University Statistics Project: Statistical Analysis with SPSS, PSY-380

This document presents a comprehensive statistical analysis project utilizing SPSS, addressing various research questions through different statistical tests. The project begins by examining the relationship between exercise hours and life satisfaction, using descriptive statistics, correlation, and linear regression. It then investigates the effects of relaxation therapy versus pharmaceutical treatment on insomnia patients through an independent samples t-test. Further analysis explores the impact of communication levels on daughters' life satisfaction via a one-way ANOVA. Additionally, the project examines the independence of gender and handedness using a Chi-square test. Finally, a two-way ANOVA is employed to study the combined effects of diet and exercise on mental acuity. Each analysis includes the interpretation of SPSS output, supporting conclusions with statistical evidence and relevant tables.

1STATISTICAL ANALYSIS USING SPSS
STATISTICAL ANALYSIS USING SPSS
Name of the Student:
Name of the University:
Author Note:

2STATISTICAL ANALYSIS USING SPSS
Answer 1
Here the data consists of hours of exercise per week and life satisfaction (rank in 1-10
scale) of 20 individuals. The main objective is to find a relationship between the workout hours
and happiness.
a. From the table1, it can be seen that the mean hours of physical activity per week is
8.85. This means that on average, an individual workout for 8.85 hours per week.
Table 1:
Descriptive Statistics
N Mean Std. Deviation Variance
Hours_of_Exercise 20 8.85 3.660 13.397
Valid N (listwise) 20
b. Table 1 shows that the variance and standard deviations of exercise hours are 13.40
and 3.66 respectively. Hence the variance shows that the practice hours of the
individuals do not have a wide range of variation. Moreover, standard deviation
implies that in most of the cases, the gym times are nearby the mean time.
c. Table 2 shows the linear association between the two variables- exercise duration and
happiness. Exercise duration and pleasures in life are negatively correlated, r= -0.103,
p=0.664.

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Table 2:
Correlations
Hours_of_Exercise Life_Satisfaction
Hours_of_Exercis
e
Pearson
Correlation
1 -.103
Sig. (2-tailed) .664
N 20 20
Life_Satisfaction Pearson
Correlation
-.103 1
Sig. (2-tailed) .664
N 20 20
d. Table 3 shows that the R square value for the linear regression model is 0.011. Hence,
it can be concluded that 1.1% variation in life satisfaction can be explained by the
workout hours per week of an individual.
Table 3:
Model Summaryb
Model R
R
Square
Adjusted R
Square
Std. Error of the
Estimate
Change Statistics
Durbin-
Watson
R Square
Change
F
Change df1 df2
Sig. F
Change
1 .103a .011 -.044 2.535 .011 .195 1 18 .664 2.052
a. Predictors: (Constant), Hours_of_Exercise
b. Dependent Variable: Life_Satisfaction
e. The linear regression formula taking exercise time as independent and life satisfaction
as dependent variable can be written as (Table 4),
Life Satisfaction=5.671−0.07 Hours of Exercise

4STATISTICAL ANALYSIS USING SPSS
that means change in the duration of physical activity time has inverse impact on a
person’s life enjoyment.
Table 4:
Coefficientsa
Model
Unstandardized Coefficients
Standardized
Coefficients
t Sig.B Std. Error Beta
1 (Constant) 5.671 1.516 3.740 .001
Hours_of_Exercise -.070 .159 -.103 -.441 .664
a. Dependent Variable: Life_Satisfaction

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Answer 2
Here the data is based on the patients suffering from insomnia. Researchers wanted to find
the effect of relaxation therapy and pharmaceutical treatment to the patients. For this, 60 patients
within 18 to 40 years age group with no health hazard were taken and randomly assigned to one
of this two therapy. The main aim is to find whether the relaxation remedy has more effect than
the medicinal treatment. Hence, the null hypothesis will be-
There is no significant dissimilarity in the effects of relaxation and pharmaceutical remedies.
For this, an independent sample t-test is performed. The outputs are given below,
Table 5:
Group Statistics
Treatment N Mean Std. Deviation Std. Error Mean
Score Relaxation 30 74.70 28.950 5.285
Pharmaceutical 30 102.30 53.596 9.785
Table 6:
Independent Samples Test
Levene's Test
for Equality of
Variances t-test for Equality of Means
F Sig. t df
Sig. (2-
tailed)
Mean
Difference
Std. Error
Difference
95% Confidence
Interval of the
Difference
Lower Upper
Scor
e
Equal
variances
assumed
19.734 .000 -2.482 58 .016 -27.600 11.121 -49.862 -5.338
Equal
variances
not
assumed
-2.482 44.595 .017 -27.600 11.121 -50.005 -5.195
Patients having medicinal treatment for insomnia have higher scores( M=74.7,SD=28.95)
than those who are having relaxation therapy (M=102.3, SD= 53.6), t(58)=-2.482, p=0.016.

