Question

doctors developed an intensive intervention program for obese patients with heart disease. subjects with a bmi of 30 kg/m² 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 hearts ejection fraction may be between 50 and 70 percent, higher numbers indicate a healthier heart. assume lvefs are normally distributed.
find the 90% confidence interval for the mean of the differences. round answers to at least 4 decimal places.
<μd<
did the intensive intervention program significantly increase the lvef measurements? ?

Explanation:

Step1: Calculate the differences

Let \(d = \text{After}-\text{Before}\).
For the first row: \(d_1=59 - 45=14\)
For the second row: \(d_2=56 - 46 = 10\)
For the third row: \(d_3=63 - 48=15\)
For the fourth row: \(d_4=60 - 52 = 8\)
For the fifth row: \(d_5=57 - 45=12\)
For the sixth row: \(d_6=64 - 44=20\)
For the seventh row: \(d_7=54 - 44 = 10\)

The differences are \(d=\{14,10,15,8,12,20,10\}\)

Step2: Calculate the mean of the differences \(\bar{d}\)

\(\bar{d}=\frac{\sum_{i = 1}^{n}d_i}{n}\)
\(\sum_{i=1}^{7}d_i=14 + 10+15+8+12+20+10=89\)
\(n = 7\)
\(\bar{d}=\frac{89}{7}\approx12.7143\)

Step3: Calculate the standard deviation of the differences \(s_d\)

\(s_d=\sqrt{\frac{\sum_{i = 1}^{n}(d_i-\bar{d})^2}{n - 1}}\)
\((d_1-\bar{d})^2=(14 - 12.7143)^2\approx1.6049\)
\((d_2-\bar{d})^2=(10 - 12.7143)^2\approx7.3673\)
\((d_3-\bar{d})^2=(15 - 12.7143)^2\approx5.2245\)
\((d_4-\bar{d})^2=(8 - 12.7143)^2\approx22.2245\)
\((d_5-\bar{d})^2=(12 - 12.7143)^2\approx0.5102\)
\((d_6-\bar{d})^2=(20 - 12.7143)^2\approx53.0765\)
\((d_7-\bar{d})^2=(10 - 12.7143)^2\approx7.3673\)
\(\sum_{i = 1}^{7}(d_i-\bar{d})^2=1.6049+7.3673 + 5.2245+22.2245+0.5102+53.0765+7.3673=97.3752\)
\(s_d=\sqrt{\frac{97.3752}{6}}\approx4.0203\)

Step4: Find the critical value \(t_{\alpha/2}\)

The confidence level is \(90\%\), so \(\alpha=1 - 0.90 = 0.10\) and \(\alpha/2=0.05\)
The degrees of freedom \(df=n - 1=7-1 = 6\)
From the \(t\)-distribution table, \(t_{0.05,6}=1.943\)

Step5: Calculate the margin of error \(E\)

\(E=t_{\alpha/2}\frac{s_d}{\sqrt{n}}\)
\(E = 1.943\times\frac{4.0203}{\sqrt{7}}\)
\(E=1.943\times\frac{4.0203}{2.6458}\approx2.9674\)

Step6: Calculate the confidence interval

The confidence interval for \(\mu_d\) is \(\bar{d}-E<\mu_d<\bar{d}+E\)
\(12.7143- 2.9674<\mu_d<12.7143 + 2.9674\)
\(9.7469<\mu_d<15.6817\)

Answer:

\(9.7469<\mu_d<15.6817\)
Since the entire confidence interval is above \(0\), the intensive intervention program significantly increased the LVEF measurements.