UGC NET / JRF Unit 5: Business Statistics and Research Methods PYQ’s 25th Dec 2021 Shift 1
Question No.1
For a given set of paired data, the correlation and regression coefficients have been calculated as being equal to r, bxy and byx respectively. Now, each of the values of the x series is divided by 5. What effect does it have on each of these co-efficients?
- The three co-efficients remain unchanged
- There is no change in r but by x changes to byx/5 and bxy changes to 5 bxy
- Each of the co-efficients will be reduced to one fourth of its value.
- There is no change in r but by x changes to 5byx and bxy changes to bxy/5
Solutions:
The correct answer is There is no change in r but by x changes to 5byx and bxy changes to bxy/5
Key Points
Correlation coefficient= r
Regression coefficients= bxy and byx
As per the question
Series x is divided by 5
- There will be no change in the value of the correlation coefficient because the correlation coefficient is independent of scale and origin both.
- Change of origin means some value has been added or subtracted whereas the change in scale means some value is multiplied or divided.
- When x series is divided by 5, it will not affect the value of r.
- The regression coefficients are independent of change of origin but not of scale.
- So, the value of regression coefficients will change when x series is divided by 5.
- The value of regression coefficients bxy changes to bxy/5 and byx changes to 5byx.
Byx is in r and bay will change in byx/5 = r = bxy.byx−−−−−−−√bxy.byx
Question No.2
Which of the following statement is correct?
- Coefficient of variance = σˉX × 100
- If values in a series are negative, the standard deviation is also negative
- If value in a series multiplied by σ, the variance would be multiplied by 36
- Standard deviation is equal to square of variance.
Solutions:
The correct answer is Option 1
Coefficient of variance = σˉX × 100
Important Points
Coefficient of Variance:
- The coefficient of variation (CV) is the ratio of the standard deviation to the mean.
- The higher the dispersion around the mean, the higher the coefficient of variation.
- In most cases, it’s expressed as a percentage.
- It facilitates comparison between distributions of values whose measurement scales are not comparable without the use of units.
Formula:
CV = σˉX × 100
where,
σ = Standard Deviation
ˉX = Mean
let consider a dataset where the mean (ˉX
Unknown node type: span
Solution:
The coefficient of variation (CV) formula is:
CV=(σ‾X)×100
Given:
- Mean (‾X) = 25
- Standard Deviation (σ) = 5
Substitute the values into the formula:
CV=(525)×100
CV=0.2×100
CV=20%
Therefore, the coefficient of variation for the dataset is 20%.
Additional Information
If the value in a series is negative, the standard deviation is also negative
- The Standard Deviation formula is computed using squares of the numbers. The square of a number cannot be negative. Hence, Standard deviation cannot be negative. Therefore, option 2 is incorrect.
If the value in a series multiplied by σ, the variance would be multiplied by 36.
- This statement is incorrect as variance is the square of Standard deviation. So if the data set is multiplied by standard deviation, the variance would be multiplied by the Square of the standard deviation.
Standard deviation is equal to the square of variance.
- This statement is also incorrect as the variance is equal to the square of the standard deviation or the standard deviation is the square root of the variance.
Question No.3
For a poisson distribution variable X, P(X = 0) = 2P(X = 1), its standard deviation would be:
- 2–√2
- 0.5
- 0.5−−−√0.5
- 2
Solutions:
The correct answer is – √0.5
Key Points
- Poisson distribution
- The probability mass function (PMF) is given by: P(X = x) = (e−λ λx) / x!
- Given: P(X = 0) = 2 × P(X = 1)
- Using the PMF:
- P(X = 0) = e−λ, and
- P(X = 1) = λe−λ
- From the condition:
- e−λ = 2λe−λ ⟹ 1 = 2λ ⟹ λ = 0.5
- Standard Deviation of Poisson distribution is √λ = √0.5
Additional Information
- Mean and Variance of Poisson Distribution
- Both are equal to λ (mean = variance = λ)
- Applications of Poisson Distribution
- Used for modeling rare events like arrival of calls, number of accidents, etc.
