ASTATISTICSPROJECT 3
Author’sname
Institutionof affiliation
Project3
Answereach question completely, showing all your work.

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 ( 1 being the lowest and 10 the highest).
x 
y 
xy 
x^{2} 
y^{2} 

1 
3 
1 
3 
9 
1 
2 
14 
2 
28 
196 
4 
3 
14 
4 
56 
196 
16 
4 
14 
4 
56 
196 
16 
5 
3 
10 
30 
9 
100 
6 
5 
5 
25 
25 
25 
7 
10 
3 
30 
100 
9 
8 
11 
4 
44 
121 
16 
9 
8 
8 
64 
64 
64 
10 
7 
4 
28 
49 
16 
11 
6 
9 
54 
36 
81 
12 
11 
5 
55 
121 
25 
13 
6 
4 
24 
36 
16 
14 
11 
10 
110 
121 
100 
15 
8 
4 
32 
64 
16 
16 
15 
7 
105 
225 
49 
17 
8 
4 
32 
64 
16 
18 
8 
5 
40 
64 
25 
19 
10 
4 
40 
100 
16 
20 
5 
4 
20 
25 
16 
Total 
177 
101 

Mean 
r (correlationcoefficient):

Find the mean hours of exercise per week by the participants.

Find the variance of the hours of exercise per week by the participants.

Determine if there is a linear relationship between the hours of exercise per week and the life satisfaction by using the correlation coefficient.
Thereis a linear relationship between the hours of exercise per week andthe life satisfaction given that the therefore the hoursof exercise per week and the life satisfaction are negativelycorrelated.

Describe the amount of variation in the life satisfaction ranking that is due to the relationship between the hours of exercise per week and the life satisfaction.
Thecorrelation of0.0.1 means 0.01^{2}x100= 0.01% of the variance hoursof exercise per week predictedby the lifesatisfaction of the participants.

Develop a model of the linear relationship using the regression line formula.
Theregressionmodel equation is given by the formula
Andthus,
Butand
Thereforethe regression model equation,

Insomnia has become an epidemic in the United States. Much research has been done in the development of new pharmaceuticals to aide those who suffer from insomnia. Alternatives to the pharmaceuticals are being sought by sufferers. A new relaxation technique has been tested to see if it is effective in treating the disorder. Sixty insomnia sufferers between the ages of 18 to 40 with no underlying health conditions volunteered to participate in a clinical trial. They were randomly assigned to either receive the relaxation treatment or a proven pharmaceutical treatment. Thirty were assigned to each group. The amount of time it took each of them to fall asleep was measured and recorded. The data is shown below. Use the appropriate ttest to determine if the relaxation treatment is more effective than the pharmaceutical treatment at a level of significance of 0.05.
x 
y 
x^{2} 
y^{2} 

1 
98 
20 
9604 
400 
2 
117 
35 
13689 
1225 
3 
51 
130 
2601 
16900 
4 
28 
83 
784 
6889 
5 
65 
157 
4225 
24649 
6 
107 
138 
11449 
19044 
7 
88 
49 
7744 
2401 
8 
90 
142 
8100 
20164 
9 
105 
157 
11025 
24649 
10 
73 
39 
5329 
1521 
11 
44 
46 
1936 
2116 
12 
53 
194 
2809 
37636 
13 
20 
94 
400 
8836 
14 
50 
95 
2500 
9025 
15 
92 
161 
8464 
25921 
16 
112 
154 
12544 
23716 
17 
71 
75 
5041 
5625 
18 
96 
57 
9216 
3249 
19 
86 
34 
7396 
1156 
20 
92 
118 
8464 
13924 
21 
75 
41 
5625 
1681 
22 
41 
145 
1681 
21025 
23 
102 
148 
10404 
21904 
24 
24 
117 
576 
13689 
25 
96 
177 
9216 
31329 
26 
108 
119 
11664 
14161 
27 
102 
186 
10404 
34596 
28 
35 
22 
1225 
484 
29 
46 
61 
2116 
3721 
30 
74 
75 
5476 
5625 
Total 

Mean 
Thusthe relaxationtreatment is not more effective than the pharmaceutical treatment,the variables are all effective at a level of significance of 0.05.

