#### Answer

Reject the null hypothesis.

#### Work Step by Step

Null hypothesis:$\mu_1=\mu_2$, alternative hypothesis:$\mu_1\ne\mu_2$. Hence the value of the test statistic: $t=\frac{(\overline{x_1}-\overline{x_2})-(\mu_1-\mu_2)}{\sqrt{s_1^2/n_1+s_2^2/n_2}}=\frac{(6.1927-5.6393)-(0)}{\sqrt{0.087^2/40+0.0619^2/40}}=32.78.$ The degree of freedom: $min(n_1-1,n_2-1)=min(40-1,40-1)=39.$ The corresponding P-value by using the table: p is less than 0.005. If the P-value is less than $\alpha$, which is the significance level, then this means the rejection of the null hypothesis. Hence:P is less than $\alpha=0.05$, because it is less than 0.005, hence we reject the null hypothesis.