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## Here's the Solution to this Question

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

$F=\dfrac{s_1^2}{s_2^2}=\dfrac{3000^2}{2500^2}=1.44$

The critical values for $df_1=100-1=99=df_2, \alpha=0.025$ are $F_L = 0.6353,$ $F_U=1.574,$ and since $F = 1.44,$ then the null hypothesis of equal variances is not rejected.

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1=\mu_2$

$H_1:\mu_1\not=\mu_2$

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Based on the information provided, the significance level is $\alpha = 0.025,$ and the degrees of freedom are $df=n_1-1+n_2-1 = 100-1+100-1=198,$ assuming that the population variances are equal.

Hence, it is found that the critical value for this two-tailed test for $\alpha=0.025,df=198$ degrees of freedom is $t_c=2.258576.$

The rejection region for this two-tailed test is $R = \{t: |t| > 2.258576\}.$

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$t=\dfrac{\bar{x}_1-\bar{x}_2}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$

$=\dfrac{20600-19700}{\sqrt{\dfrac{(100-1)3000^2+(100-1)2500^2}{100+100-2}(\dfrac{1}{100}+\dfrac{1}{100})}}$

$=2.304664$

Since it is observed that $|t| = 2.304664 > 2.258576=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for $df=198$ degrees of freedom, $t=2.304664$ is $p = 0.022222,$ and since $p=0.022222<0.025=\alpha,$ it is concluded that the null hypothes is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu_1$ is different than $\mu_2,$ at the $\alpha = 0.025$ significance level.

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1\leq\mu_2$

$H_1:\mu_1>\mu_2$

This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Based on the information provided, the significance level is $\alpha = 0.025,$ and the degrees of freedom are $df=n_1-1+n_2-1 = 100-1+100-1=198,$ assuming that the population variances are equal.

Hence, it is found that the critical value for this right-tailed test for $\alpha=0.025,df=198$ degrees of freedom is $t_c=1.972017.$

The rejection region for this two-tailed test is $R = \{t: t > 1.972017\}.$

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$t=\dfrac{\bar{x}_1-\bar{x}_2}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$

$=\dfrac{20600-19700}{\sqrt{\dfrac{(100-1)3000^2+(100-1)2500^2}{100+100-2}(\dfrac{1}{100}+\dfrac{1}{100})}}$

$=2.304664$

Since it is observed that $t = 2.304664 >1.972017=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for $df=198$ degrees of freedom, $t=2.304664$ is $p = 0.011111,$ and since $p=0.011111<0.025=\alpha,$ it is concluded that the null hypothec is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu_1$ is greater than $\mu_2,$ at the $\alpha = 0.025$ significance level.