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.

## Expert Answer

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 (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 (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 (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.**