Question

4.21 the accompanying data are a measure of variability. the data are consistent with sum - mary statistics that appeared in the paper: \shape of glass and amount of alcohol poured: comparative study of effect of practice and concentration\ (british medical journal 2005: 1512 - 1514). data represent the actual amount poured (in ml) into a tall, slender glass for individuals who were asked to pour 44.3 ml (1.5 ounces). calculate and interpret the values of the mean and standard deviation.
44.0 49.6 62.3 28.4 39.1 39.8 60.5 73.0 57.5
56.0 56.2 57.7 73.5 66.4 32.7 40.4 21.4
4.22

Explanation:

Step1: Calculate the mean

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $x_{i}$ are the data - points and $n$ is the number of data - points.
First, sum up all the data values: $44.0+49.6 + 62.3+28.4+39.1+39.8+60.5+73.0+57.5+65.0+56.2+57.7+73.5+66.4+32.7+40.4+21.4 = 871.5$.
There are $n = 17$ data - points. So, $\bar{x}=\frac{871.5}{17}\approx51.26$.

Step2: Calculate the variance

The variance $s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}$.
For each data - point $x_{i}$, calculate $(x_{i}-\bar{x})^{2}$:
$(44.0 - 51.26)^{2}=(-7.26)^{2}=52.7076$, $(49.6 - 51.26)^{2}=(-1.66)^{2}=2.7556$, etc.
Sum up these values: $\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=2174.97$.
Then, $s^{2}=\frac{2174.97}{16}\approx135.94$.

Step3: Calculate the standard deviation

The standard deviation $s=\sqrt{s^{2}}$. So, $s=\sqrt{135.94}\approx11.66$.

Step4: Interpretation of the mean

The mean of approximately $51.26$ ml represents the average amount of alcohol poured by the individuals into the tall, slender glass.

Step5: Interpretation of the standard deviation

The standard deviation of approximately $11.66$ ml measures the amount of variation or dispersion in the amounts of alcohol poured. A relatively large standard deviation indicates that there is a wide spread in the amounts of alcohol that different individuals poured.

Answer:

Mean: $\approx51.26$ ml; Standard deviation: $\approx11.66$ ml. The mean represents the average amount of alcohol poured, and the standard deviation represents the amount of variation in the amounts poured.