A researcher is interested to learn if there is a relationship between the level of interaction a women in her 20s has with her mother and her life satisfaction ranking. Below is a list of women who fit into each of four level of interaction. Conduct a OneWay ANOVA on the data to determine if a relationship exists.
No Interaction 
Low Interaction 
Moderate Interaction 
High Interaction 

X 
X 
X 
X 

2 
4 
3 
9 
3 
9 
9 
81 

4 
16 
3 
9 
10 
100 
10 
100 

4 
16 
5 
25 
2 
4 
8 
64 

4 
16 
1 
1 
1 
1 
5 
25 

7 
49 
2 
4 
2 
4 
8 
64 

8 
64 
2 
4 
3 
9 
4 
16 

1 
1 
7 
49 
10 
100 
9 
81 

1 
1 
8 
64 
8 
64 
4 
16 

8 
64 
6 
36 
4 
16 
1 
1 

4 
16 
5 
25 
3 
9 
8 
64 

247 
226 
316 
512 

Thecorrection for mean = CM
CalculatingSS total:
Calculatingthe treatment sum of squares, SST
Calculatingthe error sum of squares, SSE
CalculatingMST, MSE and their ratio F.
MST= the mean square of the treatments,
MSEmean square of errors denoted as
Theresulting ANOVA table is
Source 
SS 
DF 
MS 
F 
Treatments 
38.275 
3 
12.76 
1.57 
Error 
36 
8.125 

Total (corrected) 
330.775 

Correction Factor 
970.225 
Fromthe table, we have and thus the test statistic is much lower than the critical value wereject the alternate hypothesis of relationshipbetween the level of interaction a women in her 20s has with hermother and her life satisfaction and conclude that there is norelationship.

Is there a relationship between handedness and gender? A researcher collected the following data in hopes of discovering if handedness and gender are independent (Ambidextrous individuals were excluded from the study). Use the ChiSquare test for independence to explore this at a level of significance of 0.05.
LeftHanded x 
RightHanded y 

men 
13 
22 
women 
27 
18 
Theexpected count =in cellijunder the model of independenceand the observed count
Accordingto Preacher(2001) thechisquare test statistics formula is given by:
Thepvalue is less than 0.05, is 0.025 thus there is a relationshipbetween handedness and gender

A researcher is interested in studying the effect that the amount of fat in the diet and amount of exercise has on the mental acuity of middleaged women. The researcher used three different treatment levels for the diet and two levels for the exercise. The results of the acuity test for the subjects in the different treatment levels are shown below.
Diet 
  
  

Exercise 
<30% fat 
30% – 60% fat 
>60% fat 

<60 minutes 
4 
16 
3 
9 
2 
4 

4 
16 
1 
1 
2 
4 

2 
4 
2 
4 
2 
4 

4 
16 
2 
4 
2 
4 

3 
9 
3 
9 
1 
1 

60 minutes or more 
6 
36 
8 
64 
5 
25 

5 
25 
8 
64 
7 
49 

4 
16 
7 
49 
5 
25 

4 
16 
8 
64 
5 
25 

5 
25 
6 
36 
6 
36 


Perform a twoway analysis of variance and explain the results. (Show all work to receive full credit)
N=30
SSA,Sum of squares for factor A
=,n=5+5=10
SSA=6.2
SSB,Sum of squares for factor B =,n=5+5+5=15
SSB= 90.13

Find the effect size for each factor and the interaction and explain the results. (Show all work to receive full credit)
SSBN,SUM OF SQUARES BETWEEN GROUPS=
n=5
SSAXB,INTERACTION
SSWN,WITHIN
SSTOT,TOTAL=
DEGRESSOF freedom, for factor A
=
DEGRESSOF freedom for factor B
=
Toget the degree of interaction
=
DEGRESSOF freedom OF WITHIN, DFWN
DEGREEOF FREEDOM OF DSSTOT TOTAL=N1=301=29
FA=4.8443.32 IS SIGNIFICANT
FB=9.774.17 IS SIGNIFICANT
CRITICALVALUE= IS INSIGNIFICANT
References
Preacher,K. J. (2001). Calculation for the chisquare test: An interactivecalculation tool for chisquare tests of goodness of fit andindependence [Computer software].