Analyzing Data using SPSS

1
Analyzing Data using SPSS
2
Testing for difference
3
Parametric Test
4
t-test

• Is used in a variety of situations involving
  interval and ratio variables.
• Independent Samples
• Dependent - Samples

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Independent-Samples T-Test

• What it does The Independent Samples T Test
  compares the mean scores of two groups on a given
  variable.

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• Where to find it Under the Analyze menu, choose
  Compare Means, the Independent Samples T Test.
  Move your dependent variable into the box marked
  "Test Variable." Move your independent variable
  into the box marked "Grouping Variable." Click on
  the box marked "Define Groups" and specify the
  value labels of the two groups you wish to
  compare.

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• Assumptions-The dependent variable is normally
  distributed. You can check for normal
  distribution with a Q-Q plot.-The two groups
  have approximately equal variance on the
  dependent variable. You can check this by looking
  at the Levene's Test. See below.-The two groups
  are independent of one another

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• HypothesesNull The means of the two groups are
  not significantly different.Alternate The means
  of the two groups are significantly different.

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SPSS Output

• Following is a sample output of an independent
  samples T test. We compared the mean blood
  pressure of patients who received a new drug
  treatment vs. those who received a placebo (a
  sugar pill).

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• First, we see the descriptive statistics for the
  two groups. We see that the mean for the "New
  Drug" group is higher than that of the "Placebo"
  group. That is, people who received the new drug
  have, on average, higher blood pressure than
  those who took the placebo.

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                                 Our

• Finally, we see the results of the Independent
  Samples T Test. Read the TOP line if the
  variances are approximately equal. Read the
  BOTTOM line if the variances are not equal. Based
  on the results of our Levene's test, we know that
  we have approximately equal variance, so we will
  read the top line

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• Our T value is 3.796.
• We have 10 degrees of freedom.
• There is a significant difference between the two
  groups (the significance is less than .05).
• Therefore, we can say that there is a significant
  difference between the New Drug and Placebo
  groups. People who took the new drug had
  significantly higher blood pressure than those
  who took the placebo.

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Example Independent samples t test

• A study to determine the effectiveness of an
  integrated statistics/experimental methods course
  as opposed to the traditional method of taking
  the two courses separately was conducted.
• It was hypothesized that the students taking the
  integrated course would conduct better quality
  research projects than students in the
  traditional courses as a result of their
  integrated training.
• Ho there is no difference in students
  performance as a result of the integrated versus
  traditional courses.
• H1 students taking the integrated course would
  conduct better quality research projects than
  students in the traditional courses

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Output SPSS

• Students taking the integrated course would
  conduct better
• quality research projects than students in the
  traditional courses

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Exercise1

• The following data were obtained in an experiment
  designed to check whether there is a systematic
  difference in the weights (in grams) obtained
  with two different scales.

Rock specimen Scale I Scale II
1 2 3 4 5 6 7 8 9 10 12.13 17.56 9.33 11.40 28.62 10.25 23.37 16.27 12.40 24.78 12.17 17.61 9.35 11.42 28.61 10.27 23.42 13.26 12.45 24.75
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• Use the 0.01 level of significance to test
  whether the difference between the means of the
  weights obtained with the two scales is
  significant
• Ho there is no significant difference between
  the means of the weight obtained with the two
  scales.
• H1 there is significant difference between the
  means of the weight obtained with the two scales.

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Exercise 2

• The following are the scores for random samples
  of size ten which are taken from large group of
  trainees instructed by the two methods.
• Method 1 teaching machine as well as some
  personal attention by an instructor
• Method 2 straight teaching-machine instruction

Method 1 81 71 79 83 76 75 84 90 83 78
Method 2 69 75 72 69 67 74 70 66 76 72
What we can conclude about the claim that the
average amount by which the personal attention
of an instructor will improve trainees score.
Use ?5.
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Paired samples t-test
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Paired Samples T Test

• What it does The Paired Samples T Test compares
  the means of two variables. It computes the
  difference between the two variables for each
  case, and tests to see if the average difference
  is significantly different from zero.

