Research Methodology and Statistics in Psychology Questions and Answers

Answer: D

The PCA will result in a general factor.

Principal component analysis extracts components in order of the variance they account for, so the first unrotated component is the single dimension that captures the most variance shared by all eight variables. Most variables load on it, which makes it a general factor; rotation is then used to break it into more interpretable factors.

The other options are wrong. PCA components are orthogonal, not oblique, before rotation. PCA extracts the maximum possible variance, more than principal axis factoring, which analyses only the common variance. And a full PCA solution, with as many components as variables, accounts for 100% of the variance, not less.

Answer: D

The first unrotated component explains 40% of the variance.

In PCA of a correlation matrix each standardized variable contributes a variance of 1, so the total variance equals the number of variables, 8, which is also the sum of all the eigenvalues. A component's share is its eigenvalue divided by that total:

\[ \begin{aligned} \text{total} &= 3.2 + 2.4 + 0.54 + 0.42 \\ &\quad + 0.41 + 0.39 + 0.37 + 0.27 = 8 \\ \text{share} &= \frac{3.2}{8} = 0.40 = 40\% \end{aligned} \]

12.5% is one variable's share, 1/8, and 32% comes from treating the eigenvalue 3.2 as a share of 10 instead of 8. 10% has no basis in these figures.

Answer: C

The correct statement is that KMO uses correlations and partial correlations.

The Kaiser-Meyer-Olkin measure of sampling adequacy compares the sizes of the ordinary correlations with the sizes of the partial correlations among the variables. If the variables share common factors, the partial correlations (each pair with the others controlled) will be small compared with the correlations, and KMO approaches 1.

KMO cannot exceed 1, so 'low since it is below one' is wrong. A value of .86 is rated 'meritorious' by Kaiser's guidelines, so the data are well suited to factor analysis, and 1 - KMO being near zero is a sign of a high value. KMO is a descriptive index, not a significance test, so it needs no p-value.

Answer: B

Two factors can be retained.

The Guttman-Kaiser criterion retains components with eigenvalues greater than 1, the variance of a single standardized variable. Only the first two eigenvalues, 3.2 and 2.4, exceed 1; the third is 0.54.

The scree criterion looks for the point where the plot of eigenvalues levels off. Here the values fall steeply from 3.2 to 2.4 and then drop to 0.54, after which they stay nearly flat (0.42, 0.41, 0.39, 0.37, 0.27), so the elbow also comes after the second component. Both criteria agree, so 'one', 'six' and 'insufficient information' are wrong.

Answer: B

Varimax rotation gives identical pattern and structure matrices.

Varimax is an orthogonal rotation: the factors stay uncorrelated, so the factor correlation (phi) matrix is the identity matrix. The pattern matrix holds the regression weights of variables on factors, and the structure matrix holds their correlations with factors; when the factors are uncorrelated the two are the same, so a single loading matrix is reported.

Different pattern and structure matrices, and a phi matrix that departs from identity, are results of an oblique rotation such as the promax rotation also used here. Higher-order factor analysis needs correlated first-order factors, which an orthogonal rotation does not give.

Answer: A

The result shows that the population correlation matrix is not an identity matrix.

Bartlett's test of sphericity tests the null hypothesis that, in the population, the variables are uncorrelated, which means the population correlation matrix is an identity matrix (ones on the diagonal, zeros elsewhere). With \(p \lt .001\) the null hypothesis is rejected, so there are real correlations among the variables and factor analysis is worthwhile.

Like any significance test, it draws a conclusion about the population, not the sample: a sample correlation matrix is virtually never exactly an identity matrix. 'Population is identity matrix' is the hypothesis the test rejected.

Answer: C

The VIF measures multicollinearity.

The variance inflation factor of a predictor is \(1/(1 - R_j^{2})\), where \(R_j^{2}\) comes from regressing that predictor on all the others. It shows how much the variance of its regression coefficient is inflated by its overlap with the other predictors. A VIF of 1 means no overlap; values above about 4 or 5 are usually taken as a warning, so SPM's VIF of 4.2 signals that it is strongly correlated with the other predictors, probably CCFT.

The proportion of variance explained is \(R^{2}\). Heteroscedasticity is unequal spread of the residuals, checked with residual plots or tests such as Breusch-Pagan. The residual standard error is the typical size of a residual. None of these is what VIF measures.

Answer: A

The conclusion ignores linear dependence among predictors, that is, multicollinearity.

SPM and CCFT both measure intelligence, and SPM's VIF of 4.2 shows it is heavily predictable from the other predictors. When two predictors overlap this much, the variance they share is credited to only one of them in the coefficients, and the other's coefficient can come out non-significant even though it predicts the criterion well on its own. A non-significant b for SPM therefore does not show that SPM cannot predict creativity.

Linear dependence of the criterion on the predictors is what the regression is meant to find, so it is not ignored. There is no evidence of heteroscedasticity in the report, and the small gap between R² (.65) and adjusted R² (.63) has nothing to do with the judgement about SPM.

Answer: D

The degrees of freedom are 8 and 191.

Count the predictors, k: SPM, CCFT, the five big-five domains and the training dummy.

\[ \begin{aligned} k &= 1 + 1 + 5 + 1 = 8 \\ df_{reg} &= k = 8 \\ df_{res} &= n - k - 1 \\ &= 200 - 8 - 1 = 191 \end{aligned} \]

The extra 1 subtracted in the residual degrees of freedom is for the intercept, which the paragraph says the model includes.

'7 and 199' and '8 and 199' use \(n - 1\) as if it were the residual figure, which is in fact the total degrees of freedom, and 7 forgets one predictor. '9 and 198' counts the intercept as a regressor.

Answer: B

Adjusted R² corrects for increasing regressors.

Ordinary \(R^{2}\) can only rise when a predictor is added, even a useless one, because each new variable soaks up some chance variance. The adjusted value penalises every added predictor:

\[ R^{2}_{adj} = 1 - (1 - R^{2}) \frac{n - 1}{n - k - 1} \]

It rises only if a new predictor improves the fit by more than chance would. Sample size enters the formula, but only through its ratio to the number of predictors; the adjustment exists because predictors are added. It has nothing to do with the scale of measurement, and Spearman's rank correlation is unrelated.