Doctors developed an intensive intervention program for obese patients with heart disease. Subjects with a $\mathrm{BMI}$ of $30 \mathrm{~kg} / \mathrm{m}^{2}$ or more, with heart disease were assigned to a three-month lifestyle change of diet and exercise, Patients' Left Ventricle Ejection Fraction (LVEF) are measured before and after intervention. A normal heart's ejection fraction may je between 50 and 70 percent, higher numbers indicate a healthier heart. Assume LVEFs are normally distributed. \begin{tabular}{|r|r|} \hline Before & After \\ \hline 41 & 53 \\ \hline 44 & 68 \\ \hline 51 & 65 \\ \hline 56 & 68 \\ \hline 46 & 54 \\ \hline 42 & 51 \\ \hline 48 & 61 \\ \hline \end{tabular} Find the $90 \%$ confidence interval for the mean of the differences. Round answers to at least 4 decimal places. $<\mu_{d}<$ Did the intensive intervention program significantly increase the LVEF measurements? ?
Solution
Step 1 ：First, calculate the differences between the 'Before' and 'After' measurements for each patient. The differences are [12, 24, 14, 12, 8, 9, 13].
Step 2 ：Next, calculate the mean and standard deviation of these differences. The mean of the differences is approximately 13.1429 and the standard deviation is approximately 5.2418.
Step 3 ：Then, calculate the 90% confidence interval for the mean of the differences. The 90% confidence interval is approximately (9.2930, 16.9927).
Step 4 ：Finally, check if the confidence interval includes zero. In this case, the confidence interval does not include zero, which suggests that the intensive intervention program significantly increased the LVEF measurements.
Step 5 ：The final answer is: The 90% confidence interval for the mean of the differences is approximately \(\boxed{(9.2930, 16.9927)}\). The intensive intervention program significantly increased the LVEF measurements.
