Hypotheses:
-
Null Hypothesis (H₀): There is no significant difference in the mean weight before and after the diet. (\(\mu_d = 0\))
-
Alternative Hypothesis (H₁): There is a significant difference in the mean weight before and after the diet. (\(\mu_d \neq 0\))
Significance Level:
We will use a significance level (\(\alpha\)) of 0.05.
Let’s break down the detailed mathematics behind each step of the paired samples t-test for the diet program effectiveness example, using the provided weights before and after the diet.
Calculate the differences for each participant:
\[
\begin{align*}
d_1 & = 70 - 68 = 2 \\
d_2 & = 72 - 70 = 2 \\
d_3 & = 75 - 74 = 1 \\
d_4 & = 80 - 77 = 3 \\
d_5 & = 78 - 76 = 2 \\
\end{align*}
\] Differences: \(d = [2, 2, 1, 3, 2]\)
Calculate the mean difference (\(\bar{d}\)):
\[
\bar{d} = \frac{2 + 2 + 1 + 3 + 2}{5} = \frac{10}{5} = 2 \text{ kg}
\]
Calculate the standard deviation of the differences (\(s_d\)):
First, calculate the squared deviations from the mean: \[
\begin{align*}
(2 - 2)^2 & = 0 \\
(2 - 2)^2 & = 0 \\
(1 - 2)^2 & = 1 \\
(3 - 2)^2 & = 1 \\
(2 - 2)^2 & = 0 \\
\end{align*}
\] Sum of squared deviations: \[
0 + 0 + 1 + 1 + 0 = 2
\] Now, calculate \(s_d\): \[
s_d = \sqrt{\frac{2}{4}} = \sqrt{0.5} = 0.707 \text{ kg}
\]
Calculate the t-statistic:
Use the formula for the t-statistic with \(n = 5\) (number of participants): \[
t = \frac{\bar{d}}{s_d / \sqrt{n}} = \frac{2}{0.707 / \sqrt{5}} = \frac{2}{0.707 / 2.236} = \frac{2}{0.316} = 6.324
\]
Degrees of freedom (\(df\)):
\[
df = n - 1 = 5 - 1 = 4
\]
Compare the calculated t-statistic to the critical t-value:
The critical t-value for \(df = 4\) and a two-tailed test with \(\alpha = 0.05\) is approximately 2.776 (from t-distribution tables).
Interpretation
Since the calculated t-statistic (6.324) is significantly greater than the critical t-value (2.776), we reject the null hypothesis. This indicates a statistically significant decrease in weight due to the diet, confirming the effectiveness of the nutritionist’s program. The precise calculation steps and their results provide strong mathematical evidence for this conclusion.