In: Statistics and Probability
Participant |
Hours of Exercise |
Life Satisfaction |
1 |
3 |
1 |
2 |
14 |
2 |
3 |
14 |
4 |
4 |
14 |
4 |
5 |
3 |
10 |
6 |
5 |
5 |
7 |
10 |
3 |
8 |
11 |
4 |
9 |
8 |
8 |
10 |
7 |
4 |
11 |
6 |
9 |
12 |
11 |
5 |
13 |
6 |
4 |
14 |
11 |
10 |
15 |
8 |
4 |
16 |
15 |
7 |
17 |
8 |
4 |
18 |
8 |
5 |
19 |
10 |
4 |
20 |
5 |
4 |
Problem statement: Based on the observed data points try to answer a set of questions.
Given: Hours of exercise per week and life satisfaction observed of twenty individuals.
Solution:
(a):Mean hours of exercise per week for all participants is given by : summit = 8.85 Hours.
(b): Variance and standard deviation of exercise hours is given by : Variance== 13.39
standard deviation= square root(Variance)= 3.66
(c): Correlation : This provides the strength of linear relationship along with the direction (positive or negative) between weekly exercise and life satisfaction.
Correlation= Covariance ( Weekly exercise and life satisfaction)/(standard deviation(weekly exercise)*standard deviation(life satisfaction)= -0.10346. With this correlation value , we can conclude that there is no linear relationship weekly exercise hours and life satisfaction.
(d) A simple linear regression model is built using life satisfaction as target variable and weekly exercise as predictor variable.
regression equation
life satisfaction= 1*(weekly exercise)+0
We observe that the P-value of the regression model is 0.664, implying that weekly exercise is not a good predictor variable to compute life satisfaction.
The R2 obtained for the model is 0.01. This implies 1% variation in the life satisfaction is attributed to weekly exercise hours.
(e) : Model equation
life satisfaction= -0.07* (weekly exercise in hours)+ 5.67
Since the p-value for weekly exercise is>0.05, the variable weekly exercise is not adding any information to the regression model.