SPSS LAB # 5
ANOVA, Chi-square Analysis, Correlation Analysis
Open the file Remingtonsag.sav
What are the major factors customers have when selecting a restaurant?
A-DS-F
Variables select X1-X6
Stats box select mean
OK
What are your conclusions?___Food quality and Speed of Service are the most important factors for
customers selecting a restaurant followed by Prices
Next try to understand the perceptions of the three competitors?
Conduct an ANOVA analysis
A-CM-One way
Dependent variables X1-X6
Factor box X22
Options select descriptive
OK
Page 526
Draw a performance chart for Remingtons (Perceptual map) The two dimensions are: Variables and
Rating
More important
Excellent
Poor
Less important
How does Remington’s compare on the Variables? Better or worse?
X1________slightly better___________________________________
X2________better__________________________________
X3_____________better______________________________
X4__________________average__________________________
X5__________________worse_________________________
X6__________________better__________________________
What areas should Remington’s improve and why? Service because it is important
_____________________________________________________________________
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Run a follow up test using the Scheffe approach
A-CM-One way
Dependent variables X1-X6
Factor box X22
Post Hoc select Scheffe
Options select descriptive
OK
Examine X1
What is the mean value for X1 for each of the competitors?_____
X1 -- Large Portions
Scheffe
X22 -- Competitor
Most Familiar With
Longhorn
Subset for alpha = .05
N
1
65
Remington's
49
Outback
86
Sig.
2
4.48
5.02
5.27
1.000
.218
Means for groups in homogeneous subsets are displayed.
a Uses Harmonic Mean Sample Size = 63.264.
b The group sizes are unequal. The harmonic mean of the group sizes is used. Type I error levels are not guaranteed.
_________
is there any significant differences between the competitors? ___yes_____________
If so explain between who?______outback and longhorn and remingtons and longhorn __Scheffe test
significant at .05_____________________________________
What is the mean value of X2 for each of the competitors?_
X2 -- Competent Employees
Scheffe
X22 -- Competitor
Most Familiar With
Outback
Subset for alpha = .05
N
1
86
Longhorn
65
Remington's
49
2
3
1.85
3.75
4.51
Sig.
1.000
1.000
1.000
Means for groups in homogeneous subsets are displayed.
a Uses Harmonic Mean Sample Size = 63.264.
b The group sizes are unequal. The harmonic mean of the group sizes is used. Type I error levels are not guaranteed.
_________________________
Is there any significant differences between the competitors?____yes _____________________
Explain__________the are all significantly different_from each other__
_________________________________
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Page 551 Chi-square (X2) analysis allows us to test for significance between the frequency distributions for
two or more nominally scaled variables in a cross-tabulation to determine if there is any associations
Chi-square analysis assumes that no association exists between the nominal-scaled variables being
examined
Warning: The Chi-square results will be distorted if more than 20% of the cells have and expected count
of less than 5, or if any cell has an expected count of less than 1.
Open the file Santafegrill2ag
Do male customers travel farther than females to get to the Santa Fe Restaurant
What is the null Hypothesis? ___________________________________
What is the alternative Hypothesis? ______________________________
A-DS-C
Select x30 for the row variable
Select X32 for the column variable
Statistics button select chi-square box
Cells button select expected frequencies under counts
OK
x30 -- Distance Driven * X32 -- Gender Crosstabulation
X32 -- Gender
Males
x30 -- Distance
Driven
Less than 1 mile
Count
95
183
108.0
75.0
183.0
58
40
98
57.8
40.2
98.0
90
29
119
Expected Count
70.2
48.8
119.0
Count
236
164
400
236.0
164.0
400.0
Count
Expected Count
More than 3 miles
Total
Females
88
Expected Count
1 -- 3 miles
Total
Count
Expected Count
Compare the observed to the expected frequencies
Chi-Square Tests
Pearson Chi-Square
Likelihood Ratio
Linear-by-Linear
Association
N of Valid Cases
Value
22.616(a)
23.368
22.343
2
2
Asymp. Sig.
(2-sided)
.000
.000
1
.000
df
400
a 0 cells (.0%) have expected count less than 5. The minimum expected count is 40.18.
3
What is the Pearson Chi-Square value?______22.61_________
Is it significant?______yes________________
Should we reject the null hypothesis based on a criteria of .05 ?_________yes___________
Male customers drive farther than female to get to the Santa Fe Grill
Page 553
Relationships between variables can be described in several ways:
1. presence
2. Direction
3. Strength
4. Type
Correlation Analysis
Pearson correlation coefficient measures degree of linear association between two variables
It varies between -1.00 and 1.00 0 means there is no association
Several assumptions are made about the data you are analyzing:
1. two variables have been measured using interval or ratio scaled measures
2. the relationship is linear
3. variables come from bivariate normally distributed population
Determine if the relationship between satisfaction and likelihood to recommend the restaurant is
significant and positive.
A-Correlate-B
Transfer x22 and x24 into the variables box
Options select Means & Standard Deviations
OK
Descriptive Statistics
Mean
Std. Deviation
N
X22 -- Satisfaction
4.65
.955
400
X24 -- Likely to
Recommend
3.46
.930
400
Correlations
X22 -Satisfaction
X22 -- Satisfaction
Pearson Correlation
1
X24 -- Likely to
Recommend
.672(**)
Sig. (2-tailed)
N
X24 -- Likely to
Recommend
Pearson Correlation
.000
400
400
.672(**)
1
Sig. (2-tailed)
.000
N
400
400
** Correlation is significant at the 0.01 level (2-tailed).
Is there a relationship between the variables? ____yes______
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Is it positive or negative?___positive_______
Is it significant?_____________yes_at .01 level__
What is the Pearson Correlation coefficient?_____.672___________
Satisfaction is positively related to likely to recommend
When the correlation is weak there are two possibilities:
1. there simply is no systematic relationship between the two variables
2. the association exists but it is not linear
When you square the correlation coefficient you get the coefficient of determination r2
For the example above r2 = .672^2 =.452 meaning that approximately 45.2 percent of the variation in
likelihood to recommend is associated with satisfaction.
What if the correlation coefficient was .3 and What is the coefficient of determination?____
Would this be a meaningful result?________________
Spearman rank order correlation coefficient can be used when the variables are measured using ordinal
scales or Nominal Scales.
Management would like to determine if food quality is significantly more important selection factor than
service X26 and X29
Since this data is ordinal data use the Spearmen correlation statistic
A-C-B
Select the Spearman statistic
Correlations
Spearman's rho
X29 -- Service
Correlation Coefficient
Sig. (2-tailed)
N
X27 -- Food Quality
Correlation Coefficient
1.000
-.130(**)
.
.009
400
400
-.130(**)
1.000
Sig. (2-tailed)
.009
.
N
400
400
** Correlation is significant at the 0.01 level (2-tailed).
Is there is a relationship? ___yes___Customers who rank food quality as important tend to rank service
significantly lower____________Is it significant?_________yes_but very small__________
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