#### Answer

Fail to 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: $t=\frac{(\overline{x_1}-\overline{x_2})-(\mu_1-\mu_2)}{\sqrt{s_1^2/n_1+s_2^2/n_2}}=\frac{(0.8168-0.7848)-(0)}{\sqrt{0.0075^2/36+0.0044^2/36}}=22.081.$ The degree of freedom: $min(n_1-1,n_2-1)=min(36-1,36-1)=35.$ The corresponding P-value by using the table: p is more than 0.2. 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 more than $\alpha=0.05$, because it is more than 0.2, hence we fail to reject the null hypothesis. Hence there doesn't seem to be difference in the mean weights.