Question

1. A researcher is interested to learn if there is a linear relationship between the hours in a week spent exercising and a person's life satisfaction. The researchers collected the following data from a random sample, which included the number of hours spent exercising in a week and a ranking of life satisfaction from 1 to 10 being the lowest and 10 the highest). Hours of Exercise Life Satisfaction Participant 3 14 14 10 51 71 81 918 107 9 12115 136 14 11 8 16 171 18 191 10 20 15

Answer 1

Participant	Hours of exercise	Life satisfaction	(Exer - 8.85)^2
1	3	1	34.2225
2	14	2	26.5225
3	14	4	26.5225
4	14	4	26.5225
5	3	10	34.2225
6	5	5	14.8225
7	10	3	1.3225
8	11	4	4.6225
9	8	8	0.7225
10	7	4	3.4225
11	6	9	8.1225
12	11	5	4.6225
13	6	4	8.1225
14	11	10	4.6225
15	8	4	0.7225
16	15	7	37.8225
17	8	4	0.7225
18	8	5	0.7225
19	10	4	1.3225
20	5	4	14.8225
Average	8.85	Sum	254.55

a. Mean exercise = 8.85 hours

b.

Sample Variance Sample Standard Deviation -2 _{(x – 7) n-1 s= {(x – F)2 V n-1

So,

Var = 254.55/19 = 13.4

SD = 13.4^0.5 = 3.66

c. Correlation using excel:

X f r =CORREL(C2:C21,02:021) 3 14 Participantſ Hours of exercise Life satisfaction (Exer -8.85)^2 1 | 1 34.2225 2 26.5225 14 4 26.5225 26.5225 3 34.2225 5 14.8225 =CORREL(C2:C21,02:021) 10 1.3225 CORREL (array1, array) 11 4.6225 8 0.7225 7 3.4225 8.1225 4.6225 13 8.1225 14 11 16 15 4.6225 0.7225 37.8225 0.7225 0.7225 1.3225 14.8225 254.55 18 8 19 10 2015 Average 8.85 Sum

Value = -0.1

It is near to 0. Hence there isn't any strong relationship between exercise and life satisfaction.

d. Regression output:

Regression Statistics
Multiple R	0.103455017
R Square	0.010702941
Adjusted R Square	-0.044258007
Standard Error	2.535287526
Observations	20
ANOVA
	df	SS	MS	F
Regression	1	1.251708898	1.251709	0.194737
Residual	18	115.6982911	6.427683
Total	19	116.95
	Coefficients	Standard Error	t Stat	P-value
Intercept	5.670595168	1.51628432	3.739797	0.0015
Hours of exercise	-0.070123748	0.158906137	-0.44129	0.664258

R^{2 =} 0.01. Hence 1% of variation in satisfaction is explained by number of exercise hours

e. Using above regression output the equation is:

Satisfaction = 5.67 - 0.07 * number Hours of exercise.