Read the following paragraph and answer the questions that follow : A researcher has data (sample n = 500) on eight personality variables i.e. Activity, Sociability, Talkativeness, Warmth, Anxiety, Worry, Anger, and Depression. The correlation matrix was evaluated for factorability by Bartlett’s Test of Sphericity (p < .001) and KMO = .86. The eigenvalues were 3.2, 2.4, 0.54, 0.42, 0.41, 0.39, 0.37 and 0.27. She used principal component analysis instead of PAF. Using Guttman and Scree criteria retained factors. The varimax and promax rotations were used.
What is the percentage of variance first unrotated PCA shall explain ?
40%
In PCA on a correlation matrix, each standardized variable has variance 1, so the total variance equals the number of variables. Here there are eight personality variables, so the total variance is 8.
The eigenvalue of a component is the amount of total variance it explains. The proportion explained by the first unrotated component is its eigenvalue divided by the number of variables:
\(\dfrac{\lambda_1}{p} \times 100 = \dfrac{3.2}{8} \times 100 = 40\%\)
As a check, the eigenvalues sum to 3.2 + 2.4 + 0.54 + 0.42 + 0.41 + 0.39 + 0.37 + 0.27 = 8.0, confirming total variance of 8. (12.5% would be the share of a single variable, 1/8.)
Hence, the answer is 40%.
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Important Questions from Data Analysis - Teaching
Given below are two statements:
Statement I: In two tail test, 2.5% of the area of the normal curve falls on the right of 1.96 standard deviations above the mean and 2.5% to the left of 1.96 standard deviations below the mean.
Statement II: In one tail test, for testing the difference between large sample means, a z score of 1.64 is taken as Yardstick at the 5% level of significance.
In light of the above statements, choose the most appropriate answer from the options given below:
In a traditional extract–transform–load (ETL) workflow, data scrubbing performed during the transformation stage is:
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Temperature (°C) Number of stations 0–10 1 10–20 2 20–30 3 30–40 4 40–50 5 50–60 4 60–70 3 70–80 2 80–90 1 Using class midpoints, what is the estimated arithmetic mean temperature?
Temperature (°C) Number of stations 0–10 1 10–20 2 20–30 3 30–40 4 40–50 5 50–60 4 60–70 3 70–80 2 80–90 1 Using the standard grouped-data interpolation formula, what is the estimated median temperature?
