Business Statistics and Research Methods Questions and Answers

Q: Suppose \(\sigma_{1}\) = standard deviation of {1, 2, 3, 4, 5}\(\sigma_{2}\) = standard deviation of {11, 12, 13, 14, 15}then ...............

The two standard deviations are equal: \(\sigma_{1} = \sigma_{2}\).The second set is the first set with 10 added to every value. Adding a constant shifts the mean by the same constant, so every deviation from the mean is unchanged, and so is the standard deviation. Both sets have deviations \(-2, -1, 0, 1, 2\), giving\[ \sigma = \sqrt{\frac{4+1+0+1+4}{5}} = \sqrt{2} \]Option D confuses the standard deviation with the mean, which does rise by 10.

Q: Given that :Rank of X12345Rank of Y54321The Spearman's rank correlation is ................

Spearman's rank correlation is -1.The ranks are exactly reversed. The differences \(d = R_{X} - R_{Y}\) are \(-4, -2, 0, 2, 4\), so \(\sum d^{2} = 40\) and \(n = 5\):\[ \begin{aligned} \rho &= 1 - \frac{6 \sum d^{2}}{n(n^{2} - 1)} \\ &= 1 - \frac{6 \times 40}{5 \times 24} \\ &= 1 - 2 = -1 \end{aligned} \]A value of 1 would need identical ranks, and 0 or 1/2 would mean little or moderate agreement; complete reversal gives perfect negative correlation.

Q: Which of the following statements is always true for Poisson distribution ?

For a Poisson distribution, mean = variance.A Poisson distribution with parameter \(\lambda\) has mean \(\lambda\) and variance \(\lambda\), so the two are always equal. This is its distinguishing property, and it is used to check whether count data fit a Poisson model.Mean greater than variance is the binomial case, mean less than variance is typical of the negative binomial, and the standard deviation is \(\sqrt{\lambda}\), which equals the mean only when \(\lambda = 1\).

Q: In one-way analysis of variance suppose there are k treatments and n is the total number of observations then the degree of freedom for error sum of squares will be ............

The error degrees of freedom are n - k.In one-way ANOVA the total degrees of freedom, \(n - 1\), are split between treatments and error. Treatments take \(k - 1\), so error takes\[ (n - 1) - (k - 1) = n - k \]\(n - 1\) is the total, \(k - 1\) is for treatments, and \(nk - 1\) would apply only if \(n\) were the number of observations per treatment.

Q: As sample size increases sampling errors ...............

As sample size increases, sampling errors decrease.Sampling error arises because a sample is only part of the population. The standard error of the mean is\[ SE = \frac{\sigma}{\sqrt{n}} \]so it falls as \(n\) rises, and becomes zero when the whole population is covered. Non-sampling errors, such as mistakes in recording, may grow with a larger survey, but sampling error does not, so the other options are wrong.

Q: "................." means the practice of taking someone else's work or idea and passing them as one's own.

The practice is Plagiarism.Plagiarism is presenting someone else's words, ideas or work as one's own without acknowledgement. It is treated as serious research misconduct, and in India the UGC's 2018 regulations on academic integrity set penalties for it in theses and papers.A source is where information comes from, a script is a written text, and integrity is the value that plagiarism violates, so none of them names the practice.

Q: The advantages of ............. is that it avoids reinventing the wheel of research work.

The answer is Review of literature.A literature review shows what earlier studies have already found and how. The researcher can then build on that work, identify the gap to be filled and avoid repeating research already done, which is what 'avoiding reinventing the wheel' means.Sampling selects respondents, and objectives and hypotheses state what the study will do; none of them tells the researcher what has already been done.

Q: If for 10 observations \(\sum (x - 20) = 100\), then the arithmetic mean is ........ .

The arithmetic mean is 30.Expand the sum of deviations from 20 over the 10 observations:\[ \begin{aligned} \sum (x - 20) &= \sum x - 10 \times 20 = 100 \\ \sum x &= 300 \\ \bar{x} &= 300/10 = 30 \end{aligned} \]This is the assumed-mean method: the mean equals the assumed value plus the average deviation, 20 + 100/10 = 30. 10 is the average deviation alone, 20 is the assumed mean alone, and 120 adds the total deviation to 20 without dividing by the number of observations.

Q: State which of the following statements is true ?

The true statement is that the arithmetic mean is proper representative of the group if the standard deviation is smaller.The standard deviation measures how far the observations are scattered around the mean. A small standard deviation means the values are close together, so the mean is typical of them; a large one means the values are spread widely, and the mean may be far from most of them.So the mean is a poorer representative when the standard deviation is larger, and its representativeness does depend on dispersion. The size of the standard deviation says nothing about the size of the mean itself: two groups can have the same mean and very different spreads.

21.

Suppose \(\sigma_{1}\) = standard deviation of {1, 2, 3, 4, 5}
\(\sigma_{2}\) = standard deviation of {11, 12, 13, 14, 15}

then ...............

A.\(\sigma_{1} \gt \sigma_{2}\)
B.\(\sigma_{1} \lt \sigma_{2}\)
C.\(\sigma_{1} = \sigma_{2}\)
D.\(\sigma_{2} = \sigma_{1} + 10\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The two standard deviations are equal: \(\sigma_{1} = \sigma_{2}\).

The second set is the first set with 10 added to every value. Adding a constant shifts the mean by the same constant, so every deviation from the mean is unchanged, and so is the standard deviation. Both sets have deviations \(-2, -1, 0, 1, 2\), giving

\[ \sigma = \sqrt{\frac{4+1+0+1+4}{5}} = \sqrt{2} \]

Option D confuses the standard deviation with the mean, which does rise by 10.

Discuss in Forum

22.

