Question

In this Exercise, we have provided summary statistics for independent simple random samples from two populations. In each case, use the pooled $t$-lest and the pooled$t$-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval.
${\overline{x}}_1=20,s_1=4,n_1=20,{\overline{x}}_2=24,s_2=5,n_2=15$

a. Left-tailed test, $\alpha =0.05$

b. $90\%$confidence interval

Short answer

(a) The presented data provide adequate evidence to reject null hypotheses at a significance level of $5\%$.

(b) The difference between the means of two populations is somewhere between $-4.578$and $-3.422$, and one may be $90\%$confident about it.

Step-by-step solution

Part(a) Step 1: Given Information

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

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

${\overline{x}}_2=24,s_2=5,n_2=15$

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

Part(a) Step 2: Explanation

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

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

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

Define null and alternate hypotheses.

Null hypotheses:$H_0:{\mu }_1\ge {\mu }_2$

Alternate hypotheses: $H_a:{\mu }_1<{\mu }_2$

Hypotheses is left-tailed.

We get significance level as $5\%$.

Part(a) Step 3: Calculation

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

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

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

$\Rightarrow s_p=4.4518$

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

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

$\Rightarrow t_0=-2.6306$

We need to find the critical values

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

$\Rightarrow \mathrm{df}=33$

Using table IV when$df=33$

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

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.

Part(b) Step 1: Given Information

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

${\overline{x}}_1=20,s_1=4,n_1=20$

${\overline{x}}_2=24,s_2=5,n_2=15$

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

Part(b) Step 2: Explanation

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

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

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

Define null and alternate hypotheses.

Null hypotheses:$H_0:{\mu }_1\ge {\mu }_2$

Alternate hypotheses:$H_a:{\mu }_1<{\mu }_2$

Hypotheses is left-tailed.

Part(b) Step 3: Calculation

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

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

$\Rightarrow \mathrm{df}=33$

$\alpha =0.10$with a $90\%$confidence level.

Using table IV, when $df=33$

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

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

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