Fifteen participants had their BMI measured before and after a training program. Before: 19.24, 20.50, 23.00, 31.90, 22.00, 34.30, 25.80, 23.70, 20.10, 32.40, 25.10, 25.00, 27.67, 23.48, 28.49. After: 18.50, 19.50, 21.90, 27.00, 21.70, 33.70, 24.80, 22.50, 19.80, 31.20, 23.93, 23.00, 28.99, 24.30, 28.92. Test at the 5% level whether the program changed mean BMI.

Step-by-step solution

1. 1

   Recognise that the two columns are not independent samples. Each row is one person measured twice, so the "before" and "after" numbers are linked in pairs. A two-sample $t$-test would treat them as $30$ unrelated observations and would be swamped by the huge between-person spread (BMI ranges from $19$ to $34$ here). Working with the $15$ differences cancels each person's baseline out and leaves only the effect of the program, which is why the paired test is far more sensitive.

2. 2

   Form the differences $d_i=\text{before}-\text{after}$.

   $0.74,\;1.00,\;1.10,\;4.90,\;0.30,\;0.60,\;1.00,\;1.20,\;0.30,\;1.20,\;1.17,\;2.00,\;-1.32,\;-0.82,\;-0.43$

   Twelve of the fifteen are positive (BMI went down), three are negative. The hypotheses are stated about the mean difference $\mu_d$:

   $H_0:\mu_d=0,\qquad H_1:\mu_d\neq 0$

3. 3

   Compute the mean and standard deviation of the differences. The differences sum to $382.68-369.74=12.94$, so

   $\bar d=\frac{12.94}{15}=0.8627$

   $s_d=\sqrt{\frac{\sum (d_i-\bar d)^2}{n-1}}=\sqrt{\frac{27.9933}{14}}=1.4140$

   Use $n-1=14$ in the denominator, not $n$: the sample mean has already been fitted to the data, so only $14$ of the deviations are free to vary.

4. 4

   Turn the mean difference into a $t$ statistic. The standard error of $\bar d$ is

   $s_{\bar d}=\frac{s_d}{\sqrt n}=\frac{1.4140}{\sqrt{15}}=0.3651$

   $t=\frac{\bar d-0}{s_{\bar d}}=\frac{0.8627}{0.3651}=2.363$

   with $df=n-1=14$. The statistic is just "how many standard errors is the observed drop away from zero".

5. 5

   Compare with the critical value and read off the $p$-value. For a two-tailed test at $\alpha=0.05$ with $14$ degrees of freedom the critical value is $t^{*}=2.145$.

   $|t|=2.363>2.145$

   The corresponding two-tailed $p$-value is $p\approx 0.033$, which is below $0.05$. Reject $H_0$: the mean BMI change is statistically significant.

6. 6

   Report a confidence interval, because a $p$-value alone hides the size of the effect.

   $\bar d\pm t^{*}s_{\bar d}=0.8627\pm 2.145(0.3651)=0.8627\pm 0.7832$

   $95\%\text{ CI for }\mu_d:\;(0.0795,\;1.6458)$

   The interval excludes $0$, which is the same conclusion as the test. But it also shows the drop could plausibly be as small as $0.08$ BMI points — statistically detectable, yet of little clinical meaning at the low end. Note too that a single group measured twice has no control group, so the change cannot be attributed to the program alone.