Given that :

Rank of X	1	2	3	4	5
Rank of Y	5	4	3	2	1

The Spearman's rank correlation is ................

A.1
B.0
C.-1
D.\(\frac{1}{2}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

Spearman's rank correlation is -1.

The ranks are exactly reversed. The differences \(d = R_{X} - R_{Y}\) are \(-4, -2, 0, 2, 4\), so \(\sum d^{2} = 40\) and \(n = 5\):

\[ \begin{aligned} \rho &= 1 - \frac{6 \sum d^{2}}{n(n^{2} - 1)} \\ &= 1 - \frac{6 \times 40}{5 \times 24} \\ &= 1 - 2 = -1 \end{aligned} \]

A value of 1 would need identical ranks, and 0 or 1/2 would mean little or moderate agreement; complete reversal gives perfect negative correlation.

Discuss in Forum

23.

Which of the following statements is always true for Poisson distribution ?

A.mean \(\gt\) variance
B.mean = variance
C.mean \(\lt\) variance
D.mean = standard deviation

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

For a Poisson distribution, mean = variance.

A Poisson distribution with parameter \(\lambda\) has mean \(\lambda\) and variance \(\lambda\), so the two are always equal. This is its distinguishing property, and it is used to check whether count data fit a Poisson model.

Mean greater than variance is the binomial case, mean less than variance is typical of the negative binomial, and the standard deviation is \(\sqrt{\lambda}\), which equals the mean only when \(\lambda = 1\).

Discuss in Forum

24.

For testing equality of two population proportions we use ................. test.

A.F
B.Chi-square
C.t
D.z

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

Equality of two population proportions is tested with the z test.

For large samples the difference between two sample proportions is approximately normally distributed, so the test statistic is

\[ \begin{aligned} z &= \frac{p_{1} - p_{2}}{\sqrt{PQ \, S}}, \\ S &= \frac{1}{n_{1}} + \frac{1}{n_{2}} \end{aligned} \]

where \(P\) is the pooled proportion. The F test compares variances, the t test compares means in small samples, and the chi-square test is used for goodness of fit and independence, though it can also compare proportions in a contingency table.

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25.

In one-way analysis of variance suppose there are k treatments and n is the total number of observations then the degree of freedom for error sum of squares will be ............

A.n - 1
B.k - 1
C.nk - 1
D.n - k

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

The error degrees of freedom are n - k.

In one-way ANOVA the total degrees of freedom, \(n - 1\), are split between treatments and error. Treatments take \(k - 1\), so error takes

\[ (n - 1) - (k - 1) = n - k \]

\(n - 1\) is the total, \(k - 1\) is for treatments, and \(nk - 1\) would apply only if \(n\) were the number of observations per treatment.

Discuss in Forum

26.

As sample size increases sampling errors ...............

A.Increase
B.Decrease
C.Do not change
D.May increase or decrease

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

As sample size increases, sampling errors decrease.

Sampling error arises because a sample is only part of the population. The standard error of the mean is

\[ SE = \frac{\sigma}{\sqrt{n}} \]

so it falls as \(n\) rises, and becomes zero when the whole population is covered. Non-sampling errors, such as mistakes in recording, may grow with a larger survey, but sampling error does not, so the other options are wrong.

Discuss in Forum

27.

"................." means the practice of taking someone else's work or idea and passing them as one's own.

A.Plagiarism
B.Source
C.Script
D.Integrity

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The practice is Plagiarism.

Plagiarism is presenting someone else's words, ideas or work as one's own without acknowledgement. It is treated as serious research misconduct, and in India the UGC's 2018 regulations on academic integrity set penalties for it in theses and papers.

A source is where information comes from, a script is a written text, and integrity is the value that plagiarism violates, so none of them names the practice.

Discuss in Forum

28.

The advantages of ............. is that it avoids reinventing the wheel of research work.

A.Sampling
B.Review of literature
C.Objectives of the study
D.Hypothesis of the study

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The answer is Review of literature.

A literature review shows what earlier studies have already found and how. The researcher can then build on that work, identify the gap to be filled and avoid repeating research already done, which is what 'avoiding reinventing the wheel' means.

Sampling selects respondents, and objectives and hypotheses state what the study will do; none of them tells the researcher what has already been done.

Discuss in Forum

29.

If for 10 observations \(\sum (x - 20) = 100\), then the arithmetic mean is ........ .

A.20
B.30
C.10
D.120

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The arithmetic mean is 30.

Expand the sum of deviations from 20 over the 10 observations:

\[ \begin{aligned} \sum (x - 20) &= \sum x - 10 \times 20 = 100 \\ \sum x &= 300 \\ \bar{x} &= 300/10 = 30 \end{aligned} \]

This is the assumed-mean method: the mean equals the assumed value plus the average deviation, 20 + 100/10 = 30. 10 is the average deviation alone, 20 is the assumed mean alone, and 120 adds the total deviation to 20 without dividing by the number of observations.

Discuss in Forum

30.

State which of the following statements is true ?

A.Arithmetic mean is proper representative of the group if the standard deviation is smaller
B.Arithmetic mean is proper representative of the group if the standard deviation is larger
C.Arithmetic mean is proper representative of the group irrespective the standard deviation
D.Smaller the standard deviation, smaller will be the arithmetic mean

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The true statement is that the arithmetic mean is proper representative of the group if the standard deviation is smaller.

The standard deviation measures how far the observations are scattered around the mean. A small standard deviation means the values are close together, so the mean is typical of them; a large one means the values are spread widely, and the mean may be far from most of them.

So the mean is a poorer representative when the standard deviation is larger, and its representativeness does depend on dispersion. The size of the standard deviation says nothing about the size of the mean itself: two groups can have the same mean and very different spreads.

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