LEARN SPSS (Statistical Package for the Social Sciences) RESEARCH GRADE 9

Basics of SPSS

How to do statistical tests in SPSS:
T-TEST
Z-TEST
CHI-SQUARE
-GOODNESS OF FIT
-TEST OF INDEPENDENCE
ONE WAY ANOVA
TWO WAYANOVA
PEARSON CORRELATION COEFFICIENT
SPEARMAN

RESEARCH GRADE 9 CRSHS

1. GRADE 9
SPSS

2. INTRODUCTION
SPSS means “Statistical Package for the Social Sciences” and
was first launched in 1968.
SPSS is mainly used in the following areas like healthcare,
marketing, and educational research, market researchers,
health researchers, survey companies, education research,
government, marketing organizations, data miners, and many
others.
WHAT IS SPSS?

3. BASICS
There are two view modes in SPSS: Data View and Variable
View
Data View- is where we can create variables, enter data and
carry out statistical functions.
Each row represents one participant, or subject, or case.
Each column is dedicated to a single variable, or type of
measurement.

4. BASICS
There are two view modes in SPSS: Data View and Variable
View
Variable view- is used to create and define various variables.
Each row represent individual variable or define variable
Each Column represent the specific characteristic of
variable like Name, Type, Width, Decimals, Label, Missing,
Align, Measure etc.

5. BASICS
Vairable View:
In naming variables, you must be reminded of this things:
1. No spaces
You can instead use camel case
[HowToBasics] or underscores
[How_to_Basics]
2. It cannot begin with a special
character.
[#$%&]
3. It cannot begin with a number
Must begin with a letter
4. Each variable name must be
unique
5. 64 characters or less

6. VARIABLE VIEW
TYPE
The two basic types of variable that you will use
are numeric and string. This column enables you
to specify the type of variable.

7. VARIABLE VIEW
Decimals
It specifies how many decimal places will be
shown.

8. VARIABLE VIEW
Label
You can specifiy the details of the variable.
Up to 256 characters only.
It is like a mini description of the variable

9. VARIABLE VIEW
Value
Numbers represent which categories when
the variable represents a category

10. VARIABLE VIEW
Measure
It specifies the scale of
measurement that you
will be using.
Types of scales:
Nominal- categories that cannot be ranked
nationality, race, gender, ID number
Ordinal- categories that can be ranked
knowledge, skill, education level
Scale- Interval, Ratio
age, income, speed, weight, height

11. T-TEST

12. PROBLEM
Let's hire 8 people to join a spiritual retreat. We had them
answer a religiosity measure before the retreat and after it.

13. STEP 1: INPUT YOUR DATA
Input your values, variables in
the data view.

14. STEP 2: RUNNING T-TEST
Click the "Analyze" > "Compare Mean" > "Paired-Sample T Test"
O
Since our problem is a paired-
samples we will click on "Paried-
Sample T-test "

15. NOTES
ONE SAMPLE T-TEST
INDEPENDENT SAMPLE T-TEST
PAIRED-SAMPLE T-TEST

16. STEP 3
Click the "Post_test"
and then click the
"Arrow Sign", repeat to
"Pretest"
STEP 4
Click "Ok"

17. STEP 5: INTERPRETING RESULTS
Compare the Sig. [pvalue] to your level of significance in our case .05
p= .001 < .05
reject null hypothesis

18. Z-TEST

20. CODE [FOR ONE SAMPLE]
data list list / n sample_mean population_mean population_sd.
begin data
n x-bar Mu σ
end data.
Compute mean_difference = sample_mean - population_mean.
Compute square_root_n =SQRT(n).
Compute standard_difference = population_sd/square_root_n.
Compute z_statistic = mean_difference/standard_difference.
Compute chi_square = z_statistic*z_statistic.
Compute p_value = SIG.CHISQ(chi_square, 1).
Compute cohens_d = mean_difference/population_sd.
EXECUTE.
Formats z_statistic p_value cohens_d (f8.5).
LIST z_statistic p_value cohens_d.
This part of the code is
where you input your data
n= number sample
x-bar= sample mean
Mu= population mean
σ= population standard deviation

21. STEP 1
Open the syntax window and input the code.
STEP 2
Input your numbers/value in the code.
STEP 3
When finished, click run all.

