The acceptable level for insect filth in a certain food item is 5 insect fragments (larvae, eggs, body parts, and so on) per 10 grams. A simple random sample of 50 ten-gram portions of the food item is obtained and results in a sample mean of $\bar{x}=5.8$ insect fragments per ten-gram portion. What might we conclude? A. Since this result is unusual, it is not reasonable to conclude that the population mean is higher than 5. B. Since this result is not unusual, it is reasonable to conclude that the population mean is higher than 5. C. Since this result is unusual, it is reasonable to conclude that the population mean is higher than 5. D. Since this result is not unusual, it is not reasonable to conclude that the population mean is higher than 5 .

Step 1 ：We are given that the acceptable level for insect filth in a certain food item is 5 insect fragments per 10 grams. A simple random sample of 50 ten-gram portions of the food item is obtained and results in a sample mean of 5.8 insect fragments per ten-gram portion.

Step 2 ：We need to perform a hypothesis test. The null hypothesis is that the population mean is 5, and the alternative hypothesis is that the population mean is greater than 5. We can use a one-sample t-test to test this hypothesis.

Step 3 ：We know that the sample mean is 5.8, but we don't know the sample standard deviation. However, we can assume that the standard deviation is small because the sample size is large (50).

Step 4 ：We can calculate the t-score using the formula: \(t = \frac{{\text{{sample mean}} - \text{{population mean}}}}{{\text{{standard deviation}} / \sqrt{{\text{{sample size}}}}}}\)

Step 5 ：If the t-score is greater than the critical value for a certain significance level (let's say 0.05), we can reject the null hypothesis and conclude that the population mean is greater than 5.

Step 6 ：Given that the sample mean is 5.8, the population mean is 5, the sample size is 50, and the standard deviation is 0.1, we calculate the t-score to be 56.56854249492379.

Step 7 ：The critical value for a significance level of 0.05 is 1.6765508919142629.

Step 8 ：Since the t-score is much larger than the critical value, we can reject the null hypothesis. This means that it is reasonable to conclude that the population mean is greater than 5.

Step 9 ：This result is unusual if we assume that the population mean is 5, so the correct answer is \(\boxed{\text{C. Since this result is unusual, it is reasonable to conclude that the population mean is higher than 5.}}\)