Step 1 :Given that the population mean (μ) is 109 hours, the population standard deviation (σ) is 6 hours, the sample size (n) is 210 students, and the sample mean (X) is 110 hours.
Step 2 :We are conducting a one-sample z-test. The test statistic (z) can be calculated using the formula: \(z = \frac{X - μ}{σ/\sqrt{n}}\).
Step 3 :Substituting the given values into the formula, we get: \(z = \frac{110 - 109}{6/\sqrt{210}}\), which simplifies to \(z = 2.4152294576982394\).
Step 4 :The P-value is the probability that we would observe a result as extreme as the test statistic, assuming the null hypothesis is true. In this case, the null hypothesis is that the mean training time is 109 hours. We are testing the alternative hypothesis that the mean training time is greater than 109 hours.
Step 5 :Using the calculated z-score, we find that the P-value is 0.007862649877252714.
Step 6 :Rounding to four decimal places, the final answer is: \(\boxed{0.0079}\).