- It is a discrete distribution defined for x = 0, 1, 2, …
- Standard Deviation
- Formula: SD = √λ, where λ is the average rate of success over a fixed interval
- In this case, since λ = 0.5, SD = √0.5
Question No.4
A test that contains a fair sample of the tasks and skills actually needed for the job in question is:
- Construct validity
- Content validity
- Test validity
- Criterion validity
Solutions:
The correct answer is Content Validity
Key Points Content Validity:
A test that is content valid is one that contains a fair sample of the tasks and skills actually needed for the job in question.
Example : Selecting students for dental school, many schools give applicants chunks of chalk, and ask them to carve something that looks like a tooth. If the content you choose for the test is a representative sample, then the test is content valid.
Additional Information Construct validity: Construct validity refers to how well the measure ‘behaves’ in accordance with theoretical hypotheses, and it measures how well the instrument’s results reflect the theoretical construct.
Test validity: The degree to which a test (such as a chemical, physical, or scholastic test) accurately measures what it is designed to assess is known as test validity.
Criterion validity: A measure’s criterion validity is an estimate of how well it agrees with a gold standard (i.e., an external criterion of the phenomenon being measured). The lack of gold standards in criterion validity assessment for questionnaire-based measures is a key issue.
Question No.5
Which of the following statement is correct about Mann-Whitney U-test?
(A) It is non parametric test
(B) It requires that samples are independent
(C) It can be used when populations involved are normally distributed
(D) It is used to test the null hypothesis that the two populations are involved are identical.
(E) It is always a two-tailed test.
Choose the correct answer from the options given below:
- A, B, D only
- A, C, D only
- A, D, E only
- A, B, E only
Solutions:
The correct answer is A, B, D only
Key Points Mann-Whitney U-test:
- The non-parametric Mann-Whitney U test is an alternative to the independent sample t-test. It’s a non-parametric test that compares two sample means from the same population and determines whether they’re equal.
- When the data is ordinal or the t-test assumptions are not met, the Mann-Whitney U test is usually utilised.
- Since, the results are reported as group rank differences rather than group mean differences, it might be difficult to interpret the Mann-Whitney U.
- In hypothesis testing, Statistical tests are used to check whether the null hypothesis is rejected or not rejected. It is used to test the null hypothesis that the two populations are involved are identical. Hence, D is true.
Important Points Mann-Whitney U test is a non-parametric test, so it does not assume any assumptions related to the distribution of scores. Hence, A is true:
There are, however, some assumptions that are assumed:
- The sample drawn from the population is random.
- Independence within the samples and mutual independence is assumed. That means that an observation is in one group or the other (it cannot be in both). Hence, B is true.
- Ordinal measurement scale is assumed.
Question No.6
Which of the following are correct about complementary events?
(A) Mutually exclusive
(B) Independent
(C) Such that their probabilities add up to 1.
(D) Collective exhaustive.
Choose the most appropriate answer from the options given below:
- A, B, C only
- A, B, D only
- B, C, D only
- A, C, D only
Solutions:
The correct answer is A, C, D only
Key Points
Complementary Events:
- Complementary events are two occurrences that occur only if and only if the other does not occur.
- Two events must be mutually exclusive and exhaustive in order to be characterised as complimentary.
Important Points
- The sum of probabilities of complementary events must be equal to 1.
P(A) + P(A’) = 1 - Complementary events can take place only when there are exactly two outcomes.
- The complement and the event are mutually exclusive. This signifies that no results are shared between the event and its complement.
- An event and its complement are also collective exhaustive. This signifies that the event and its complement contain all the sample space’s outcomes.
