Question

In this Exercise, we have provided summary statistics for independent simple random samples from two populations. In each case, use the pooled $t$-test and the pooled $t$-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval.

${\overline{x}}_1=20,s_1=4,n_1=10,{\overline{x}}_2=23,s_2=5,n_2=15$

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

b. $90\%$confidence interval

Short answer

(a) The presented data do not provide adequate evidence to reject null hypotheses at a significance level of $5\%$.

(b) One can be $90\%$confident that the difference between two populations' means is between $-3.8446$and $2.1554$.

Step-by-step solution

Part(a) Step 1: Given Information

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

${\overline{x}}_1=20,s_1=4,n_1=10$

${\overline{x}}_2=23,s_2=5,n_2=15$

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

Part(a) Step 2: Explanation

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

Population $2:{\overline{x}}_2=23,s_2=5,n_2=15$.

Firstly 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(a) Step 3: Calculation

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

$\Rightarrow s_p=\sqrt{\frac{(10-1)(4{)}^2+(15-1\left)\right(5{)}^2}{10+15-2}}$ $\Rightarrow s_p=\sqrt{\frac{9\left(16\right)+14\left(25\right)}{23}}$$\Rightarrow s_p=4.6345$

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{20-23}{4.6345\sqrt{\frac{1}{10}+\frac{1}{15}}}$$\Rightarrow t_0=-1.5856$

We have to determine the critical values

Here,$$\begin{gathered} df=n_1+n_2-2 \\ =10+15-2 \\ =23 \end{gathered}$$

$\Rightarrow \mathrm{df}=23$

Using table IV we get the critical values, When $df=23$

$$\begin{gathered} \text{Critical value,}-t_{\alpha }=-t_{0.05} \\ =-1.714 \end{gathered}$$

From above, $t_0=-1.5856$, indicating that the test statistic does not fall into the rejection zone of the left-tailed hypotheses test. As a result, null hypotheses are not ruled out.

Part(b) step 1: Given Information

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

${\overline{x}}_1=20,s_1=4,n_1=10$

${\overline{x}}_2=23,s_2=5,n_2=15$

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

Part(b) Step 2: Explanation

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

Population $2:{\overline{x}}_2=23,s_2=5,n_2=15$.

The main goal is to calculate a $90\%$confidence interval for the difference between two population means, $\mu 1$and $\mu 2$.

Firstly, 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 $1-\alpha$.

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

$$\begin{gathered} \text{df}=n_1+n_2-2 \\ =(10+15-2) \\ =23 \end{gathered}$$

When $df=23$use table IV for important values.

$$\begin{gathered} \text{Critical value,}{t}_{\alpha /2}=t_{0.10/2} \\ =t_{0.05} \\ =2.069 \end{gathered}$$

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

$$\begin{gathered} Confidenceinterval=(20-23)\pm 2.069\sqrt{\frac{1}{10}+\frac{1}{15}} \\ =-3\pm 0.8446 \\ =-3.8446to2.1554 \end{gathered}$$