91Ó°ÊÓ

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 $6$.

Population 1

${\stackrel{¯}{x}}_1=7.9$

$s_1=5.4$

$n_1=3$

Population 2

${\stackrel{¯}{x}}_2=7.1$

$s_2=4.6$

$n_2=6$

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

${\mathrm{μ}}_{{\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2}={\mathrm{μ}}_1-{\mathrm{μ}}_2$

$=7.9-7.1$

$=0.8$

Then the standard deviation is

${\mathrm{σ}}_{{\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2}=\sqrt{\frac{{\mathrm{σ}}_1^2}{n_1}+\frac{{\mathrm{σ}}_2^2}{n_2}}$

$⇒{\mathrm{σ}}_{{\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2}=\sqrt{\frac{5.4^2}{3}+\frac{4.6^2}{6}}$

$⇒{\mathrm{σ}}_{{\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2}=\sqrt{\frac{29.16}{3}+\frac{21.16}{6}}$

$⇒{\mathrm{σ}}_{{\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2}=3.64$.

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 $6$.

From the given information

Population 1

${\stackrel{¯}{x}}_1=7.9$

$s_1=5.4$

$n_1=3$

Population 2

${\stackrel{¯}{x}}_2=7.1$

$s_2=4.6$

$n_2=6$

From part (a)

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

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

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 $6$.

From the given data

The confidence interval is $-3to4$

Pooled standard deviation is

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

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

$⇒s_p=4.74$

Confidence interval $=\left({\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2\right)Â\pm t_{\mathrm{Î\pm }/2}·s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$

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

$⇒0.0Â\pm 3.5=0.8Â\pm t_{\mathrm{Î\pm }/2}×2.65$

$⇒t_{\mathrm{Î\pm }/2}=1.328$

Critical value

$t_{\mathrm{Î\pm }/2}=1.328$

$⇒t_{\mathrm{Î\pm }/2}=t_{0.10}$

$⇒\mathrm{Î\pm }=0.20$

$⇒1-\mathrm{Î\pm }=0.80$

Confidence level$=80\%.$