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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.

The PCA shall :

This question was previously asked in
MH SET 2020 Paper 2 Philosophy Question Paper (27-Dec-2020)
The correct answer is

result in general factor.

Principal component analysis (PCA, Hotelling, 1933) extracts components in order of the variance they explain. The first unrotated component is the linear combination of all variables that captures the maximum possible variance, so every variable tends to load substantially on it; later components explain the remaining variance under the condition that they are uncorrelated with the earlier ones.

Because the first component absorbs the variance shared by all eight variables, the unrotated PCA solution typically shows a general factor on which all variables load, followed by bipolar components. In the passage the first eigenvalue (3.2) is clearly the largest, consistent with such a general component. This is why rotation (varimax or promax) is applied afterwards to obtain a simple, interpretable structure.

The other statements do not describe PCA. PCA components are extracted as orthogonal (uncorrelated), not oblique; obliqueness arises only from an oblique rotation such as promax. PCA analyses total variance (common plus unique), so it extracts more variance than principal axis factoring (PAF), which analyses only common variance; it is not second best in extracting variance. And a full PCA solution with as many components as variables reproduces 100% of the variance, not less.

Hence, the answer is result in general factor.

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Important Questions from Data Analysis - Teaching

  1. 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:

  2. In a traditional extract–transform–load (ETL) workflow, data scrubbing performed during the transformation stage is:

  3. Temperature (°C)Number of stations
    0–101
    10–202
    20–303
    30–404
    40–505
    50–604
    60–703
    70–802
    80–901

    Using the standard grouped-data interpolation formula with position 3N/4, estimate the third quartile Q₃ of the temperature distribution.

  4. Temperature (°C)Number of stations
    0–101
    10–202
    20–303
    30–404
    40–505
    50–604
    60–703
    70–802
    80–901

    Using class midpoints, what is the estimated arithmetic mean temperature?

  5. Temperature (°C)Number of stations
    0–101
    10–202
    20–303
    30–404
    40–505
    50–604
    60–703
    70–802
    80–901

    Using the standard grouped-data interpolation formula, what is the estimated median temperature?

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