A clinical trial is run to assess the effects of different forms of regular exercise on HDL levels in persons between the ages of 18 and 29. Participants in the study are randomly assigned to one of three exercise groups - Weight training, Aerobic exercise or Stretching/Yoga - and instructed to follow the program for 8 weeks. Their HDL levels are measured after 8 weeks and are summarized below.
| Exercise Group | N | Mean | Std Dev |
|---|---|---|---|
| Weight Training | 20 | 49.7 | 10.2 |
| Aerobic Exercise | 20 | 43.1 | 11.1 |
| Stretching/Yoga | 20 | 57.0 | 12.5 |
Suppose that in the aerobic exercise group we also measured the number of hours of aerobic exercise per week and the mean is 5.2 hours with a standard deviation of 2.1 hours. The sample correlation is -0.42.
a) Estimate the equation of the regression line that best describes the relationship between number of hours of exercise per week and HDL cholesterol level (Assume that the dependent variable is HDL level).
b) Estimate the HDL ievel for a person who exercises 7 hours per week.
c) Estimate the HDL level for a person who does not exercise.
Solution
a) Let $X$ denote the number of hours of aerobic exercise per week and $Y$ denote the HDL cholesterol level for aerobic exercise group.
Given that $n = 20$. The mean and sd of $X$ are $5.2$ and $2.1$ respectively. The mean and sd of $Y$ are $43.1$ and $11.1$ respectively. The correlation coefficient between $X$ and $Y$ is -0.42.
The regression equation is $Y = a+ b*X$, where
$$ \begin{aligned} b &= r \frac{s_y}{s_x}\\ &= -0.42 \big(\frac{11.1}{2.1}\big)\\ &=-2.22 \end{aligned} $$
$$ \begin{aligned} a&=\overline{Y}-b*\overline{X}\\ &=43.1 --2.22* 5.2\\ &=54.644 \end{aligned} $$
The estimated equation of the line that best describe the association between number of hours of exercise per week and HDL cholesterol level is
$$ Y = 54.644 + (-2.22)*X $$
b) Estimate of the HDL level for a person who exercises 7 hours per week is
$$ \begin{aligned} \hat{Y} &=54.644 + (-2.22) * 7\\ &= 39.104 \end{aligned} $$
c) Estimate of the HDL level for a person who does not exercise is
$$ \begin{aligned} \hat{Y} &=54.644 + (-2.22) * 0\\ &= 54.644 \end{aligned} $$
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
