91Ó°ÊÓ

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

${\stackrel{¯}{x}}_1=20,s_1=4,n_1=10;$

${\stackrel{¯}{x}}_2=18,s_2=5,n_2=15$

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

Population $1:{\stackrel{¯}{x}}_1=20,s_1=4,n_1=10$;

Population $2:{\stackrel{¯}{x}}_2=18,s_2=5,n_2=15$.

The most important goal is to perform a right-tailed hypothesis test.

First, state the null and alternate hypotheses.

Null hypotheses: $H_0:{\mathrm{μ}}_1≤{\mathrm{μ}}_2$

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

Hypotheses is right-tailed.

Secondly, Determine the level of relevance

The significance level is set at $5$percent, or $\mathrm{Î\pm }=0.05$.

Compute the value of test statistics

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

localid="1651216842049" $$\begin{gathered} ⇒s_p=\sqrt{\frac{(10-1)(4{)}^2+(15-1\left)\right(5{)}^2}{10+15-2}} \\ ⇒s_p=\sqrt{\frac{9\left(16\right)+14\left(25\right)}{23}} \\ ⇒s_p=4.6345 \end{gathered}$$

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

localid="1651216847575" $$\begin{gathered} ⇒t_0=\frac{20-18}{4.6345\sqrt{\frac{1}{10}+\frac{1}{15}}} \\ ⇒t_0=1.0571 \end{gathered}$$

Identify critical values.

Here,

localid="1651216857506" $$\begin{gathered} df=n_1+n_2-2 \\ =10+15-2 \\ =23 \end{gathered}$$

$⇒df=23$

When localid="1651155040608" $df=23$using table IV for important values.

localid="1651216866513" $$\begin{gathered} \text{Critical value,}{t}_{\mathrm{Î\pm }}=t_{0.05} \\ =1.714 \end{gathered}$$

From above,$t_0=1.057$i.e. the test statistic does not fall into the right-tailed hypotheses test rejection zone. As a result, null hypotheses are not ruled out.

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

${\stackrel{¯}{x}}_1=20,s_1=4,n_1=10$

${\stackrel{¯}{x}}_2=18,s_2=5,n_2=15$

The hypotheses test is right-tailed and the confidence level is$90\%$.

Population $1$: ${\stackrel{¯}{x}}_1=20,s_1=4,n_1=10$;

Population $2$: ${\stackrel{¯}{x}}_2=18,s_2=5,n_2=15$.

The main goal is to calculate a $90\%$confidence interval for the difference between two population means, $\mathrm{μ}\left[1\right]$and $\mathrm{μ}\left[2\right]$.

First, state the null and alternate hypotheses.

Null hypotheses: $H_0:{\mathrm{μ}}_1≤{\mathrm{μ}}_2$

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

Hypotheses is right-tailed.

For a confidence level of $1-\mathrm{Î\pm }$, use Table IV to find $t_{\mathrm{Î\pm }/2}$with localid="1651156291948" $df=n_1+n_2-2$

For $90\%$confidence level, $\mathrm{Î\pm }=0.10$.

localid="1651216881955" $$\begin{gathered} \text{Critical value,}{t}_{\mathrm{Î\pm }/2}=t_{0.10/2} \\ =t_{0.05} \\ =2.069 \end{gathered}$$

Find the confidence interval's endpoints.

localid="1651216886998" $\left({\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2\right)Â\pm t_{\mathrm{Î\pm }/2}×\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$localid="1651216894071" $$\begin{gathered} Confidenceinterval=(20-18)Â\pm 2.069\sqrt{\frac{1}{10}+\frac{1}{15}} \\ =2Â\pm 0.8446 \\ =2.8446to1.1554 \end{gathered}$$