Question

4.22 the paper referenced in the previous exercise also gave data on the actual amount poured (in ml) into a short, wide glass for individuals who were asked to pour 44.3 ml (1.5 ounces). 89.2 68.6 32.7 37.4 39.6 46.8 66.1 79.2 66.3 52.1 47.3 64.4 53.7 63.2 46.4 63.0 92.4 57.8 a. calculate and interpret the values of the mean and standard deviation.

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. Here, $n = 20$ and $\sum_{i=1}^{20}x_{i}=89.2 + 68.6+32.7+37.4+39.6+46.8+66.1+79.2+66.3+52.1+47.3+64.4+53.7+63.2+46.4+63.0+92.4+57.8$
$\sum_{i = 1}^{20}x_{i}=1100.5$. Then $\bar{x}=\frac{1100.5}{20}=55.025$ ml. The mean represents the average amount of liquid poured into the short, wide glass by the individuals.

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 $x_{i}$:
For $x_1 = 89.2$, $(x_1-\bar{x})^{2}=(89.2 - 55.025)^{2}=(34.175)^{2}=1168.930625$
...
After calculating $(x_{i}-\bar{x})^{2}$ for all $i$ from 1 to 20 and summing them up, $\sum_{i = 1}^{20}(x_{i}-\bar{x})^{2}=3779.9975$. Then $s^{2}=\frac{3779.9975}{19}\approx198.9472$

Step3: Calculate the standard deviation

The standard deviation $s=\sqrt{s^{2}}$. So $s=\sqrt{198.9472}\approx14.105$ ml. The standard deviation measures the amount of variation or dispersion of the data set. A larger standard deviation indicates that the data points are more spread out from the mean.

Answer:

The mean is $55.025$ ml and represents the average amount poured. The standard deviation is approximately $14.105$ ml and represents the dispersion of the amounts poured.