In each of Exercises 10.75-10.80, we have provided summary statistics for independent simple random samples from two 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,$$s_1=2,$$n_1=30,$$x_2=18,$$s_2=5,$$n_2=40$.

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

b. $90\%$confidence interval.

Using the non-pooled test and the non-pooled interval technique, find the specified confidence interval.

The hypothesis test to be carried out is as follows:

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

The test is run using a significance level of localid="1651414790125" $a=0.05$, which is alocalid="1651414793966" $5\%$significance level. The test statistic's value is calculated as follows:

$t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\left(\frac{s_1^2}{n_1}\right)+\left(\frac{s_2^2}{n_1}\right)}}$

$=\frac{20-18}{\sqrt{\left(\frac{2^2}{30}\right)+\left(\frac{5^2}{40}\right)}}$

$=2.297$T

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

The value of $df$is determined as follows:

$df=\frac{{\left[\left(\frac{s_1^2}{n_1}\right)+\left(\frac{s_2^2}{n_2}\right)\right]}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{{\left(\frac{s_2^2}{n_2}\right)}^2}+\frac{\left[n_2-1\right.}{n_1-1}}$

$=\frac{{\left[\left(\frac{2^2}{30}\right)+\left(\frac{5^2}{40}\right)\right]}^2}{\frac{{\left(\frac{2^2}{30}\right)}^2}{30-1}+\frac{{\left(\frac{5^2}{40}\right)}^2}{40-1}}$

$=54$

For localid="1651414806427" $df=54$see the distribution table.

,$t_{0.05}=1.674$is the critical value obtained.

The graph is depicted in the diagram below.

The value of the test statistic $t=2.297$appears to be in the rejection region in figure 1.

At the $5\%$level, the results are significant.

The $t$statistic has$df=\Delta$.

The probability of seeing a value localid="1651414828078" $t=2.297$is represented by the localid="1651414836717" $P$value.

The value of localid="1651414845409" $P$obtained using technological means is localid="1651414851473" $0.0128$.

Use the localid="1651414857870" $t$table with localid="1651414865291" $df=54$to refer to the figure.

Below is a graph of the data.

Figure 2 shows that the p value does not exceed the $0.05$significance level.

Using the non-pooled test and the non-pooled interval technique, find the specified confidence interval.

For $df=54$see the distribution table.

$t_{0.05}=1.674$is the crucial value obtained.

The confidence interval's end point for ${\mu }_1-{\mu }_2$is calculated as,

Simplify,

$\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\alpha /2}\times \sqrt{\left(s_1^2/n_1\right)+\left(s_2^2/n_2\right)}=(20-18)\pm 1.674\times \sqrt{\left(\frac{2^2}{30}\right)+\left(\frac{5^2}{40}\right)}$

$=(0.543,3.458)$