Question

In each of Exercises 10.75-10.80, we have provided summary statistics for independent simple random samples from non populations. In each case, use the non pooled $t-$test and the non pooled $t-$interval procedure to conduct the required hypothesis test and obtain the specified confidence interval.

${\overline{x}}_1=15,s_1=2,n_1=15,{\overline{x}}_2=12,s_2=5\text{and}{n}_2=15.$

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

b. $95\%$confidence interval.

Short answer

a. As a result, the statistics give adequate evidence to infer that the two population means are equivalent at a significance level of $5\%$.

b. As a result, the selected interval's end points are$(0.079,5.921).$

Step-by-step solution

Part (a) Step 1: Given Information 

Two tailed test with$\alpha =0.05,{\overline{x}}_1=15,s_1=2,n_1=15,{\overline{x}}_2=12,s_2=5\text{and}{n}_2=15.$

Part (a) Step 2: Explanation 

The hypothesis test to be carried out is as follows:

$H_{\mathrm{a}}:{\mu }_1\ne {\mu }_2$

The test is carried out at a significance level of $\alpha =0.05$or $5\%$.

The test statistic's value is computed as follows:

$$\begin{gathered} t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\left(\frac{s_1^2}{n_1}\right)+\left(\frac{s_2^2}{n_1}\right)}} \\ =\frac{15-12}{\sqrt{\left(\frac{2^2}{15}\right)+\left(\frac{5^2}{15}\right)}} \\ =2.16 \end{gathered}$$

The two-tailed test's critical value is $\pm t_{\alpha /2}$with $df=\Delta$.

Part (a) Step 3: Explanation 

The significance of$df$is determined as follows:

$$\begin{gathered} df=\frac{{\left[\left(\frac{s_1^2}{n_1}\right)+\left(\frac{s_2^2}{n_2}\right)\right]}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{s_2^2}{n_2}\right)}^2}{n_2-1}} \\ =\frac{\left[\left(\frac{2^2}{15}\right)+{\left(\frac{5^2}{15}\right)}^2\right.}{\frac{{\left(\frac{2^2}{15}\right)}^2}{15-1}+\frac{{\left(\frac{5^2}{15}\right)}^2}{15-1}} \\ =18.36817 \\ =18 \end{gathered}$$

For$df=18$, see the distribution table.

The obtained critical value is,$\pm t_{0.025}=\pm 2.101$.

Part (a) Step 4: Explanation 

The given figure shows that the test statistic $t=2.16$falls within the rejection zone. As a result, the results are significant at the $5\%$level.

The ($P$-Value Approach) is a method of calculating the value of a variable.

$df=∆$is the t statistic.

The probability of seeing a value of $t=2.16$is known as the $P$value.

The value of $P$derived using technological means is $0.0447$.

Using the $t$table with $df=18$, refer to the figure.

The graph is depicted in the diagram below.

Part (a) Step 5: Explanation 

The given figure shows that the $P$value is greater than the predefined significance threshold of $0.05$. As a result, the findings are statistically significant at the $5\%$level.

As a result, the statistics give adequate evidence to infer that the two population means are equivalent at a significance level of $5\%$.

Below is a graph of the data.

Part (b) Step 1: Given Information 

Two tailed test with$\alpha =0.05,{\overline{x}}_1=15,s_1=2,n_1=15,{\overline{x}}_2=12,s_2=5\text{and}{n}_2=15\text{.}$

Part (b) Step 2: Explanation 

Refer to the distribution table; the critical value found for$df=18$is,

$\pm t_{0.025}=\pm 2.101$

The confidence interval's end point for${\mu }_1-{\mu }_2$is computed as follows:

Substitute,

$$\begin{gathered} \left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\alpha /2}\times \sqrt{\left(s_1^2/n_1\right)+\left(s_2^2/n_2\right)}=(15-12)\pm 2.101\times \sqrt{\left(\frac{2^2}{15}\right)+\left(\frac{5^2}{15}\right)} \\ =(0.079,5.921) \end{gathered}$$

As a result, the selected interval's end points are$(0.079,5.921)$.