22. CHI-SQUARE
GOODNESS OF FIT

23. PROBLEM
A clothing manufacturer wants to
determine whether customers prefer any
specific color over other colors in shirts.
He selects a random sample of 102 shirts
sold and notes the color. The table shows
the results.
At α = 0.10, is there a color preference?
H0: Customers have no color preference.
Ha: Customers show a color preference.

24. STEP 1: INPUT YOUR DATA
Input your values, variables in the data view and change
necessary characteristics in the variable view.

25. STEP 2: WEIGHTING CASES
Before calculating let us weight our variables so the
program know how to handle these data.
First go to Data> Weight cases

26. STEP 3: WEIGHTING CASES
Click on "Weight cases by" and select your variable [scale]
and place it in the Frequency Variable. Then click ok.

27. STEP 4: CALCULATING
Click on Analyze > Non Parametric Tests > Legacy Dialogs > Chi-Square

28. STEP 4: CALCULATING
Next move the your variables into
the Test Variable List box.
Check the "all categories equal"

29. STEP 4: CALCULATING
Click on OPTIONS, and select
descriptive.
Click Continue
Then click ok.

30. STEP 5: INTERPRETING RESULTS
In Descriptive Statistics, make sure that the data looks right [with
the word problem]
And since we are calculating a nominal data the mean and std.
deviation is pointless since the mean of the color of shirts doesnt
tell us anything.

31. Frequencies
Here we see the observed values that
we entered and the expected values.
Each expected value is the same
because we assume that the
occurrences are random and there is no
preference for any color of shirts.
The largest residual value is 26 for the
color white. Meaning that it differs the
most from the expected
STEP 5: INTERPRETING RESULTS

32. Test Statistics
We see the chi-square value is 59.765
and the probability of getting that chi-
square if the null hypothesis is true is
<0.001 which is less than α = 0.10.
So in conclusion, there is preference for
one of the color shirts. [Reject Null
Hypothesis.]
STEP 5: INTERPRETING RESULTS

33. Step 1: Analyze > Nonparametric Tests>
One Sample
Step 2: Click Run.
ANOTHER EASIER WAY!

34. So in this method, it tells you the pvalue and the conclusion.
You can use this method to check your answers!
ANOTHER EASIER WAY!

35. CHI-SQUARE
TEST OF INDEPENDENCE

36. PROBLEM
In the Star Trek TV series, Captain Kirk and the crew wear different
colored uniforms to identify the crewmember’s work area. Those who
wear red shirts have the unfortunate reputation of dying more often
than those who wear gold or blue shirts. Captain Kirk wants to know
whether or not the color of uniforms is associated with the deaths.
Null hypothesis: There are no relationships between the categorical
variables.
Alternative hypothesis: There are relationships between the
categorical variables.

37. STEP 1
Input your values, variables in the data view and change
necessary characteristics in the variable view.

38. Before calculating let us weight our variables so the
program know how to handle these data.
First go to Data> Weight cases
STEP 2: WEIGHTING CASES

39. STEP 2: WEIGHTING CASES
Click on "Weight cases by" and select your variable [scale]
and place it in the Frequency Variable. Then click ok.

40. STEP 3: CALCULATING
Analyze > Descriptive statistics > Cross tops

41. STEP 3: CALCULATING
Place your nominal
variables in their
coresponding boxes.
Row- independent
Column- dependent
dont worry abt putting the
wrong variables because it
doesnt matter!

42. STEP 3: CALCULATING
Click Statistics, and
select Chi-square.
Select also "Phi and
Cramer's V"
Click Continue.

43. STEP 3: CALCULATING
Click Cells, and select
Expected
Click Continue.
Click ok.

44. STEP 4: INTERPRETING RESULTS
The first table displays the number of missing cases in the
dataset. We can see that there are 0 missing cases in this
example.