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Paired Samples T Test

• Where to find it Under the Analyze menu, choose
  Compare Means, then choose Paired Samples T Test.
  Click on both variables you wish to compare, then
  move the pair of selected variables into the
  Paired Variables box.

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Paired Samples T Test

• Assumption-Both variables should be normally
  distributed. You can check for normal
  distribution with a Q-Q plot.

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Paired Samples T Test

• HypothesisNull There is no significant
  difference between the means of the two
  variables.Alternate There is a significant
  difference between the means of the two variables

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SPSS Output

• Following is sample output of a paired samples T
  test. We compared the mean test scores before
  (pre-test) and after (post-test) the subjects
  completed a test preparation course. We want to
  see if our test preparation course improved
  people's score on the test.

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First, we see the descriptive statistics for both
variables.

• The post-test mean scores are higher than
  pre-test scores

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Next, we see the correlation between the two
variables

• There is a strong positive correlation. People
  who did well on the pre-test also did well on the
  post-test.

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• Finally, we see the results of the Paired Samples
  T Test. Remember, this test is based on the
  difference between the two variables. Under
  "Paired Differences" we see the descriptive
  statistics for the difference between the two
  variables

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To the right of the Paired Differences, we see
the t, degrees of freedom, and significance.
The t value -2.171 We have 11 degrees of
freedom Our significance is .053 If the
significance value is less than .05, there is a
significant difference.If the significance value
is greater than. 05, there is no significant
difference. Here, we see that the significance
value is approaching significance, but it is not
a significant difference. There is no difference
between pre- and post-test scores. Our test
preparation course did not help!
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Example

• Twenty first-grade children and their parents
  were selected for a study to determine whether a
  seminar instructing on inductive parenting
  techniques improve social competency in children.
  The parents attended the seminar for one month.
  The children were tested for social competency
  before the course began and were retested six
  months after the completion of the course.

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Hypothesis

• Ho there is no significant difference between
  the means of pre and post seminar social
  competency scores
• In other words, the parenting seminar has no
  effect on child social competency scores

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• There is a strong positive correlation. children
  who did well on the pre-test also did well on the
  post-test.

There is significant difference between pre- and
post-test scores. the parenting seminar has
effect on child social competency scores!
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Exercise 3

• The table below shows the number of words per
  minute readings of 20 student before and after
  following a particular method that can improve
  reading.

Student Pre Post
11 50 64
12 56 62
13 75 87
14 49 62
15 66 62
16 86 90
17 90 84
18 58 62
19 41 40
20 82 77
Student Pre Post
1 48 57
2 89 102
3 78 81
4 50 61
5 70 74
6 98 100
7 78 83
8 98 86
9 58 67
10 61 71
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• Using a 0.05 level of significance, test the
  claim that the method is effective in improve
  reading.

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Exercise 4

• The table below shows the weight of seven
  subjects before and after following a particular
  diet for two months
• Subject A B C D E F G
• After 156 165 196 198 167 199 164
• Before 149 156 194 203 153 201 152
• Using a 0.01 level of significance, test the
  claim that the diet is effective in reducing
  weight.

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One-WayANOVA

• Similar to a t-test, in that it is concerned with
  differences in means, but the test can be applied
  on two or more means.
• The test is usually applied to interval and ratio
  data types. For example differences between two
  factors (1 and 2).
• The test can be undertaken using the Analyze -
  Compare Means - One-Way ANOVA menu items, then
  select for appropriate variables.
• You will observe the One-Way ANOVA for factor 1
  and factor 2

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Procedure

• 1. You will need one column of group codes
  labelling which group your data belongs to. The
  codes need to be numerical, but can be labelled
  with text.
• 2. You will also need a column containing the
  data points or scores you wish to analyze.
• 3. Select One-way ANOVA from the Analyze and
  Compare Means menus.
• 4. Click on your dependent variables (data
  column) and click on the top arrow so that the
  selected column appears in the dependent list
  box.
• 5. Click on your code column (your condition
  labels) and click on the bottom arrow so that the
  selected column appears in the factor box.

