Question

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.

Answer 1

A. There is enough evidence that p < 0.05 that the mean distance between ball one and two as measured by his computed tomography machine is greater than 16.02mm.

Explanation:

n= 5

x bar= 16.02

Standard Error= 0.02236

NB:

The null hypothesis is referred to as a characteristic arithmetic theory that gives a statement that suggests that no statistical relationship and significance exists in a set of given observed variables between two sets of observed data and also, measured phenomena. The hypotheses play a crucial role in testing the significance of differences in experiments and between observations. H o represents the null hypothesis of no difference.

The alternative hypothesis is generally represented as H1. It makes a statement that suggests a potential result or an outcome that a researcher may expect.

Standard error measures how far the sample mean of the data is most likely to be from the true population mean.

We want to test the null and alternative hypothesis

H o : μ = 15.96 vs H1 : μ > 15.96

1. Test statistics

Z= xbar - μ / Standard error

Z= 16.02 - 15.96 / 0.002236

Z= 2.6834

2. P value= 0.0036

3. Yes, there is enough evidence that p < 0.05 that the mean distance between ball one and two as measured by his computed tomography machine is greater than 16.02mm.

Therefore, option A is the correct answer.