Aantekeningen SPSs
Confidence interval outcome= If we were to repeat this experiment a large number of times, and
compute the CI for each one, 95% of them would contain the true mean of the population. For this
reason we are 95% confident that our interval [26.59; 27.22] may contain the true mean of the
population.
Corretation test dichotomous and continuous variables:
i), since vegetable intake is a continuous variable, and greentea consumption (although actually an
count variable, but has a sufficient number of categories). For situation ii) it is not sensible to
compute a correlation coefficient, as they are both dichotomous variables. The appropriate
alternative would be to do a chi-square test or fishers’s exact-test on the 2 by 2 contingency table.
Another alternative here, could be the 2-sample z-test.
A 2 by 2 table can be analyzed using a Chi-square test, which follows a Chi-square
distribution with 1 degree of freedom as the degrees of freedom is given by: (r-1)*(c-1),
where r stands for number of rows, and c for number of columns.
Now use SPSS to conduct the same test (hint: analyze>descriptives>crosstabs, and check
out ‘statistics’). Remember to paste the syntax.
The adjusted R2 is 0.071, meaning that we can expect to explain 7.1% of the individual variation in
processing speed in the population based on fish oil consumption
Graphs: ‘Most importantly; they do not indicate strong skewness.’
Regression WG2:
β=0
The slope is not constant (≠0), so there appears to be an association. However, we’d still need to
test if this slope is statistically significant, before we can say there is a significant association). The
intercept is 2.55, and the regression coefficient is 0.02.
Pearson correlation: [Analyze Correlate Bivariate]
Divide the unstandardized regression coefficient (B) by the unstandardized regression
coefficient’s standard error= de T-test
, In the Shapiro-Wilk test, the null hypothesis H0 is that the data are normally distributed,
while the alternative hypothesis Ha is that they are not normally distributed. If you run
such a test and find P-value < 0.05, does this mean your data violate the assumption?
Yes, it means that the conclusion of the test is to reject the H 0 that the residuals are normally
distributed.:
Shapiro wilk test significant look better at data non parametric?
The H0 of both the Kolmogorov-Smirnov and the Shapiro-Wilk is: the data are normally
distributed, with Ha being: the data are not normally distributed.
As all p-values, of both the Kolmorgorov-Smirnov and the Shapiro-Wilks test, are > .05,
H0 is not rejected and the data distributions in all groups can thus be considered normal.
Again: Officially one can never state this. Yet it is often done so, and it is justifiable. The
statement ”We consider the data normal” means in fact, “We consider the data not non-
normal, so let’s continue!”.
ANOVA WG 3:
Simple: Each group is compared to the first (or last)
1 vs 2, 1 vs 3, 1 vs 4
Repeated: each group is compared to the next group
1 vs 2, 2 vs 3, 3 vs 4
Helmert: each group is compared to the mean of all subsequent groups
1 vs 2+3+4, 2 vs 3+4, 3 vs 4
Levene test:
H0 of the Levene test is “all variances are equal”, Ha is “not all variances are equal”.
p value higher than 0.05 NS equal variance
o The test is not significant: F(3,86)=1.11, ns, so we conclude that the variances
can be considered equal.
Significant? Welch test/ brown test
Welch F(3,45.00)=1.11, ns
Brown-Forsythe F(3,69.19)=.66, ns
These tests indicate that the groups did not differ with respect to mean IQ.
the differences in IQ were not statistically significant, i.e., we were not able to establish a
difference in IQ between the 4 groups
Confidence interval outcome= If we were to repeat this experiment a large number of times, and
compute the CI for each one, 95% of them would contain the true mean of the population. For this
reason we are 95% confident that our interval [26.59; 27.22] may contain the true mean of the
population.
Corretation test dichotomous and continuous variables:
i), since vegetable intake is a continuous variable, and greentea consumption (although actually an
count variable, but has a sufficient number of categories). For situation ii) it is not sensible to
compute a correlation coefficient, as they are both dichotomous variables. The appropriate
alternative would be to do a chi-square test or fishers’s exact-test on the 2 by 2 contingency table.
Another alternative here, could be the 2-sample z-test.
A 2 by 2 table can be analyzed using a Chi-square test, which follows a Chi-square
distribution with 1 degree of freedom as the degrees of freedom is given by: (r-1)*(c-1),
where r stands for number of rows, and c for number of columns.
Now use SPSS to conduct the same test (hint: analyze>descriptives>crosstabs, and check
out ‘statistics’). Remember to paste the syntax.
The adjusted R2 is 0.071, meaning that we can expect to explain 7.1% of the individual variation in
processing speed in the population based on fish oil consumption
Graphs: ‘Most importantly; they do not indicate strong skewness.’
Regression WG2:
β=0
The slope is not constant (≠0), so there appears to be an association. However, we’d still need to
test if this slope is statistically significant, before we can say there is a significant association). The
intercept is 2.55, and the regression coefficient is 0.02.
Pearson correlation: [Analyze Correlate Bivariate]
Divide the unstandardized regression coefficient (B) by the unstandardized regression
coefficient’s standard error= de T-test
, In the Shapiro-Wilk test, the null hypothesis H0 is that the data are normally distributed,
while the alternative hypothesis Ha is that they are not normally distributed. If you run
such a test and find P-value < 0.05, does this mean your data violate the assumption?
Yes, it means that the conclusion of the test is to reject the H 0 that the residuals are normally
distributed.:
Shapiro wilk test significant look better at data non parametric?
The H0 of both the Kolmogorov-Smirnov and the Shapiro-Wilk is: the data are normally
distributed, with Ha being: the data are not normally distributed.
As all p-values, of both the Kolmorgorov-Smirnov and the Shapiro-Wilks test, are > .05,
H0 is not rejected and the data distributions in all groups can thus be considered normal.
Again: Officially one can never state this. Yet it is often done so, and it is justifiable. The
statement ”We consider the data normal” means in fact, “We consider the data not non-
normal, so let’s continue!”.
ANOVA WG 3:
Simple: Each group is compared to the first (or last)
1 vs 2, 1 vs 3, 1 vs 4
Repeated: each group is compared to the next group
1 vs 2, 2 vs 3, 3 vs 4
Helmert: each group is compared to the mean of all subsequent groups
1 vs 2+3+4, 2 vs 3+4, 3 vs 4
Levene test:
H0 of the Levene test is “all variances are equal”, Ha is “not all variances are equal”.
p value higher than 0.05 NS equal variance
o The test is not significant: F(3,86)=1.11, ns, so we conclude that the variances
can be considered equal.
Significant? Welch test/ brown test
Welch F(3,45.00)=1.11, ns
Brown-Forsythe F(3,69.19)=.66, ns
These tests indicate that the groups did not differ with respect to mean IQ.
the differences in IQ were not statistically significant, i.e., we were not able to establish a
difference in IQ between the 4 groups
