91Ó°ÊÓ

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:

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

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:

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

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

$=2.297$T

The right-tailed test's critical value is $t_{\mathrm{Î\pm }}$with$df=\mathrm{Δ}$.

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$df=\mathrm{Δ}$.

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 ${\mathrm{μ}}_1-{\mathrm{μ}}_2$is calculated as,

Simplify,

$\left({\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2\right)Â\pm t_{\mathrm{Î\pm }/2}×\sqrt{\left(s_1^2/n_1\right)+\left(s_2^2/n_2\right)}=(20-18)Â\pm 1.674×\sqrt{\left(\frac{2^2}{30}\right)+\left(\frac{5^2}{40}\right)}$

$=(0.543,3.458)$