91Ó°ÊÓ

The following table shows sample data for separate simple random sampling from two populations.

${\stackrel{¯}{x}}_1=20,s_1=4,n_1=20$;

${\stackrel{¯}{x}}_2=24,s_2=5,n_2=15$

The hypotheses test is left-tailed, with a significance level of $5\%$.

Population $1:{\stackrel{¯}{x}}_1=20,s_1=4,n_1=20$;

Population $2:{\stackrel{¯}{x}}_2=24,s_2=5,n_2=15$.

The main goal is to do a left-tail hypothesis test.

Define null and alternate hypotheses.

Null hypotheses:$H_0:{\mathrm{μ}}_1â‰\yen {\mathrm{μ}}_2$

Alternate hypotheses: $H_a:{\mathrm{μ}}_1<{\mathrm{μ}}_2$

Hypotheses is left-tailed.

We get significance level as $5\%$.

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

$⇒s_p=\sqrt{\frac{(20-1)(4{)}^2+(15-1\left)\right(5{)}^2}{20+15-2}}$

$⇒s_p=\sqrt{\frac{19\left(16\right)+14\left(25\right)}{33}}$

$⇒s_p=4.4518$

Test statistic,$t_0=\frac{{\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

$⇒t_0=\frac{20-24}{4.4518\sqrt{\frac{1}{20}+\frac{1}{15}}}$

$⇒t_0=-2.6306$

We need to find the critical values

Here, $$\begin{gathered} df=n_1+n_2-2 \\ =20+15-2 \\ =33 \end{gathered}$$

$⇒\mathrm{df}=33$

Using table IV when$df=33$

$$\begin{gathered} \text{Critical value,}-t_{\mathrm{Î\pm }}=-t_{0.05} \\ =-2.035 \end{gathered}$$

From above, $t_0=-2.6306$, i.e. the left-tailed hypotheses test statistic falls in the rejection zone. As a result, null hypotheses are rejected.

The following table shows sample data for separate simple random sampling from two populations.

${\stackrel{¯}{x}}_1=20,s_1=4,n_1=20$

${\stackrel{¯}{x}}_2=24,s_2=5,n_2=15$

The hypotheses test is left-tailed, with a significance level of$5\%$.

Population $1:{\stackrel{¯}{x}}_1=20,s_1=4,n_1=20$;

Population $2:{\stackrel{¯}{x}}_2=24,s_2=5,n_2=15$.

The main goal is to calculate a $90\%$confidence interval for the difference between two population means, ${\mathrm{μ}}_1$and ${\mathrm{μ}}_2$.

Define null and alternate hypotheses.

Null hypotheses:$H_0:{\mathrm{μ}}_1â‰\yen {\mathrm{μ}}_2$

Alternate hypotheses:$H_a:{\mathrm{μ}}_1<{\mathrm{μ}}_2$

Hypotheses is left-tailed.

Table IV may be used to find $t_{\mathrm{Î\pm }/2}$with $df=n_1+n_2-2$for a confidence level of $1-\mathrm{Î\pm }$.

Here, $$\begin{gathered} df=n_1+n_2-2 \\ =20+15-2 \\ =33 \end{gathered}$$

$⇒\mathrm{df}=33$

$\mathrm{Î\pm }=0.10$with a $90\%$confidence level.

Using table IV, when $df=33$

$$\begin{gathered} \text{Critical value,}{t}_{\mathrm{Î\pm }/2}=t_{0.10/2} \\ =t_{0.05} \\ =1.692\text{.} \end{gathered}$$

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

Confidence interval $$\begin{gathered} =(20-24)Â\pm 1.692\sqrt{\frac{1}{20}+\frac{1}{15}} \\ =-4Â\pm 0.578 \\ =-4.578to-3.422 \end{gathered}$$