Fifteen participants were weighed before and after a training programme. Before (kg): 55.60, 56.80, 64.80, 100.50, 72.10, 100.40, 73.80, 77.70, 64.00, 108.60, 76.00, 77.00, 84.75, 67.85, 87.24. After (kg): 54.20, 55.30, 62.10, 100.30, 71.40, 98.70, 71.10, 73.00, 63.10, 104.80, 72.45, 69.00, 88.78, 70.24, 88.57. Find the mean and sample standard deviation of each set.

Step-by-step solution

1. 1

   Add each column and divide by n = 15. The two totals are $1167.14$ kg and $1143.04$ kg, so

   $\bar x_{\text{before}}=\frac{1167.14}{15}=77.809\ \text{kg},\qquad \bar x_{\text{after}}=\frac{1143.04}{15}=76.203\ \text{kg}$

2. 2

   Choose the sample formula. These fifteen people are a sample, so the squared deviations are divided by $n-1=14$:

   $s=\sqrt{\frac{\sum\left(x_i-\bar x\right)^{2}}{n-1}}$

   Using $n$ instead would understate the spread — the error most often seen in write-ups of tables like this one.

3. 3

   Compute the before-group spread. The squared deviations range from $(55.60-77.809)^2=493.3$ up to $(108.60-77.809)^2=948.1$, and total

   $\sum\left(x_i-\bar x\right)^{2}=3556.77$

   so

   $s_{\text{before}}=\sqrt{\frac{3556.77}{14}}=\sqrt{254.055}=15.939\ \text{kg}$

4. 4

   Compute the after-group spread. Here $\sum\left(x_i-\bar x\right)^{2}=3668.24$, so

   $s_{\text{after}}=\sqrt{\frac{3668.24}{14}}=\sqrt{262.017}=16.187\ \text{kg}$

5. 5

   Report the four numbers.

   $\boxed{\text{Before: }\bar x=77.81\ \text{kg},\ s=15.94\ \text{kg};\qquad \text{After: }\bar x=76.20\ \text{kg},\ s=16.19\ \text{kg}}$

6. 6

   Interpret, and note what these numbers cannot show. The mean fell by $1.61$ kg while the spread rose slightly, so the group did not become more uniform. Crucially, these are descriptive statistics only: because the measurements are paired (same people twice), deciding whether the $1.61$ kg drop is statistically significant requires a paired $t$-test on the fifteen differences, not a comparison of these two standard deviations.