The National Institute of Standards and Technology (NIST) offers Standard Reference Materials to aid in the calibration of measurement instruments and verify the accuracy of measurements. Suppose a medical researcher wants to verify that the distances he measures with his computed tomography machine are not too large, so he orders a Dimensional Standard for Medical Computed Tomography from NIST. This standard consists of 18 plastic balls held in place by a plastic support structure. The certified distance between the balls numbered 1 and 2 is 15.96 mm, and NIST determines that measurement errors are normally distributed with a standard deviation of 0.05 mm. Upon receiving the Standard Reference Material, the researcher measured the distance between balls 1 and 2 five times. Take these five measurements to be a simple random sample of all measurements made on this device under carefully controlled conditions. The researcher then calculated the summary statistics, where is the mean distance between balls 1 and 2, as measured by his computed tomography machine. Test of =15.96 vs >15.96The assumed standard deviation=0.05Samplesize Samplemean Standarderror x⎯⎯⎯ SE5 16.02 0.02236(1). Complete the analysis by calculating the value of the one-sample z-statistic, the p-value, and making the decision. First, calculate z to at least two decimal places. z=(2).Use software or a table of standard normal critical values to determine the p‑value. Give your answer precise to at least four decimal places. p=(3). Should the researcher reject his null hypothesis if his significance level is =0.05?A. Yes, there is enough evidence (p<0.05) that the mean distance between balls 1 and 2, as measured by his computed tomography machine, is greater than 16.02 mm. B. No, there is not enough evidence (p>0.05) that the mean distance between balls 1 and 2, as measured by his computed tomography machine, is greater than 16.02 mm. C. No, there is not enough evidence (p<0.05) that the mean distance between balls 1 and 2, as measured by his computed tomography machine, is greater than 16.02 mm. D. Yes, there is enough evidence (p>0.05) that the mean distance between balls 1 and 2, as measured by his computed tomography machine, is greater than 16.02 mm.