A manufacturer must test that his bolts are $4.00 \mathrm{~cm}$ long when they come off the assembly line. He must recalibrate his machines if the bolts are too long or too short. After sampling 196 randomly selected bolts off the assembly line, he calculates the sample mean to be $4.14 \mathrm{~cm}$. He knows that the population standard deviation is $0.83 \mathrm{~cm}$. Assuming a level of significance of 0.05 , is there sufficient evidence to show that the manufacturer needs to recalibrate the machines? Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.

Step 1 ：First, we need to set up our null and alternative hypotheses. The null hypothesis (H0) is that the mean bolt length is 4.00 cm, and the alternative hypothesis (H1) is that the mean bolt length is not 4.00 cm.

Step 2 ：Next, we calculate the test statistic using the formula for a z-test: \(Z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}}\), 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 3 ：Substituting the given values into the formula, we get: \(Z = \frac{4.14 - 4.00}{\frac{0.83}{\sqrt{196}}}\).

Step 4 ：Solving the equation, we get: \(Z = \frac{0.14}{\frac{0.83}{14}} = 2.36\).

Step 5 ：Therefore, the value of the test statistic is \(\boxed{2.36}\).