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:

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

The test is run at a significance level oflocalid="1651413915402" $a=0.05$, which equals 5%. The test statistic's value is calculated as follows:

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

$=\frac{20-24}{\sqrt{\left(\frac{6^2}{20}\right)+\left(\frac{2^2}{15}\right)}}$

$=-2.78$

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

The value of$df$is determined as follows:

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

$=\frac{{\left[\left(\frac{6^2}{20}\right)+\left(\frac{2^2}{15}\right)\right]}^2}{\frac{{\left(\frac{6^2}{20}\right)}^2}{20-1}+\frac{{\left(\frac{2^2}{15}\right)}^2}{15-1}}$

$=24$

Refer to the distribution table; the critical value found for localid="1651413932060" $df=22$is,

$t_{0.05}=-1.711$.

As indicated in the diagram below, the graph is plotted.

Figure 1 shows that the test statistic $t=-2.78$falls into the rejection zone. As a result, the results are significant at the $5\%$level.

$df=\mathrm{Δ}$in the $t$statics

The probability of detecting a value is represented by the localid="1651413970022" $P$value localid="1651413975778" $t=-2.78$

The value of localid="1651413983768" $P$obtained using technological means is localid="1651413990466" $0.0052$.

Use the localid="1651413997998" $t$table with localid="1651414004029" $df=22$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 threshold. At the $5\%$level, the results are statistically significant.

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

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

$t_{0.05}=1.711$

The confidence interval's end point for ${\mathrm{μ}}_1-{\mathrm{μ}}_2$is determined as

$\left({\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2\right)Â\pm t_{\mathrm{Î\pm }/2}×\sqrt{\left(s_1^2/n_1\right)+\left(s_2^2/n_2\right)}=(20-24)Â\pm 1.711×\sqrt{\left(\frac{6^2}{20}\right)+\left(\frac{2^2}{15}\right)}$

$=(-6.46,-1.54)$