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Quarters are currently minted with weights having a mean of 5.146 and a standard deviation of 0.066 . New equipment is being tested in an attempt to improve quality by reducing variation. A simple random sample of 29 quarters is obtained from those manufactured with the new equipment, and this sample has a standard deviation of 0.049 . Use a 0.01 significance level to test the claim that quarters manufactured with the new equipment have weights with a standard deviation less than 0.066 . Does the new equipment appear to be effective in reducing the variation of weights?
What can be said about the effectiveness of the new equipment?
A. Since $\mathrm{H}_{0}: \sigma< 0.066$ is rejected, the new equipment is more effective.
B. Since $\mathrm{H}_{0}: \sigma=0.066$ is not rejected, the new equipment is not more effective.
C. Since $\mathrm{H}_{0}: \sigma=0.066$ is rejected, the new equipment is more effective.
D. Since $\mathrm{H}_{0}: \sigma< 0.066$ is not rejected, the new equipment is not more effective.
Answer
\(\boxed{\text{D. Since } H_{0}: \sigma=0.066 \text{ is not rejected, the new equipment is not more effective.}}\)
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 ：\(\boxed{\text{D. Since } H_{0}: \sigma=0.066 \text{ is not rejected, the new equipment is not more effective.}}\)
