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Psychological Testing Questions and Answers

Direction for [ Question No: 1 To 5 ] :

Read the following paragraph and answer the five questions following it :

A psychologist wanted to develop a numerical ability test for the student population. She wrote eighty multiple-choice items, each item with five alternatives. The item analysis was carried out by finding discrimination index and item-remainder correlation for each item. Sixty items were retained in item analysis. The new version, with finally retained sixty items was administered to a new sample (N = 400) twice with a time gap of eight weeks. The test-retest correlation was found to be 0.22, significant at .01 level. The data obtained at the first administration of the sixty item version was also used for computing split-half reliability. The correlation between scores based on odd items and the scores based on even items was found to be 0.60. The split-half reliability coefficient was obtained after applying Spearman-Brown correction.

The scores obtained by the students were correlated with their marks in Mathematics in their annual examination which took place three months after the data collection. The product-moment correlation of 0.6 was obtained between test scores and mathematics. Norms were developed for the test. The normative sample yielded a mean of 40 and standard deviation of 8. The data were found to be normally distributed. As norms, percentile ranks were obtained.

1.

Which of the following correlation is suitable while computing item-remainder correlation in the above analysis ?

Exam Year(s)

UGC NET - 2014,

Answer: B

The suitable correlation is the point-biserial. In the item-remainder correlation each item, scored right or wrong, which is a true dichotomy, is correlated with the total score on the remaining items, a continuous variable. Dichotomous against continuous calls for the point-biserial coefficient. The phi coefficient needs two dichotomous variables, Spearman's rank difference needs two ranked sets, and multiple correlation relates one criterion to several predictors, none of which fits.

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Answer: A

The test has unsatisfactory test-retest reliability. The coefficient found after eight weeks was only 0.22, which is far below the 0.7 to 0.9 expected of an acceptable ability test, even though it is statistically significant at the .01 level with N = 400. Significance shows that the value is not zero, not that it is high. So the options claiming satisfactory test-retest reliability are wrong, and the split-half figure is a different, internal consistency matter.

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3.

What would be the percentile rank for a score of 48 ?

Exam Year(s)

UGC NET - 2014,

Answer: D

The percentile rank is 84. With a mean of 40 and a standard deviation of 8, a score of 48 is one standard deviation above the mean: \[ z = \frac{48 - 40}{8} = 1 \] In a normal distribution about 84 percent of scores fall below z = 1. Since the data were normal, the percentile rank is 84. A rank of 68 is the area within one standard deviation either side, and 16 is the area below z = -1.

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4.

What would be the split-half reliability coefficient after Spearman-Brown correction ?

Exam Year(s)

UGC NET - 2014,

Answer: C

The corrected split-half reliability is 0.75. The odd-even correlation of 0.60 is for half the test, so the Spearman-Brown formula is used to estimate the reliability of the full-length test: \[ r_{xx} = \frac{2r}{1 + r} = \frac{2(0.60)}{1.60} \] \[ r_{xx} = 0.75 \] The value 0.60 is the uncorrected figure, and 0.90 and 0.40 are not given by the formula.

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5.

What evidence has been obtained for the validity of the above test ?

Exam Year(s)

UGC NET - 2014,

Answer: A

The test shows satisfactory predictive validity. The scores were correlated with the students' mathematics marks in the annual examination, which took place three months after the data were collected. A criterion measured later is the mark of predictive validity, and the correlation of 0.6 is moderate to good and is satisfactory. Concurrent validity needs the criterion to be measured at about the same time as the test, so the options that mention it are wrong.

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Direction for [ Question No: 6 To 10 ] :

Read the following paragraph and answer the five questions which follow :

The notion of true score is at the heart of classical reliability theory. This model proposes that the obtained score on a variable is a sum of true score and the random error of measurement, the two components being independent. Using this model, various ways of computing reliability have been developed. One of the way of examining the reliability of a test is to test the homogeneity of test items or the internal consistency. The most widely used method of estimating the reliability is split-half technique, also called as odd-even reliability. In this context, Spearman Brown correction is important. Kuder-Richardson reliability and Chronbach Alpha are the two other methods of studying internal consistency.

Another important type of reliability is the interscorer reliability. This reliability is not useful for objective tests, but for certain types it is quite important. There are also other types of reliability such as parallel form reliability and test-retest reliability, the latter concerned with temporal stability of the test results.

Answer: A

Reliability is the ratio of true score variance to total variance.

In classical test theory the obtained score is the true score plus random error. Reliability is the proportion of the observed variance that is due to the true scores:

\[ r_{xx} = \frac{\sigma^{2}_{T}}{\sigma^{2}_{X}} \]

The second option, error variance over total variance, is the unreliability, and the other two ratios do not have a meaning of this kind.

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7.

For a newly developed test, the scores on odd and even part correlated by 0.4. After applying Spearman-Brown formula

Exam Year(s)

UGC NET - 2013,

Answer: C

The corrected reliability would be greater than 0.4 and less than 0.8.

The Spearman-Brown formula corrects the half-test correlation to the reliability of the full test:

\[ r_{full} = \frac{2r}{1+r} = \frac{2(0.4)}{1.4} \approx 0.57 \]

That value lies between 0.4 and 0.8. It is not 0.8, which would arise by simply doubling 0.4, and the correction always raises the value, so it cannot be 0.4 or less.

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Answer: C

For tests with dichotomous items Cronbach alpha and Kuder-Richardson reliability are equal.

The Kuder-Richardson formula 20 is a special case of Cronbach alpha for items scored one or zero, and when the items are dichotomous the two give the same value.

So alpha is neither greater nor smaller, and it does not vary from one to the other, which excludes the other three statements.

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9.

For which of the following types of tests interscorer reliability is relevant ?

1. Objective personality inventories
2. Projective tests
3. Achievement tests with descriptive items
4. Intelligence tests
Codes :

Exam Year(s)

UGC NET - 2013,

Answer: D

Interscorer reliability matters for projective tests and for achievement tests with descriptive items.

Where the scoring needs the judgment of the scorer, as with responses to inkblots or answers written in full sentences, different scorers may give different scores, so their agreement must be checked. That is statements 2 and 3.

Objective personality inventories and standard intelligence tests have fixed scoring keys, so the scorer makes no difference and interscorer reliability is not relevant to them.

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10.

If two groups are administered the same test with a time gap of eight weeks, we would be able to compute

Exam Year(s)

UGC NET - 2013,

Answer: D

None of the reliabilities named can be computed.

Test-retest reliability needs the same test given twice to the same group. Here two groups take the test with a gap of eight weeks, so there is no pair of scores for any individual and the test-retest correlation cannot be found.

Parallel form reliability needs two different forms, and interscorer reliability needs scores by two scorers. Neither is available either, so the last option is right.

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