Fifteen participants were measured before and after a treatment. Before: 9.7, 10.8, 16.0, 34.7, 10.9, 45.6, 23.4, 22.9, 17.6, 43.7, 25.4, 17.0, 35.7, 24.7, 36.3. After: 7.8, 8.8, 14.8, 33.2, 9.2, 42.5, 21.8, 21.4, 16.4, 41.5, 22.8, 13.0, 37.8, 26.5, 38.2. Find the mean and sample standard deviation of each set.

Step-by-step solution

1. 1

   Compute the two means from the totals. The before values sum to $374.4$ and the after values to $355.7$, so with $n=15$

   $\bar x_{\text{before}}=\frac{374.4}{15}=24.96,\qquad \bar x_{\text{after}}=\frac{355.7}{15}=23.71\overline{3}$

   Correction to the printed table: the worksheet lists the before-mean as $24.99$, but the fifteen listed values average $24.96$. The after-mean of $23.71$ does check out. Always re-add the column rather than trusting the printed total.

2. 2

   Choose the sample (n − 1) formula. These fifteen people are a sample from a larger population, so divide the squared deviations by $n-1=14$:

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

   Using $n$ instead would understate the spread — a mistake that has been found repeatedly in worked solutions to this kind of table.

3. 3

   Accumulate the before deviations. With $\bar x=24.96$, the squared deviations run from $(9.7-24.96)^2=232.87$ down to $(24.7-24.96)^2=0.0676$ and up to $(45.6-24.96)^2=426.01$. Their total is

   $\sum(x_i-\bar x)^{2}=1951.816$

4. 4

   Finish the before standard deviation.

   $s_{\text{before}}=\sqrt{\frac{1951.816}{14}}=\sqrt{139.4154}=11.807$

5. 5

   Repeat for the after data. With $\bar x=23.713$, the squared deviations total $2119.197$, so

   $s_{\text{after}}=\sqrt{\frac{2119.197}{14}}=\sqrt{151.3712}=12.303$

6. 6

   State and interpret the results.

   $\boxed{\bar x_{\text{before}}=24.96,\ s_{\text{before}}=11.81;\qquad \bar x_{\text{after}}=23.71,\ s_{\text{after}}=12.30}$

   The mean fell by about $1.25$ percentage points, but the spread rose slightly, so the treatment did not make the group more uniform. (For reference, the population standard deviations, dividing by $15$, would be $11.41$ and $11.89$.)

7. 7

   Check with the computational formula. For the before data $\sum x=374.4$ and $\sum x^{2}=11296.84$, so $\sum x^{2}-\tfrac{(\sum x)^2}{n}=11296.84-9345.024=1951.816$ — matching the deviation sum found above, which confirms both the mean and the squared-deviation total.