Activity 3.8 Precision Accuracy Measurement
Procedure
Your instructor will choose two dial calipers. One caliper should be labeled #1 and the other #2. Each student in the class will measure a gauge block with each of the two dial calipers and record the measurements to the appropriate number of significant figures independently (without looking at another student’s recorded measurements). Be sure to keep the measurements for each dial caliper separate.
Conclusion
1. Why is it important to know the accuracy and precision of a measuring device?
So that the measurements of your project are precise and accurate.
2. Do you think that the dial caliper manufacturer’s claim that the accuracy of the instrument is ±.001 is appropriate? Why or why not?
No, because there can only positive measurements of an instrument if it hasn't been changed.
3. Do you think that either of the dial calipers needs to be adjusted in order to accurately display measurements? Explain.
Yes because if your dial caliper isn't set to 0 then it will give you inaccurate measurements.
( Part 1)
a. Which student’s data is more accurate? Explain.
Student A's data is more accurate, because it has significant numbers
b. Looking at the dot plot, which student’s measurement instrument do you feel is more precise? Is this result expected given the measuring instruments used by each student? Explain.
Student A's measurement instrument is more accurate of course, by the end result of his measurement. Yes the result expected from the measuring instrument what was most likely expected because the plastic ruler does not really have have the advantage as the micrometer does when grams,length, and mass is given.
c. Find the mean of each set of measurements and enter each value in the appropriate cell in the table. Then answer the following questions.
Mean Student A: 85.1 Mean Student B: 85.
· Use the standard deviation to make a statement about the actual length of the credit card at the 68% confidence level using Student A’s data. Give your answer in both plus/minus notation and using a compound inequality.
85.1 + -0.08 = 85.02 85.1 + 0.08 = 85.18
68% of the data will fall between 85.02 to 85.18
· Use the standard deviation to make a statement about the true length of the credit card at the 95% confidence level using Student A’s data. Give your answer in both plus/minus notation and using a compound inequality.
85.1 + - 0.16 = 84.94 85.1 + 0.16 = 85.26
95% of the data will fall between 84.94 to 85.26
· Use the standard deviation to indicate the precision of Student B’s measurement data at the 68% and 95% confidence levels.
68% : 85.2 + -0.0015 = 85.1985 85.2 + 0.0015 = 85.2015
85.1085 to 85.2015
95% : 85.2 + -.003 = 85,197 85.2 + .003 = 85.2
(Part 2)
b. Find the mean and sample standard deviation of each set of data.
Mean-.77569 Mean-.77299
Sample Standard Deviation- .01556 Sample Standard Deviation- .01386
c. Compare the accuracy of the two dial calipers.Your discussion should include the mean of each data set.
The comparison of the accuracy of the two dial calipers mean they were both fairly close when it comes to the trial results. Not saying that the experiment was accurate
d. Make a statement regarding the precision of Dial Caliper 1 at the 95% confidence level.
Dial Caliper 1 at 95% confidence is .74456 to .80681
e. Make a statement regarding the precision of Dial Caliper 2 at the 95% confidence level.
Dial Caliper 2 at 95% confidence is .74527 to .80071
f. Based on the accepted value of the length and the accepted value of the width, which dial caliper is more precise? Explain.
The trial result was really close to each other although the precise measurements are close together they are still far away from the intended result. Concluding a systematic error
Your instructor will choose two dial calipers. One caliper should be labeled #1 and the other #2. Each student in the class will measure a gauge block with each of the two dial calipers and record the measurements to the appropriate number of significant figures independently (without looking at another student’s recorded measurements). Be sure to keep the measurements for each dial caliper separate.
Conclusion
1. Why is it important to know the accuracy and precision of a measuring device?
So that the measurements of your project are precise and accurate.
2. Do you think that the dial caliper manufacturer’s claim that the accuracy of the instrument is ±.001 is appropriate? Why or why not?
No, because there can only positive measurements of an instrument if it hasn't been changed.
3. Do you think that either of the dial calipers needs to be adjusted in order to accurately display measurements? Explain.
Yes because if your dial caliper isn't set to 0 then it will give you inaccurate measurements.
( Part 1)
a. Which student’s data is more accurate? Explain.
Student A's data is more accurate, because it has significant numbers
b. Looking at the dot plot, which student’s measurement instrument do you feel is more precise? Is this result expected given the measuring instruments used by each student? Explain.
Student A's measurement instrument is more accurate of course, by the end result of his measurement. Yes the result expected from the measuring instrument what was most likely expected because the plastic ruler does not really have have the advantage as the micrometer does when grams,length, and mass is given.
c. Find the mean of each set of measurements and enter each value in the appropriate cell in the table. Then answer the following questions.
Mean Student A: 85.1 Mean Student B: 85.
· Use the standard deviation to make a statement about the actual length of the credit card at the 68% confidence level using Student A’s data. Give your answer in both plus/minus notation and using a compound inequality.
85.1 + -0.08 = 85.02 85.1 + 0.08 = 85.18
68% of the data will fall between 85.02 to 85.18
· Use the standard deviation to make a statement about the true length of the credit card at the 95% confidence level using Student A’s data. Give your answer in both plus/minus notation and using a compound inequality.
85.1 + - 0.16 = 84.94 85.1 + 0.16 = 85.26
95% of the data will fall between 84.94 to 85.26
· Use the standard deviation to indicate the precision of Student B’s measurement data at the 68% and 95% confidence levels.
68% : 85.2 + -0.0015 = 85.1985 85.2 + 0.0015 = 85.2015
85.1085 to 85.2015
95% : 85.2 + -.003 = 85,197 85.2 + .003 = 85.2
(Part 2)
b. Find the mean and sample standard deviation of each set of data.
Mean-.77569 Mean-.77299
Sample Standard Deviation- .01556 Sample Standard Deviation- .01386
c. Compare the accuracy of the two dial calipers.Your discussion should include the mean of each data set.
The comparison of the accuracy of the two dial calipers mean they were both fairly close when it comes to the trial results. Not saying that the experiment was accurate
d. Make a statement regarding the precision of Dial Caliper 1 at the 95% confidence level.
Dial Caliper 1 at 95% confidence is .74456 to .80681
e. Make a statement regarding the precision of Dial Caliper 2 at the 95% confidence level.
Dial Caliper 2 at 95% confidence is .74527 to .80071
f. Based on the accepted value of the length and the accepted value of the width, which dial caliper is more precise? Explain.
The trial result was really close to each other although the precise measurements are close together they are still far away from the intended result. Concluding a systematic error