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.
Which of the following is correct about KMO value ?
KMO uses correlations and partial correlations
The Kaiser-Meyer-Olkin (KMO) measure of sampling adequacy indicates whether the correlations among variables are suitable for factor analysis. It compares the size of the observed (zero-order) correlations with the size of the partial correlations (anti-image correlations) between each pair of variables after controlling for all others:
\(KMO = \dfrac{\sum\sum r_{ij}^2}{\sum\sum r_{ij}^2 + \sum\sum p_{ij}^2}\), where \(r_{ij}\) are correlations and \(p_{ij}\) are partial correlations.
If variables share common factors, partial correlations are small relative to correlations and KMO approaches 1. Kaiser's guidelines: 0.90s marvellous, 0.80s meritorious, 0.70s middling, 0.60s mediocre, 0.50s miserable, below 0.50 unacceptable. So KMO = .86 is meritorious, not low.
The other statements are wrong: KMO always lies between 0 and 1, so being below one does not make it low; 1 minus KMO being close to zero would actually mean KMO is high; and KMO is a descriptive index interpreted directly, with no p-value needed.
Hence, the answer is KMO uses correlations and partial correlations.
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Codes :
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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?
