Step 1 :Given values are: sample size \(n = 29\), sample standard deviation \(s = 0.049\), population standard deviation \(\sigma = 0.066\), and significance level \(\alpha = 0.01\).
Step 2 :Calculate the test statistic using the formula \((n-1) \times s^2 / \sigma^2\). Substituting the given values, we get the test statistic as 15.433425160697889.
Step 3 :Calculate the critical value using the chi-square distribution with \(n-1\) degrees of freedom and significance level \(\alpha\). The critical value is 13.564709754618812.
Step 4 :The test statistic is greater than the critical value. This means that we do not reject the null hypothesis.
Step 5 :Therefore, we cannot say that the new equipment is effective in reducing the variation of weights.
Step 6 :\(\boxed{\text{D. Since } H_{0}: \sigma=0.066 \text{ is not rejected, the new equipment is not more effective.}}\)