To calculate the average and the standard deviation of the mean (standard error) of the given data set, we follow these steps:

Data Set (in mL):

19.83, 20.14, 19.28, 19.75, 19.59, 19.66, 19.71, 19.42

1. Calculate the average (mean):

The average is simply the sum of all the data points divided by the number of data points.

$\text{Mean} = \frac{\sum \text{Data Points}}{\text{Number of Data Points}}$

Sum of the data points: $19.83 + 20.14 + 19.28 + 19.75 + 19.59 + 19.66 + 19.71 + 19.42 = 157.38$

Number of data points: 8

$\text{Mean} = \frac{157.38}{8} = 19.6725 \, \text{mL}$

2. Calculate the standard deviation (SD):

The standard deviation is calculated as:

$\text{SD} = \sqrt{\frac{\sum (x_i - \text{Mean})^2}{n - 1}}$

Where $x_i$ represents each individual data point, $\text{Mean}$ is the average calculated above, and $n$ is the number of data points.

Calculating squared deviations:

• $(19.83 - 19.6725)^2 = 0.02493125$
• $(20.14 - 19.6725)^2 = 0.21825625$
• $(19.28 - 19.6725)^2 = 0.15375625$
• $(19.75 - 19.6725)^2 = 0.00525625$
• $(19.59 - 19.6725)^2 = 0.00680625$
• $(19.66 - 19.6725)^2 = 0.00015625$
• $(19.71 - 19.6725)^2 = 0.00140625$
• $(19.42 - 19.6725)^2 = 0.06350625$

Sum of squared deviations = $0.473075$

Now, calculate SD:

$\text{SD} = \sqrt{\frac{0.473075}{8 - 1}} = \sqrt{\frac{0.473075}{7}} = \sqrt{0.06758214} = 0.260 \, \text{mL}$

3. Calculate the standard error of the mean (SE):

The standard error of the mean is calculated by dividing the standard deviation by the square root of the number of data points.

$\text{SE} = \frac{\text{SD}}{\sqrt{n}} = \frac{0.260}{\sqrt{8}} = \frac{0.260}{2.8284} = 0.092 \, \text{mL}$

Final Results:

• Average (Mean): 19.6725 mL
• Standard Deviation: 0.260 mL
• Standard Error (SE): 0.092 mL

Would you like more details or have any questions about the calculations?

Related Questions:

1. How is the standard deviation different from the standard error?
2. Why do we divide by $n-1$ when calculating the standard deviation?
3. How would the standard error change if the sample size increased?
4. How can outliers in a data set affect the mean and standard deviation?
5. What is the importance of calculating the standard error in experimental data?

Tip:

In Excel, you can quickly calculate the mean using =AVERAGE(range) and the standard deviation using =STDEV.S(range).