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Answer 3
Here the main objective is to find how levels of communication between mother and
daughter impacts life satisfaction of a daughter at her 20s. For this study, it is assumed that there
is no significant difference in happiness due to change in interaction level. A one-way ANOVA
test is performed and the relevant results are described below:
Table 7:
ANOVA
Ratings
Sum of Squares df Mean Square F Sig.
Between Groups 38.275 3 12.758 1.570 .213
Within Groups 292.500 36 8.125
Total 330.775 39
An analysis of variance showed that the effect of various interaction level on life satisfaction is
not significant, F(3,36)= 1.570, p=0.213. Post hoc test is not necessary for this analysis.

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Answer 4
Here the goal is to check whether gender and handedness of a person are independent or
not. Hence, for this study, it is assumed that the gender and handedness are independent
variables. The necessary outcomes of the Chi-square test are given below:
Table 8:
Chi-Square Tests
Value df
Asymptotic
Significance (2-
sided)
Exact Sig. (2-
sided)
Exact Sig. (1-
sided)
Pearson Chi-Square 4.114a 1 .043
Continuity Correctionb 3.251 1 .071
Likelihood Ratio 4.153 1 .042
Fisher's Exact Test .071 .035
N of Valid Cases 80
a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 17.50.
b. Computed only for a 2x2 table
The Chi-Square test shows that the gender and handedness of a person are dependent,
χ(2, N=80)=4.114, p=0.043<0.05.

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Answer 5
Here the objective is to study the effect of fat level in diet and physical activity on the mental
sharpness of a middle-aged woman. For this, a two-way ANOVA is performed based on the
assumption that there is no effect of these two factors on the mental acuity. The relevant outputs
are given below:
a.
Table 9:
Tests of Between-Subjects Effects
Dependent Variable: Score
Source
Type III Sum of
Squares df Mean Square F Sig.
Corrected Model 114.800a 5 22.960 34.440 .000
Intercept 529.200 1 529.200 793.800 .000
Factor_A_Diet 6.200 2 3.100 4.650 .020
Factor_B_Exercise 90.133 1 90.133 135.200 .000
Factor_A_Diet *
Factor_B_Exercise
18.467 2 9.233 13.850 .000
Error 16.000 24 .667
Total 660.000 30
Corrected Total 130.800 29
a. R Squared = .878 (Adjusted R Squared = .852)
Mental acuity was subjected to a two way analysis of variance with two factors-
Fat in diet with 3 levels( <30%, 30%-60%, >60%) and Exercise with two levels(<60 min,
60 min or more). Both the factors and the interaction between factors are statistically
significant at 5% level of significance.
The main effect of factor A(fat in diet) has the value of F-ratio as F(2,24)=4.650,
p-value=0.020<0.05, showing that different levels of fat in meals effect the mental
sharpness significantly. The main effect of factor B( exercise) gives F-ratio as

9STATISTICAL ANALYSIS USING SPSS
F(1,24)=135.2, p=0.000<0.05, leading to notable effect on mental acuity. The interaction
effect is also significant with F(2,24)=13.850,p=0.000<0.05.
b. It is already shown that both the factors are significant. Since the amount of exercise has
less than two levels, post hoc test can not be performed. Hence a post hoc test is
performed for factor A to check which pair of levels of A differ noticeably.
Table 10:
Multiple Comparisons
Dependent Variable: Score
Tukey HSD
(I) Factor_A_Diet (J) Factor_A_Diet
Mean Difference
(I-J) Std. Error Sig.
95% Confidence Interval
Lower Bound Upper Bound
<30% fat 30%-60% fat -.7000 .36515 .156 -1.6119 .2119
>60% fat .4000 .36515 .526 -.5119 1.3119
30%-60% fat <30% fat .7000 .36515 .156 -.2119 1.6119
>60% fat 1.1000* .36515 .016 .1881 2.0119
>60% fat <30% fat -.4000 .36515 .526 -1.3119 .5119
30%-60% fat -1.1000* .36515 .016 -2.0119 -.1881
Based on observed means.
The error term is Mean Square(Error) = .667.
*. The mean difference is significant at the .05 level.
Tukey HSD test shows that the diet group (30%-60%,>60%) has p-
value=0.016<0.05. Hence , it can be concluded that effect of this diet group on the mental
acuity is significant.
c. The effect size of the factor A that is amount of fat in food is 10, based on the
observations. The results are shown in table 11.

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Table 11:
Score
Tukey HSDa,b
Factor_A_Diet N
Subset
1 2
>60% fat 10 3.7000
<30% fat 10 4.1000 4.1000
30%-60% fat 10 4.8000
Sig. .526 .156
Means for groups in homogeneous subsets are displayed.
Based on observed means.
The error term is Mean Square(Error) = .667.
a. Uses Harmonic Mean Sample Size = 10.000.
b. Alpha = .05.

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Bibliography
Leech, N. L., Barrett, K. C., & Morgan, G. A. (2014). IBM SPSS for intermediate statistics: Use
and interpretation. Routledge.
Lowry, R. (2014). Concepts and applications of inferential statistics.
Roberts, M., & Russo, R. (2014). A student's guide to analysis of variance. Routledge.
Schroeder, L. D., Sjoquist, D. L., & Stephan, P. E. (2016). Understanding regression analysis:
An introductory guide (Vol. 57). Sage Publications.