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• 6. Click on Post Hoc if you wish to perform
  post-hoc tests.(optional).
• 7. Choose the type of post-hoc test(s) you wish
  to perform by clicking in the small box next to
  your choice until a tick appears. Tukey's and
  Scheffe's tests are commonly used.
• 8. Click on Dunnett to perform a Dunnett's test
  which allows you to compare experimental groups
  with a control group.Choose whether your control
  category is the first or last code entered in
  your code column.

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• The main output table is labelled ANOVA. The
  F-ratio of the ANOVA, the degrees of freedom and
  the significance are all displayed. The top value
  of the df column is the df of the factor, the
  bottom value is the df of the error term.
• Tukey's test will also try to find combinations
  of similar groups or conditions.
• In the Score table there will be one column for
  each pair of conditions that are shown to be
  'similar'. The mean of each condition within the
  pair are given in the appropriate column. The
  p-value for the difference between the means of
  each pair of groups is given at the bottom of the
  appropriate column.

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Example one-way ANOVA

• We would like to determine whether the scores on
  a test of aggression are different across 4
  groups of children (each with 5 subjects)
• Each child group has been exposes to differing
  amounts of time watching cartoons depicting toon
  violence

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At the 0.05 significance level, test the claim
that the four groups have the same mean if the
following sample results have been obtained.
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Output SPSS
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Exercise 5

• At the same time each day, a researcher records
  the temperature in each of three greenhouses. The
  table shows the temperatures in degree Fahrenheit
  recorded for one week.
• Greenhouse 1 greenhouse 2 greenhouse 3
• 73 71 61
• 72 69 63
• 73 72 62
• 66 72 61
• 68 65 60
• 71 73 62
• 72 71 59
• Use a 0.05 significance level to test the claim
  that the average temperature is the same in each
  greenhouse.

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Nonparametric Test
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Sign Test

• A sign test compares the number of positive and
  negative differences between related conditions

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Procedure

• 1. You should have data in two or more columns -
  one for each condition tested.
• 2. Select 2 Related Samples from the Analyze -
  Nonparametric Tests menu.
• 3. Click on the first variable in the pair and
  the second variable in the pair.
• The names of the variables appear in the current
  selections section of the dialogue box.
• 5. Click on the central selection arrow when you
  are happy with the variable pair selection.
• The chosen pair appairs in the Test Pair(s) List.
  
• Make sure the Sign box is ticked and remove the
  tick from the Wilcoxon box

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Example

• The data in table on the next slide are matched
  pairs of heights obtained from a random sample of
  12 male statistics students. Each student
  reported his height, then his weight was
  measured. Use a 0.05 significance level to test
  the claim that there is no difference between
  reported height and measured height.

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Reported and measured height of male statistics
student
Reported height 68 74 82.25 66.5 69 68 71 70 70 67 68 70
Measured height 66.8 73.9 74.3 66.1 67.2 67.9 69.4 69.9 68.6 67.9 67.6 68.8
Ho there is no significant difference between
reported heights and measured heights H1
there is a difference
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Output
Reject Ho. There is sufficient evidence to reject
the claim that no significant difference between
the reported and measured heights.
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Exercise 6

• Listed here are the right- and left-hand reaction
  times collected from 14 subject with right
  handed. Use 0.05 significance level to test the
  claim of no difference between the right hand-
  and left-hand reaction times.

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Right/left reaction times
Right 191 97 116 165 116 129 171 155 112 102 188 158 121 133
Left 224 171 191 207 196 165 171 165 140 188 155 219 177 174
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Wilcoxon

• The Wilcoxon test is used with two columns of
  non-parametric related (linked) data.
• Either one person has taken part in two
  conditions or paired participants (e.g. brother
  and sister) have taken part in the same
  condition.
• This is the non-parametric equivelant of the
  paired sample t-test

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Procedure

• 1. Put your data in two or more columns, one for
  each condition tested.
• 2. Select 2 Related Samples from Analyze -
  Nonparametric Tests menu.
• 3. Click on the first variable in the pair.
• 4. Click on the second variable in the pair.
• 5. Make sure the Wilcoxon box is ticked
• The Ranks table produced in the output window
  summarises the ranking process.
• In the Test Statistics table the Z statistic is
  the result of the Wilcoxon test.
• The p-value for this statistic is shown below it.
  This is the two tailed significance.

