Question

1.2. Using the appropriate /test, determine whether or not the actual volume of your pipet is different from its mo miraz/volume of 10.00 mL at the 50%, 95%, and 99% level.

Answer 1

The given measurements can be converted to volumes by subtracting the empty flask weight (i.e. 44.3142).

1. 54.2779 - 44.3142 = 9.9637

2. 64.2779 - 44.3142 = 19.9637

3. 74.1966 - 44.3142 = 29.8824

4. 84.1377 - 44.3142 = 39.8235

5. 94.0753 - 44.3142 = 49.7611

The mean of given measurements (x̄) = (9.9637 + 19.9637 + 29.8824 + 39.8235 + 49.7611)/5, i.e. 29.8789

The sample standard deviation for the above measurements (s) = {(9.9637 - 29.8789)² + (19.9637 - 29.8789)² + (29.8824 - 29.8789)² + (39.8235 - 29.8789)² + (49.7611 - 29.8789)²) / (5-1)}^1/2, i.e. 15.7252

The 50% confidence interval for the given data can be calculated as shown below.

x̄ $\pm$ (0.674*s/n^1/2), where n = no. of measurements, i.e. 5

Here, 0.674*15.7252/5^1/2 = 4.7399; 29.8789 - 4.7399 = 25.139 and 29.8789 + 4.7399 = 34.6188

Therefore, the 50% confidence interval for the given data: 25.139 to 34.6188​

The 95% confidence interval for the given data can be calculated as shown below.

x̄ $\pm$ (1.96*s/n^1/2), where n = no. of measurements, i.e. 5

Here, 1.96*15.7252/5^1/2 = 13.7837; 29.8789 - 13.7837 = 16.0952 and 29.8789 + 13.7837 = 43.6626

Therefore, the 95% confidence interval for the given data: 16.0952 to 43.6626​

The 99% confidence interval for the given data can be calculated as shown below.

x̄ $\pm$ (2.576*s/n^1/2), where n = no. of measurements, i.e. 5

Here, 2.576*15.7252/5^1/2 = 18.1158; 29.8789 - 18.1158 = 11.7631 and 29.8789 + 18.1158 = 47.9947

Therefore, the 99% confidence interval for the given data: 11.7631 to 47.9947

Conclusion: The actual volume of the pipet is different from the nominal value of 10.00 mL at all the 50%, 95% and 99% confidence intervals.