Question No.7
Arrange the following stages of processing of data in a correct sequence:
(A) Coding
(B) Editing
(C) Tabulation
(D) Classification
(E) Using Percentages
Choose the correct answer from the options given below:
- (A), (B), (C), (D), (E)
- (D), (A), (B), (C), (E)
- (C), (D), (A), (B), (E)
- (B), (A), (D), (C), (E)
Solutions:
The correct answer is (B), (A), (D), (C), (E)
Key Points Stages of processing of data:
Stage 1: Editing of Data:
- The editing of data is a process of examining the raw data to detect errors and omissions and to correct them, if possible, to ensure legibility, completeness, consistency and accuracy.
- The recorded data must be legible so that it could be coded later. An illegible response may be corrected by getting in touch with people who recorded it, or alternatively it may be inferred from other parts of the question.
Stage 2: Coding of Data:
- Coding is the process of assigning some symbols (either) alphabetical or numerals or (both) to the answers so that the responses can be recorded into a limited number of classes or categories.
- The classes should be appropriate to the research problem being studied.
- They must be exhaustive and must be mutually exclusive so that the answer can be placed in one and only one cell in a given category.
Stage 3: Classification of Data:
- Classification condenses the data, facilitates comparisons, helps to study the relationships and facilitates in statistical treatment of data.
- The classification should be unambiguous and mutually exclusive and collectively exhaustive.
- Further, it should not only be flexible but also suitable for the purpose for which it is sought.’
- Classification can either be according to attributes or numerical characteristics
Stage 4: Tabulation of Data:
- The tabulation is used for summarization and condensation of data.
- It aids in analysis of relationships, trends and other summarization of the given data.
- The tabulation may be simple or complex.
- Simple tabulation results in one-way tables, which can be used to answer questions related to one characteristic of the data.
- The complex tabulation usually results in two way tables, which give information about two interrelated characteristics of the data; three way tables which give information about three interrelated characteristics of data; and still higher order tables, which supply information about several interrelated characteristics of data.
Stage 5: Using Percentages:
The tabulated data is converted into percentages and useful results are drawn from it
Question No.8
For a very large sample size the ratio between SEX̄ and σ is 8 ∶ 40. Determine the sample size n:
- 25
- 5
- 4
- 16
Solutions:
The correct answer is 25
Important Points
Formula: SEX¯SEX¯ = σ / √n
Therefore, √n = σ / SEX¯SEX¯
√n = 40 / 8
√n = 5
n = 25
Therefore, For a very large population the ratio between SEX¯SEX¯ and σ is 8 : 40. The sample size n will be 25
Question No.9
Which of the following statements are correct about Chi-Square test?
(A) The only parameter of a Chi-Square distribution is its number of degrees of freedom.
(B) The null hypothesis in given Chi-Square test is rejected when calculated value of variable exceed its critical value.
(C) The rejection region in a goodness of fit test lies only in the right tail of the distribution.
(D) The Chi-Square test is a parametric test.
(E) At α =.05 and V = 1 the critical value of X2 is equal to Z-value at the same level of significance.
Choose the most appropriate answer from the options given below:
- B, D, E only
- A, C, E only
- A, B, C only
- B, C, D only
Solutions:
The correct answer is A, B, C only
Key Points
Chi Square Test:
- The chi-square test is a statistical test that compares observed and expected outcomes.
- The goal of this test is to figure out whether a disparity between observed and expected data is due to chance or a relationship between the variables you’re looking at.
- As a result, using a chi-square test to better understand and interpret the relationship between our two category variables is an ideal choice.
Important Points
Chi-square is a statistical test used to check the difference between categorical variables from a random sample to test the goodness of fit between expected and observed results. It was discovered by Carl Pearson.
X2=∑(Oi−Ei)2EiX2=∑(Oi−Ei)2Ei
where,
X2=X2= Chi Squared
Oi=Oi= Observed Value
Ei=Ei= Expected Value
It is never negative
Note: Range 0 to ∞
Calculated Value < Table Value – H0 is accepted
Calculated Value > Table Value – H0 is rejected
This is the only parameter of independent number of quantities, in which the null hypothesis is rejected.