Test the claim about the population mean, $\mu$, at the given level of significance using the given sample statistics. Claim: $\mu \neq 5000 ; \alpha=0.02 ; \sigma=357$. Sample statistics: $\bar{x}=4700, n=33$ Determine the outcome and conclusion of the test. Choose from the following. A. Fail to reject $\mathrm{H}_{0}$. At the $2 \%$ significance level, there is not enough evidence to support the claim. B. Fail to reject $\mathrm{H}_{0}$. At the $2 \%$ significance level, there is not enough evidence to reject the claim. C. Reject $\mathrm{H}_{0}$. At the $2 \%$ significance level, there is enough evidence to reject the claim D. Reject $\mathrm{H}_{0}$. At the $2 \%$ significance level, there is enough evidence to support the claim.

Step 1 ：Given that the claim is that the population mean is not equal to 5000, the null hypothesis is that the population mean is equal to 5000 and the alternative hypothesis is that the population mean is not equal to 5000.

Step 2 ：The level of significance is 0.02, the population standard deviation is 357, the sample mean is 4700 and the sample size is 33.

Step 3 ：We can use a z-test to test this claim because we know the population standard deviation. The formula for the z-score is: \(z = \frac{{\bar{x} - \mu}}{{\sigma / \sqrt{n}}}\), where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.

Step 4 ：The critical z-score for a two-tailed test at the 0.02 level of significance is approximately ±2.33. If the calculated z-score is less than -2.33 or greater than 2.33, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

Step 5 ：Calculating the z-score gives us -4.83, which is less than the critical z-score of -2.33. Therefore, we reject the null hypothesis.

Step 6 ：\(\boxed{\text{D. Reject } H_{0}. \text{At the } 2\% \text{ significance level, there is enough evidence to support the claim.}}\)