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Example

• Use the previous data to test the claim that
  there is no difference between reported heights
  and measured heights using Wilcoxon test at 0.05
  significance level.

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Output
Reject Ho. There is sufficient evidence to reject
the claim that no difference between reported
and measured heights.
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Mann-Whitney

• The Mann-Whitney test is used with two columns of
  independent (unrelated) non-paramteric data.This
  is the non-parametric equivalent of the
  independent samples t-test.

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Procedure

• Put all of your measured data into one column.
• 2. Make a second column that contains codes to
  indicate the group from which each value was
  obtained.
• 3. Select 2 Independent Samples from the Analyze
  - Nonparametric Tests menu.
• 4. Select the column containing the data you want
  to analyse and click the top arrow.
• 5. Select the Grouping Variable - the column
  which contains your group codes - and click the
  bottom arrow.
• 6. Make sure the Mann-Whitney U option is
  selected.

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• The output is produced in the output window.
• The top table summarises the ranking process.
• The result of the Mann-Whitney test is given at
  the top of the Test Statistics table.
• The two-tailed significance of the result is
  given in the same table.

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Example

• One study used x-ray computed tomography (CT) to
  collect data on brain volumes for a group of
  patients with obsessive-compulsive disorders and
  a control group of healthy persons. The following
  data shows sample results (in mm) for volumes of
  the right cordate.

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Volumes of the right cordate
Obsessive-compulsive patients 0.308 0.210 0.304 0.344 0.407 0.455 0.287 0.288 0.463 0.334 0.340 0.305
Control group 0.519 0.476 0.413 0.429 0.501 0.402 0.349 0.594 0.334 0.483 0.460 0.445
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Output
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Kruskal-Wallis

• examines differences between 3 or more
  independent groups or conditions.

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Procedure

• 1 Put all your measured data into one column.
• 2. Make a second column that contains codes to
  indicate the group from which each value was
  obtained.
• 3. Select K Independent Samples from the Analyze
  - Non-parametric Tests menu.
• 4. Select the grouping variable, the column that
  contains your group codes, then click on the
  bottom arrow.
• Make sure the Kruskal-Wallis box is checked
• In the output window the chi-square statistic is
  shown in the test statistic section, as is the
  P-value.

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Example

• We would like to determine whether the scores on
  a test of Spanish are different across three
  different methods of learning
• Method 1 classroom instruction and language
  laboratory
• Method 2 only classroom instruction
• Method3 only self-study in language laboratory.

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The following are the final examination scores of
samples of students from the three group
Method 1 94 88 91 74 86 97 Method 2 85 82 79
84 61 72 80 Method 3 89 67 72 76 69
At the 0.05 level of significance, test the null
hypothesis that the population sampled are
identical .
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Output SPSS
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Exercise 7

• The following are the miles per gallon which a
  test driver got in random samples of six tankfuls
  of each of three kinds of gasoline
• Gasoline 1 30 15 32 27 24 29
• Gasoline 2 17 28 20 33 32 22
• Gasoline 3 19 23 32 22 18 25
• Test the claim that there is no difference in the
  true average mileage yield of the three kinds of
  gasoline. (use 0.05 level of significance)

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Testing for Relationships
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Pearson's Correlation

• Pearson's correlation is a parametric test for
  the strength of the relationship between pairs of
  variables.

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• What it does The Pearson R correlation tells you
  the magnitude and direction of the association
  between two variables that are on an interval or
  ratio scale.

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• Where to find it Under the Analyze menu, choose
  Correlations. Move the variables you wish to
  correlate into the "Variables" box. Under the
  "Correlation Coefficients," be sure that the
  "Pearson" box is checked off.

