Step 1 :State the hypotheses. The null hypothesis is \(H_{0}: \sigma=0.007\). The alternative hypothesis is \(H_{1}: \sigma < 0.007\).
Step 2 :Calculate the test statistic using the formula \(\chi^{2} = (n-1)\left(\frac{s}{\sigma}\right)^{2}\), where \(n\) is the sample size, \(s\) is the sample standard deviation, and \(\sigma\) is the population standard deviation. Substituting the given values, we get \(\chi^{2} = (29-1)\left(\frac{0.0063}{0.007}\right)^{2} = 22.68\).
Step 3 :Use a chi-square distribution table or statistical software to find the p-value corresponding to the test statistic. The p-value is approximately 0.749.
Step 4 :Compare the p-value with the significance level \(\alpha\). If the p-value is less than \(\alpha\), reject the null hypothesis. If the p-value is greater than \(\alpha\), fail to reject the null hypothesis. In this case, the p-value (0.749) is greater than the significance level (0.10), so we fail to reject the null hypothesis.
Step 5 :Interpret the result. Since we failed to reject the null hypothesis, there is not enough evidence to conclude that the standard deviation has decreased. Therefore, the final answer is: The test statistic is \(\boxed{22.68}\) and the p-value is \(\boxed{0.749}\). Since the p-value is greater than the significance level of 0.10, we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that the standard deviation has decreased.