Question

4.21 the accompanying data are consistent with summary statistics that appeared in the paper “shape of glass and amount of alcohol poured: comparative study of effect of practice on concentration” (british medical journal 2005: 1512 - 1514). data represent the actual amount poured (in ml) into a tall, slender glass by 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 56.5 65.0 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.
$n = 20$, and $\sum_{i=1}^{20}x_{i}=44.0 + 49.6+62.3+28.4+39.1+39.8+60.5+73.0+57.5+56.5+65.0+56.2+57.7+73.5+66.4+32.7+40.4+21.4$
$\sum_{i = 1}^{20}x_{i}=1008.5$
$\bar{x}=\frac{1008.5}{20}=50.425$

Step2: Calculate the variance

The variance $s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}$.
First, calculate $(x_{i}-\bar{x})^{2}$ for each $i$:
For $x_1 = 44.0$, $(x_1-\bar{x})^{2}=(44.0 - 50.425)^{2}=(- 6.425)^{2}=41.280625$
...
After calculating $(x_{i}-\bar{x})^{2}$ for all $i$ from $1$ to $20$ and summing them up, $\sum_{i = 1}^{20}(x_{i}-\bar{x})^{2}=2379.9325$
$s^{2}=\frac{2379.9325}{19}\approx125.26$

Step3: Calculate the standard deviation

The standard deviation $s=\sqrt{s^{2}}$.
$s=\sqrt{125.26}\approx11.19$

Step4: Interpret the results

The mean of $50.425$ ml represents the average amount of alcohol poured into the tall, slender glass by the individuals. The standard deviation of approximately $11.19$ ml indicates the amount of variability or spread in the amounts poured. A relatively large standard deviation suggests that there is a wide range of amounts that individuals pour, with some pouring much more and some pouring much less than the average amount.

Answer:

Mean: $50.425$ ml, Standard deviation: approximately $11.19$ ml. The mean represents the average amount poured, and the standard deviation represents the variability in the amounts poured.