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• Assumption -Both variables are normally
  distributed. You can check for normal
  distribution with a Q-Q plot.

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• HypothesesNull There is no association between
  the two variables.Alternate There is an
  association between the two variables.

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• SPSS Output
• Following is a sample output of a Pearson R
  correlation between the Rosenberg Self-Esteem
  Scale and the Assessing Anxiety Scale.

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SPSS creates a correlation matrix of the two
variables. All the information we need is in the
cell that represents the intersection of the two
variables
SPSS gives us three pieces of information -the
correlation coefficient-the significance-the
number of cases (N
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• The correlation coefficient is a number between
  1 and -1. This number tells us about the
  magnitude and direction of the association
  between two variables.
• The MAGNITUDE is the strength of the correlation.
  The closer the correlation is to either 1 or -1,
  the stronger the correlation. If the correlation
  is 0 or very close to zero, there is no
  association between the two variables. Here, we
  have a moderate correlation (r -.378).

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• The DIRECTION of the correlation tells us how the
  two variables are related. If the correlation is
  positive, the two variables have a positive
  relationship (as one increases, the other also
  increases). If the correlation is negative, the
  two variables have a negative relationship (as
  one increases, the other decreases). Here, we
  have a negative correlation (r -.378). As
  self-esteem increases, anxiety decreases

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Example

• The following data were obtained in a study of
  the relationship between the resistance (ohms)
  and the failure time (minutes) of certain
  overloaded resistors.
• Resistance 48 28 33 40 36 39 46 40 30 42 44 48
  39 34 47
• Failure time 45 25 39 45 36 35 36 45 34 39 51 41
  38 32 45
• Test the null hypothesis that there is a
  significant correlation between resistance and
  failure time.

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Output SPSS
There is significant positive correlation between
resistance and failure time, indicating that
failure time increases as resistance increases.
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Exercise 8

• An aerobics instructor believes that regular
  aerobic exercise is related to greater mental
  acuity, stress reduction, high self-esteem, and
  greater overall life satisfaction.
• She asked a random sample of 30 adult to fill out
  a series of questionnaire.
• The result are as followstest whether there is
  significant correlation between aerobic exercise
  and high self-esteem

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Subject Exercise Self-esteem Satisfaction stress
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 10 33 9 14 3 12 7 15 3 21 2 20 4 8 0 25 37 12 32 22 31 30 30 15 34 18 37 19 33 10 45 40 30 39 27 44 39 40 46 50 29 47 31 38 25 20 10 13 15 29 22 13 20 25 10 33 5 23 21 30
Subject Exercise Self-esteem Satisfaction stress
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 17 25 2 18 3 27 4 8 10 0 12 5 7 30 14 35 39 13 35 15 35 17 20 22 14 35 20 29 40 30 42 40 30 47 28 39 32 34 41 27 35 30 30 48 45 13 10 27 9 25 7 34 20 15 35 20 23 12 14 15
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The Spearman Rho correlation
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The Spearman Rho correlation

• What it does The Spearman Rho correlation tells
  you the magnitude and direction of the
  association between two variables that are on an
  interval or ratio scale.

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The Spearman Rho correlation

• Where to find it Under the Analyze menu, choose
  Correlations. Move the variables you wish to
  correlate into the "Variables" box. Under the
  "Correlation Coefficients," be sure that the
  "Spearman" box is checked off.

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The Spearman Rho correlation

• Assumption -Both variables are NOT normally
  distributed. You can check for normal
  distribution with a Q-Q plot. If the variables
  are normally distributed, use a Pearson R
  correlation.

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The Spearman Rho correlation

• HypothesesNull There is no association between
  the two variables.Alternate There is an
  association between the two variables.

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SPSS Output

• Following is a sample output of a Spearman Rho
  correlation between the Rosenberg Self-Esteem
  Scale and the Assessing Anxiety Scale.

