In a study of the accuracy of fast food, drive-through orders one restaurant had 34 orders that were not accurate among 361 orders observed. Use a 0.10 significance level to test the claim that the rate of an accurate orders is equal to 10%. Does the accuracy rate appear to be acceptable? Identify the P-value for this hypothesis test. The P-value for this hypothesis test is (Round to three decimal places as needed.)

Step 1 ：Given that there were 361 orders observed and 34 of them were not accurate, we can calculate the sample proportion (p_hat) as \(\frac{34}{361} = 0.094\).

Step 2 ：The null hypothesis is that the proportion of accurate orders is 10% (p_0 = 0.1), and the alternative hypothesis is that the proportion is not 10%.

Step 3 ：We can use a z-test for proportions to test this hypothesis. The z-score is calculated as \(z = \frac{p_{hat} - p_{0}}{\sqrt{\frac{p_{0} * (1 - p_{0})}{n}}}\), where n is the sample size.

Step 4 ：Substituting the given values, we get \(z = \frac{0.094 - 0.1}{\sqrt{\frac{0.1 * (1 - 0.1)}{361}}} = -0.368\).

Step 5 ：The P-value is then calculated using the standard normal distribution, which gives us a P-value of 0.713.

Step 6 ：Since the P-value is greater than the significance level of 0.10, we do not reject the null hypothesis that the proportion of accurate orders is 10%.

Step 7 ：Therefore, the accuracy rate appears to be acceptable.

Step 8 ：Final Answer: The P-value for this hypothesis test is \(\boxed{0.713}\).