45. STEP 4: INTERPRETING RESULTS
The second table displays a crosstab of the total number of
individuals by uniform color and status.

46. STEP 4: INTERPRETING RESULTS
The third table shows the results of the Chi-Square Test of
Independence. The test statistic is 6.189 and the corresponding two-
sided p-value is .045

47. STEP 5: CONCLUSION
From our statistical results, the p-values are less than 0.05. We can
reject the null hypothesis and conclude there is a relationship
between shirt color and deaths.

48. ONE WAY
ANOVA

49. In a statistics course, a group of psychology students were assigned
to three different tutorial groups taught by three different tutors,
Gary, Janni and Jin. We want to see if the students taught by
different tutors did equally well at the end semester exams.
PROBLEM

50. STEP 1: INPUT YOUR DATA
Input your values, variables in the
data view and change necessary
characteristics in the variable view.

51. STEP 2: RUNNING 1-WAY ANOVA
Click Analyze > Compare Means > One-Way ANOVA...

52. STEP 2: RUNNING 1-WAY ANOVA
Now put your dependent variable into the Dependent List box
and your Independent variable into the Factor box. Click “OK”.

53. Now put your dependent variable into the Dependent List box
and your Independent variable into the Factor box.
STEP 2: RUNNING 1-WAY ANOVA

54. Click on Options
Select the Descriptive, Homogeneity
of variance test, Brown-Forsythe test,
Welch test, Means plot
Click continue and OK
STEP 2: RUNNING 1-WAY ANOVA

55. Tests of Homogeneity of Variances
From our assumptions, The variances should be equal. and this table
will show you if its equal or not.
The Sig. value should be greater than our alpha level.
STEP 3: INTERPRETING RESULTS
Sig. value > Alpha
= equal variances

56. ANOVA
Since the p-value is .241 which is greater than our level of significance
[0.05] we failed to reject the null hypothesis.
STEP 3: INTERPRETING RESULTS
Sig. value < Alpha
= Reject null
Sig. value > Alpha
= Accept null

57. Since the p-value is .241, we reject the alternative hypothesis and
conclude that the mean scores of students from different tutorials
are the same at 5% significant level.
STEP 4: CONCLUSION

58. TWO WAY
ANOVA

59. PROBLEM
A teacher has given a math test to different boys and girls at
different age groups. Ten years old, eleven years old and twelve
years old. And he wanted to know if its the gender that is
causing the variation in scores or if its the age.
Hypothesis:
Ho1: Gender will have no significant effect on students score.
Ho2: Age will have no significant effect on students score.
Ho3: Gender and age interaction will have no significant effect on students
score.

60. STEP 1: INPUT YOUR DATA
Input your values, variables in the data view.

61. STEP 2: VARIABLE VIEW
Change necessary characteristics in the variable view.

62. STEP 3: SET VARIABLES
Click on analyze at the top
then click on
General Linear Model >
Univariate.

63. STEP 3: SET VARIABLES
Now click your dependent
variable, then click the
arrow to move that to the
dependent variable box.
Next move your
independent variables to
the fixed factors box.

64. STEP 4: MODEL
Click on model, and check
that type III is set for sum of
squares.
[Go back to the Univariate
window]

65. STEP 5: EM MEANS
Click all your factors [except for OVERALL] then click the
arrow to copy them Display Means for box.
Check the box for Compare Main effects then click
continue.

66. Go back to the main Univariate
window, and click Options
Check the box for Descriptive
statistics, Homogeneity tests
Here you can change your
significance level!
STEP 6: OPTIONS

67. STEP 7: PLOTS
Go back to the main Univariate
window, and click Plots.
Here you can make the plot of the
interaction effect. [Graph]
Dont forget to click add.

68. Go back to the main Univariate window, and click OK.
STEP 8:

69. STEP 9: INTERPRETING RESULTS
Theres alot of data shown but we will only focus on Tests of
Between-Subjects Effects for hypothesis testing.

