# A STATISTICS PROJECT 3

ASTATISTICSPROJECT 3

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Project3

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     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
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].