Question

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

${\tilde{x}}_1=20,$$s_1=4,$$n_1=10,$${\tilde{x}}_2=18,$$s_2=5,$$n_2=15$.

a. Right-tailed test,localid="1651298373729" $\alpha =0.05$.

b. $90\%$confidence interval.

Short answer

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

b. As a result, the selected interval's end points are $(-1.104,5.104).$

Step-by-step solution

Part (a) Step 1: Given Information 

Right tailed test with$\alpha =0.05,{\overline{x}}_1=20,s_1=4,n_1=10,{\overline{x}}_2=18,s_2=5\text{and}{n}_2=15.$

Part (a) Step 2: Explanation 

The hypothesis test to be carried out is as follows:,

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

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

The test statistic's value is computed as follows:

localid="1651298393877" $$\begin{gathered} t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\left(\frac{s_1^2}{n_1}\right)+\left(\frac{s_2^2}{n_1}\right)}} \\ =\frac{20-18}{\sqrt{\left(\frac{4^2}{10}\right)+\left(\frac{5^2}{15}\right)}} \\ =1.11 \end{gathered}$$

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:

localid="1651298401992" $$\begin{gathered} df=\frac{{\left[\left(\frac{s_1^2}{n_1}\right)+\left(\frac{s_2^2}{n_2}\right)\right]}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{s_2^2}{n_2}\right)}^2}{n_2-1}} \\ =\frac{{\left[\left(\frac{4^2}{10}\right)+\left(\frac{5^2}{15}\right)\right]}^2}{\frac{{\left(\frac{4^2}{10}\right)}^2}{10-1}+\frac{{\left(\frac{5^2}{15}\right)}^2}{15-1}} \\ =22 \end{gathered}$$

For $df=22$, see the distribution table.

The obtained critical value is, $t_{0.05}=1.717$.

Part(a) Step 4: Explanation 

The graph is depicted in the diagram below.

Figure (1) shows that the test statistic$t=1.11$falls within the rejection zone. As a result, the results are significant at the $5\%$level.

The (localid="1651298411306" $P$-Value Approach) is a method of calculating the value of a variable.

localid="1651298420216" $df=∆$is the t statistic.

The probability of seeing a value of$t=1.11$is known as the localid="1651298431490" $P$value.

The value of $P$derived using technological means islocalid="1651298439366" $0.1402.$

Using the $t$table with $df=22$,refer to the figure.

Part (a) Step 5: Explanation 

Below is a graph of the data.

Figure (2), 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 $5\%$.

Part (b) Step 1: Given Information 

Right tailed test with$\alpha =0.05,{\overline{x}}_1=20,s_1=4,n_1=10,{\overline{x}}_2=18,s_2=5\text{and}{n}_2=15.$

Part (b) Step 2: Explanation 

Refer to the distribution table; the critical value found for $df=22$is,

$t_{0.05}=1.717$

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

Substitute,

localid="1651298462106" $$\begin{gathered} \left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\alpha /2}\times \sqrt{\left(s_1^2/n_1\right)+\left(s_2^2/n_2\right)}=(20-18)\pm 1.717\times \sqrt{\left(\frac{4^2}{10}\right)+\left(\frac{5^2}{15}\right)} \\ =(-1.104,5.104) \end{gathered}$$

As a result, the selected interval's end points are $(-1.104,5.104).$