70. STEP 9: INTERPRETING RESULTS
p= .035 < .05
reject null hypothesis
SPSS Results
1st Factor: Gender

71. STEP 9: INTERPRETING RESULTS
p= .006 < .05
reject null hypothesis
SPSS Results
2nd Factor: Age Group

72. STEP 9: INTERPRETING RESULTS
p= .556 > .05
fail to reject null hypothesis
SPSS Results
3rd Factor:
Gender_AgeGroup

73. Hypothesis:
Ho1: Gender will have no significant effect on students score. [REJECTED]
Ha1: Gender will have a significant effect on students score.
Ho2: Age will have no significant effect on students score. [REJECTED]
Ha2: Age will have a significant effect on students score.
Ho3: Gender and age interaction will have no significant effect on students
score. [FAIL TO REJECT NULL]
Ha3: Gender and age interaction will have a significant effect on students
score.
STEP 10: CONCLUSION

74. For Gender
As p value > .05 , Reject the Ho, There is a significant effect
on the student's score.
For Age Group
As p value > .05 , Reject the Ho, There is a significant effect
of on the student's score.
For Gender and Age Group Interaction
As p value < .05 , Accept the Ho, There is no significant
effect on the student's score.
STEP 10: CONCLUSION

75. PEARSON
CORRELATION
COEFFICIENT

76. Ho: There is no correlation
between participant ages and
blood total cholesterol levels.
Ha: There is a correlation
between participant ages and
blood total cholesterol levels.
PROBLEM

77. STEP 1: INPUT UR DATA
Input your values, variables in
the data view and change
necessary characteristics in the
variable view.

78. STEP 2: RUNNING PEARSON
Go to Analyze > Correlate >
Bivariate

79. STEP 2: RUNNING PEARSON
Click on your
variables and click
the arrow to move
them in the
Variables box. [all]
Make sure that
Pearson is ticked
under the title
Correlation
Coefficients

80. STEP 3: INTERPRETING RESULTS
By looking at the results in the table, it can be seen that the correlation
between age and blood cholesterol levels gave a Pearson Correlation
Coefficient (r) value of .882, which indicates a strong positive association
between the two variables.

81. STEP 3: INTERPRETING RESULTS
The P value of the association was 0.001, thus indicating a highly significant
result. Therefore, we will reject the null hypothesis.
p value < 0.05
reject null hypothesis

82. STEP 4: CONCLUSION
Therefore, there is a correlation between participant ages and blood total
cholesterol levels.

83. SPEARMAN

84. PROBLEM
The following are the grades of 7 students in English and in
Mathematics.
At 0.01 significance level, can we conclude that there is a significant
correlation between the grades of students in English and in Math?
Ho: rs = 0 (There is no correlation between grades in English and in
Mathematics.)
Ha: rs ≠ 0 (There is a correlation between grades in English and in
Mathematics.)

85. STEP 1: INPUT YOUR DATA
Input your values, variables in the data view.

86. STEP 2: VARIABLE VIEW
Change necessary characteristics in the variable view.

87. STEP 3: RUNNING SPEARMAN
Click Analyze > Correlate
> Bivariate

88. STEP 3: RUNNING SPEARMAN
Click on your variables
and click the arrow to
move them in the
variables box. [all]

89. STEP 3: RUNNING SPEARMAN
Then Check Spearman
from Correlation
coeffiecients.
Lastly, click OK

90. STEP 4: INTERPRETING RESULTS
Between english and math there is a -0.357 correlation
It means we have a negative slight correlation.

91. STEP 4: INTERPRETING RESULTS
Since our p value is .432 which is greater than our
significance level 0.01 and so we reject the null hypothesis.

92. STEP 4: INTERPRETING RESULTS

93. STEP 5: CONCLUSION
There is a significant relationship between the grades of the
students in Math and English. The correlation/relationship is slight
negative correlation.

94. THANK YOU