Question

A variable of two populations has a mean of $7.9$and a standard deviation of $5.4$for one of the populations and a mean of $7.1$and a standard deviation of $4.6$for the other population. Moreover. the variable is normally distributed in each of the two populations.

a. For independent samples of sizes $3$and $6$, respectively, determine the mean and standard deviation of ${\overline{x}}_1-{\overline{x}}_2$.

b. Can you conclude that the variable ${\overline{x}}_1-{\overline{x}}_2$is normally distributed? Explain your answer.

c. Determine the percentage of all pairs of independent samples of sizes $4$and $16$, respectively, from the two populations with the property that the difference${\overline{x}}_1-{\overline{x}}_2$ between the simple means is between $-3$and $4$.

Short answer

(a) ${\mu }_{{\overline{x}}_1-{\overline{x}}_2}=0.8$

${\sigma }_{{\overline{x}}_1-{\overline{x}}_2}=3.64$

(b) The standard deviation of ${\overline{x}}_1-{\overline{x}}_2$the formula is also predicated on the assumption that the samples are independent. Because each individual population is normally distributed, ${\overline{x}}_1-{\overline{x}}_2$is also normally distributed, and because of the central limit theorem, this holds true for samples of vast sizes.

(c) Confidence level$=80\%.$

Step-by-step solution

Part (a) Step 1: Given information

Population 1: Mean is $7.9$, the standard deviation is $5.4$, and the sample size is $3$.

Population $2$: Mean is $7.1$and the standard deviation is $4.6$, and the sample size is $6$.

Part (a) Step 2: Explanation

Population 1

${\overline{x}}_1=7.9$

$s_1=5.4$

$n_1=3$

Population 2

${\overline{x}}_2=7.1$

$s_2=4.6$

$n_2=6$

The mean of ${\overline{x}}_1-{\overline{x}}_2$is

${\mu }_{{\overline{x}}_1-{\overline{x}}_2}={\mu }_1-{\mu }_2$

$=7.9-7.1$

$=0.8$

Then the standard deviation is

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

$\Rightarrow {\sigma }_{{\overline{x}}_1-{\overline{x}}_2}=\sqrt{\frac{5.4^2}{3}+\frac{4.6^2}{6}}$

$\Rightarrow {\sigma }_{{\overline{x}}_1-{\overline{x}}_2}=\sqrt{\frac{29.16}{3}+\frac{21.16}{6}}$

$\Rightarrow {\sigma }_{{\overline{x}}_1-{\overline{x}}_2}=3.64$.

Part (b) Step 1: Given information

Population $1$: Mean is $7.9$, the standard deviation is $5.4$, and the sample size is $3$.

Population $2$: Mean is $7.1$and the standard deviation is $4.6$, and the sample size is $6$.

Part (b) Step 2: Explanation

From the given information

Population 1

${\overline{x}}_1=7.9$

$s_1=5.4$

$n_1=3$

Population 2

${\overline{x}}_2=7.1$

$s_2=4.6$

$n_2=6$

From part (a)

Regardless of the distribution of variables in an individual population, the mean and standard deviation of ${\overline{x}}_1-{\overline{x}}_2$remain constant.

The standard deviation of ${\overline{x}}_1-{\overline{x}}_2$the formula is also predicated on the assumption that the samples are independent. Because each individual population is normally distributed, ${\overline{x}}_1-{\overline{x}}_2$is also normally distributed, and because of the central limit theorem, this holds true for samples of vast sizes.

Part (c) Step 1: Given information

Population $1$: Mean is $7.9$, the standard deviation is $5.4$, and the sample size is $3$.

Population $2$: Mean is $7.1$and the standard deviation is $4.6$, and the sample size is $6$.

Part (c) Step 2: Explanation

From the given data

The confidence interval is $-3to4$

Pooled standard deviation is

$s_p=\sqrt{\frac{\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2}{n_1+n_2-2}}$

$\Rightarrow s_p=\sqrt{\frac{(4-1)(5.4{)}^2+(16-1\left)\right(4.6{)}^2}{4+16-2}}$

$\Rightarrow s_p=4.74$

Confidence interval $=\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\alpha /2}·s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$

$-3to4=$$(7.9-7.1)\pm t_{a/2}\times 4.74\sqrt{\frac{1}{4}+\frac{1}{16}}$

$\Rightarrow 0.0\pm 3.5=0.8\pm t_{\alpha /2}\times 2.65$

$\Rightarrow t_{\alpha /2}=1.328$

Critical value

$t_{\alpha /2}=1.328$

$\Rightarrow t_{\alpha /2}=t_{0.10}$

$\Rightarrow \alpha =0.20$

$\Rightarrow 1-\alpha =0.80$

Confidence level$=80\%.$