91Ó°ÊÓ

${\stackrel{¯}{x}}_1=29,{\stackrel{¯}{x}}_2=21$

$s_1=59,s_2=52$

$n_1=427,n_2=2733$

The test statistic formula is:

$t=\frac{{\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$

The null and alternative hypotheses for the provided case are:

$H_0:{\mathrm{μ}}_1={\mathrm{μ}}_2$

$H_1:{\mathrm{μ}}_1≠{\mathrm{μ}}_2$

The test statistic is computed as:

$=\frac{29-21}{\sqrt{\frac{{59}^2}{427}+\frac{{52}^2}{2733}}}$

$=2.646$

The degree of freedom is calculated as:

$$\begin{gathered} df=\min \left(n_1-1,n_2-1\right) \\ =\min (427-1,2733-1) \\ =426 \end{gathered}$$

The p-value:

$P-\text{value}=0.0042$

In this case, $P-$value $=0.00424>0.05$

The null hypothesis could not be rejected which is not showing the sufficient evidences for the claim at significance level of $5\%$.

Confidence level is$95\%$.

The $95\%$confidence interval using $\mathrm{Ti}-83$calculator is:

Confidence level is$95\%$.

The above computed confidence interval is very close to the value $0$ . Thus, no larger gain is being earned which implies that it is not worth paying for.