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.
${\overline{x}}_1=20,s_1=4,n_1=30,{\overline{x}}_2=18,s_2=5,n_2=40$

a. Right-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 $2.403$and $1.597$, according to $90\%$confidence.

Step-by-step solution

Part(a) Step 1: Given Information

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

${\overline{x}}_1=20,s_1=4,n_1=30$;

${\overline{x}}_2=18,s_2=5,n_2=40$

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

Part(a) Step 2: Explanation

Population $1:{\overline{x}}_1=20,s_1=4,n_1=30$

Population $2:{\overline{x}}_1=18,s_1=5,n_1=40$

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

Define null and alternate hypotheses.

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

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

Hypotheses is right-tailed.

Part(a) Step 3: Calculation

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

$\Rightarrow s_p=\sqrt{\frac{(30-1)(4{)}^2+(40-1\left)\right(5{)}^2}{30+40-2}}$

$\Rightarrow s_p=\sqrt{\frac{29\left(16\right)+39\left(25\right)}{68}}$

$\Rightarrow s_p=4.6002$

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

$\Rightarrow t_0=\frac{20-18}{4.6002\sqrt{\frac{1}{30}+\frac{1}{40}}}$

$\Rightarrow t_0=1.800$

We decide on critical values

Here, $$\begin{gathered} \text{df}=n_1+n_2-2 \\ =30+40-2 \\ =68 \end{gathered}$$

$\Rightarrow \mathrm{df}=68$

Using table IV, when localid="1651300372581" $df=68$

localid="1651300377806" role="math" $$\begin{gathered} \text{Critical value,}{t}_{\alpha }=t_{0.05} \\ =1.668 \end{gathered}$$

From above, localid="1651300386229" $t_0=1.800$. i.e. the test statistic is in the right-tailed hypotheses test rejection zone. As a result, null hypotheses are ruled out.

Part(b) Step 1: Given Information

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

${\overline{x}}_1=20,s_1=4,n_1=30$

${\overline{x}}_2=18,s_2=5,n_2=40$

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

Part(b) Step 2: Explanation

Population $1:{\overline{x}}_1=20,s_1=4,n_1=30$

Population $2:{\overline{x}}_2=18,s_2=5,n_2=40$

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:$H_0:{\mu }_1\le {\mu }_2$

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

Hypotheses is right-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 $1-\alpha$

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

Using table IV, when $df=68$

$$\begin{gathered} \text{Critical value,}{t}_{\alpha /2}=t_{0.10/2} \\ =t_{0.05} \\ =1.668\text{.} \end{gathered}$$

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

Confidence interval $$\begin{gathered} =(20-18)\pm 1.668\sqrt{\frac{1}{30}+\frac{1}{40}} \\ =2\pm 0.4028 \\ =2.403\text{to}1.597 \end{gathered}$$