Question

You are checking the accuracy of a volumetric flask marked 10.00 mL. Measurement| Volume (mL) You measure the volume of water contained in the flask by measuring the mass of the empty flask, the mass of the flask filled with water, taking the difference, correcting for the buoyancy factor, and dividing by the density of the water. The results of eight such measurements are given in the table 10.064 10.063 10.017 10.049 10.020 10.025 10.045 10.008 2 3 4 5 Calculate the mean and the 95% confidence interval for these measurements mean 7 8 Number Number mL+ Can you, with 95% confidence, say that the nominal value of 10.00 mL can be the true volume of the flask? O Yes O No

Answer 1

Mean (X^-) = (10.064+ 10.063 + 10.017 +10.049 + 10.020 + 10.025 + 10.045 + 10.008)/8

Mean (X^-) = 10.036 ml

-he number of data points The mean of the Xi Xi-Each of the values of the data

Substituting the values in above formula:

We get, s = 0.02

So, Mean of the measurements is:

10.036 mL ± 0.020 mL

95% confidence interval:

10.021 mL - 10.056 mL

Observed interval :

10.036 mL ± 0.020 mL = 10.016 mL - 10.056 mL

Since it contains the 95% confidence interval, we can say that the nominal value of 10.00 mL can be the true volume of the flask.

Yes.