A STATISTICS PROJECT 3

ASTATISTICSPROJECT 3

Author’sname

Institutionof affiliation

Project3

Answereach question completely, showing all your work.

  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 ( 1 being the lowest and 10 the highest).

x

y

xy

x2

y2

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&nbsp(correlationcoefficient):

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

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

  1. 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.

  1. 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&nbspof0.0.1 means 0.012x100= 0.01% of the&nbspvariance&nbsphoursof exercise per week predictedby the lifesatisfaction of the participants.

  1. 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,

  1. 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 t-test to determine if the relaxation treatment is more effective than the pharmaceutical treatment at a level of significance of 0.05.

x

y

x2

y2

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.

  1. 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 One-Way 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.

  1. 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 Chi-Square test for independence to explore this at a level of significance of 0.05.

Left-Handed x

Right-Handed y

men

13

22

women

27

18

Theexpected count =in cellijunder the model of independenceand the observed count

Accordingto Preacher(2001) thechi-square test statistics formula is given by:

Thep-value is less than 0.05, is 0.025 thus there is a relationshipbetween handedness and gender

  1. 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 middle-aged 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

&nbsp

&nbsp

Exercise

&lt30% fat

30% – 60% fat

&gt60% fat

&lt60 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

  1. Perform a two-way 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

  1. 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=N-1=30-1=29

FA=4.8443.32 IS SIGNIFICANT

FB=9.774.17 IS SIGNIFICANT

CRITICALVALUE= IS INSIGNIFICANT

References

Preacher,K. J. (2001). Calculation for the chi-square test: An interactivecalculation tool for chi-square tests of goodness of fit andindependence [Computer software].