under consideration be normally distributed on each of the two population for you to answer part (a) ? Explain your answer.

b. Can you conclude that the variable $\overline{x_1}-\overline{x_2}$is normally distributed? Explain your answer.

Short answer

(a) ${\mu }_1-{\mu }_2=0$and ${\sigma }_{\overline{x_1}-\overline{x_2}}=5$

(b) No, we cannot conclude that the variable under consideration be normally distributed .

(c) No ,we cannot conclude that the variable is normally distributed.

Step-by-step solution

Part (a)  Step 1: Given Information

Given in the question that,

Population 1: Mean is $40$, standard deviation is $12$, and sample size is $9$

Population 2: Mean is $40$and standard deviation is $6$, and sample size is$4$

we have to calculate the mean and standard deviation of$\overline{x_1}-\overline{x_2}$

Part (a) Step 2: Explanation

The mean of $\overline{x_1}-\overline{x_2}$is:

$$\begin{gathered} {\mu }_1-{\mu }_2=40-40 \\ =0 \end{gathered}$$

standard deviation is:

${\sigma }_{{\overline{x}}_1-{\overline{x}}_2}=\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}$

$\Rightarrow {\sigma }_{{\overline{x}}_1-{\overline{x}}_2}=\sqrt{\frac{{12}^2}{9}+\frac{6^2}{4}}$

$\Rightarrow {\sigma }_{{\overline{x}}_1-{\overline{x}}_2}=5$

Hence, ${\mu }_1-{\mu }_2=0$and ${\sigma }_{\overline{x_1}-\overline{x_2}}=5$

Part (b)  Step 1: Given Information

Given in the question that,

Population 1: Mean is $40$, standard deviation is $12$, and sample size is $9$

Population 2: Mean is $40$and standard deviation is $6$, and sample size is $4$

we have To calculate If the variable under consideration must be normally distributed on each of the population given.

Part (b) Step 2: Explanation

From part (a) ${\mu }_1-{\mu }_2=0and{\sigma }_{\overline{x_1}-\overline{x_2}}=5$

Regardless of the distribution of variables on an individual population, the distribution of mean and standard deviation of $\overline{x_1}-\overline{x_2}$remains the same.

The standard deviation formula for$\overline{x_1}-\overline{x_2}$is also based on the assumption that the samples are independent.

As a result, the variable under examination does not need to be normally distributed across the entire population.

Part (c)  Step 1: Given Information

Given in the question that,

Population 1: Mean is $40$, standard deviation is $12$, and sample size is $9$

Population 2: Mean is $40$and standard deviation is $6$and sample size is $4$

we have To calculate If it can be concluded that the variable $\overline{x_1}-\overline{x_2}$is normally distributed.

Part (c) Step 2: Explanation

It is unclear whether the two populations are typically distributed based on the information provided. The sample sizes of $9$and $4$ are also quite small. As a result, the variable $\overline{x_1}-\overline{x_2}$cannot be considered to be regularly distributed.