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• SPSS creates a correlation matrix of the two
  variables. All the information we need is in the
  cell that represents the intersection of the two
  variables.
• SPSS gives us three pieces of information -the
  correlation coefficient-the significance-the
  number of cases (N)

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• The correlation coefficient is a number between
  1 and -1. This number tells us about the
  magnitude and direction of the association
  between two variables.
• The MAGNITUDE is the strength of the correlation.
  The closer the correlation is to either 1 or -1,
  the stronger the correlation. If the correlation
  is 0 or very close to 0, there is no association
  between the two variables. Here, we have a
  moderate correlation (r -.392).

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• The DIRECTION of the correlation tells us how the
  two variables are related. If the correlation is
  positive, the two variables have a positive
  relationship (as one increases, the other also
  increases). If the correlation is negative, the
  two variables have a negative relationship (as
  one increases, the other decreases). Here, we
  have a negative correlation (r -.392). As
  self-esteem increases, anxiety decreases.

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Example

• The following are the numbers of hours which ten
  students studied for an examination and the
  grades which they received

Number of hour studied grade in examination
9 5 11 13 10 5 18 15 2 8 56 44 79 72 70 54 94 85 33 65
Is there any relationship between number of
our studied and grade in examination
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Output SPSS
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Exercise 9

• The following table shows the twelve weeks sales
  of a downtown department store, x, and its
  suburban branch, y
• X 71 64 67 58 80 63 69 59 76 60 66 55
• Y 49 31 45 24 68 30 40 37 62 22 35 19
• Is there any significant relationship between x
  and y?

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Two way chi-square from frequencies

• A chi-square test is a non-parametric test for
  nominal (frequency data).
• The test will calculate expected values for each
  combination of category codes based on the null
  hypothesis that there is no association between
  the two variables.

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Procedure

• 1. You will need two columns of codes. Each value
  in each column provides a code to a group or
  criteria category within the appropriate
  variable. You should have one row for each
  combination of category code.
• 2. You will also need a column giving the
  frequency that each combination of codes is
  observed.
• Before carrying out your chi-square test you
  first need to tell SPSS that the numbers in your
  frequency column are indeed frequencies. You do
  this using weight cases...
• 3. Select Weight Cases from the Data menu.
• 4. Click the Weight cases by button.
• 5. Select the column containing your frequencies
  and click on the across arrow.

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• Click Crosstabs from the Analyze - Descriptive
  Statistics menu.
• 8. Select the first variable and click on the top
  arrow to move it into the Rows box.
• 9. Select the second variable and click on the
  middle arrow to move it into the Columns box.
• Click on Statistics to choose to perform a
  chi-square test on your data.
• 11. Select the chi-square option from the
  Crosstabs Statistics dialogue box.
• 12. Click on Continue when ready.
• 13. Click on Cells to choose to output the
  chi-square expected values.
• 14. Select the top left boxes to display both the
  Observed and the Expected values

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Two way chi-square from raw data

• 1. You will need two columns of codes. Each value
  in each column provides a code to a group or
  criteria category within the appropriate
  variable.
• 2. Click Crosstabs from the Analyze - Descriptive
  Statistics menu.
• 3. Select the first variable and click on the top
  arrow to move it into the Rows box.
• 4. Select the second variable and click on the
  middle arrow to move it into the Columns box.
• Click on Statistics to choose to perform a
  chi-square test on your data.
• 6. Select the chi-square option from the
  Crosstabs Statistics dialogue box

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Example

• Suppose we want to investigate whether there is a
  relationship between the intelligence of
  employees who have through a certain job training
  program and their subsequent performance on the
  job.
• A random sample of 50 cases from files yielded
  the following results

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Performance
Poor fair
good
8 8 3
5 10 7
1 3 5
Below average Average Above average
IQ
Test at the 0.01 level of significance whether on
the job performance of persons who have gone
through the training program is independent of
their IQ
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Exercise 10

• Suppose that a store carries two different
  brands, A and B, of a certain type of breakfast
  cereal. During a one-week, 44 packages were
  purchased and the results shows below
• brand A brand B
• Men 9 6
• Women 13 16
• Test the hypothesis that the brand purchased and
  the sex of the purchaser are independent.