Sampling Distributions Questions and Answers

Q: Let \(X\) and \(Y\) be independently distributed central chi-squared random variables with degrees of freedom \(m\ (\ge 3)\) and \(n\ (\ge 3)\), respectively. If \(E\left(\frac{X}{Y}\right) = 3\) and \(m + n = 14\), then \(E\left(\frac{Y}{X}\right)\) is equal to

\(E\left(\frac{Y}{X}\right)\) is equal to \(\frac{5}{7}\).For a chi-squared variable with \(k\) degrees of freedom, \(E(\chi_{k}^{2}) = k\) and, for \(k \gt 2\), \(E\left(\frac{1}{\chi_{k}^{2}}\right) = \frac{1}{k - 2}\). By independence:\[ \begin{aligned} E\left(\tfrac{X}{Y}\right) &= \tfrac{m}{n - 2} = 3, \quad m + n = 14 \\ 14 - n &= 3n - 6 \Rightarrow n = 5,\ m = 9 \\ E\left(\tfrac{Y}{X}\right) &= \tfrac{n}{m - 2} = \tfrac{5}{7} \end{aligned} \]Taking \(E(Y/X)\) as the reciprocal \(\frac{1}{3}\) is the common mistake: the expectation of a reciprocal is not the reciprocal of the expectation, and \(\frac{1}{3}\) is not offered.

Q: Let \(\{X_{i}\}\) be a sequence of independent Poisson(\(\lambda\)) variables and let \(W_{n} = \frac{1}{n}\sum_{i=1}^{n} X_{i}\). Then the limiting distribution of \(\sqrt{n}(W_{n} - \lambda)\) is the normal distribution with zero mean and variance given by

The variance is \(\lambda\).A Poisson(\(\lambda\)) variable has mean \(\lambda\) and variance \(\lambda\). \(W_{n}\) is the sample mean of independent, identically distributed variables with finite variance, so by the central limit theorem\[ \begin{aligned} \sqrt{n}(W_{n} - \lambda) &\to N(0, \sigma^{2}) \\ &= N(0, \lambda) \end{aligned} \]The scaling by \(\sqrt{n}\) is exactly what keeps the variance at \(\sigma^{2}\) rather than shrinking it to \(\sigma^{2}/n\). \(\sqrt{\lambda}\) is the standard deviation, not the variance; \(\lambda^{2}\) squares the variance again; and 1 is the variance only after dividing by \(\sqrt{\lambda}\) as well.

Q: A combination of numbers has geometric mean as 6.0 and arithmetic mean as 6.3. These numbers are :

The numbers are 4, 6, 9.Test them against both means:\[ \begin{aligned} GM &= \sqrt[3]{4 \times 6 \times 9} = \sqrt[3]{216} = 6.0 \\ AM &= \frac{4 + 6 + 9}{3} = \frac{19}{3} \approx 6.3 \end{aligned} \]For 6, 6, 6 both means are exactly 6, since the AM equals the GM only when all values are equal. For 5, 6, 8 the AM is 6.33 but the GM is \(\sqrt[3]{240} \approx 6.21\), not 6.0. As 4, 6, 9 fits both, 'None of the above' is wrong.

Q: The sampling procedure in which an interviewer is asked to interview 25 teachers, 50 public servants and 25 farmers is called :

This is quota sampling.In quota sampling the interviewer is given a fixed number of respondents to interview from each category, here 25 teachers, 50 public servants and 25 farmers, but chooses the actual individuals himself, usually as convenient. It is a non-probability method.Stratified sampling also divides the population into groups, but selects units within each stratum at random, not by the interviewer's choice. Accidental (convenience) sampling sets no quotas, and spatial sampling selects locations in an area.

Q: Which of the following is not an example of non-sampling risk ?

The one that is not non-sampling risk is obtaining an unrepresentative sample.In auditing, sampling risk is the risk that the sample examined does not represent the population, so the auditor's conclusion would differ from what testing every item would show. An unrepresentative sample is the very definition of sampling risk.Non-sampling risk comes from everything else: using an audit procedure that does not suit the objective, failing to recognise an error in an item examined, or misjudging the results. These could happen even if every item were tested.

Q: When an investigator wants a sample containing in units which possess a rare attribute. The appropriate sampling procedure is :

The appropriate procedure is inverse sampling.When an attribute is rare, a fixed-size sample may contain few or none of the units of interest. In inverse sampling (introduced by J.B.S. Haldane) units are drawn one at a time until a fixed, predetermined number of units with the rare attribute has been found. The sample size is therefore random, and the proportion is estimated from it.Simple random, stratified and systematic sampling all fix the sample size in advance, so they can easily miss a rare attribute altogether.

Q: The variance value for an F-distribution with parameters V1 = 5 and V2 = 9 is :

The variance is 1.587.For an F-distribution with \(\nu_{1}\) and \(\nu_{2}\) degrees of freedom (here 5 and 9; the variance exists only when \(\nu_{2}\) is greater than 4):\[ \begin{aligned} V &= \frac{N}{D} \\ N &= 2\nu_{2}^{2}(\nu_{1} + \nu_{2} - 2) \\ D &= \nu_{1}(\nu_{2} - 2)^{2}(\nu_{2} - 4) \end{aligned} \]\[ \begin{aligned} V &= \frac{2(81)(12)}{5(49)(5)} \\ &= \frac{1944}{1225} = 1.587 \end{aligned} \]The mean of this distribution is \(\nu_{2}/(\nu_{2} - 2) = 9/7 = 1.286\), which none of the options gives; 1.378, 2.578 and 1.498 do not follow from the formula.

Q: The first group sample has 100 items with mean 15 and standard deviation 3.0. If the whole group has 250 items with mean 15.6 and standard deviation \(\sqrt{13.44}\). What is the standard deviation of the second group ?

The standard deviation of the second group is 4.The second group has 250 - 100 = 150 items. Its mean, from the combined total:\[ \begin{aligned} \bar{x}_{2} &= \frac{250(15.6) - 100(15)}{150} \\ &= \frac{2400}{150} = 16 \end{aligned} \]So \(d_{1} = 15 - 15.6 = -0.6\) and \(d_{2} = 16 - 15.6 = 0.4\). The combined variance is 13.44, since the combined SD is \(\sqrt{13.44}\):\[ \begin{aligned} 250(13.44) &= 100(9 + 0.36) \\ &\quad + 150(\sigma_{2}^{2} + 0.16) \\ 3360 &= 936 + 24 + 150\sigma_{2}^{2} \\ \sigma_{2}^{2} &= 16,\quad \sigma_{2} = 4 \end{aligned} \]Check the others: with a second-group SD of 2, 6 or 8 the combined variance would come out as 6.24, 25.44 or 42.24 instead of 13.44.

Which of the given options best describes the truthfulness of the following statements? i) In the probability theory, an outcome is the result of a random experiment. ii) The purpose of statistical inference is to estimate the uncertainty or sample to sample variation. iii) A sampling error is the difference between the true population values (called parameters) and the measured sample values (called statistics).

A.i-True, ii-True, iii-True
B.i-False, ii-True, iii-True
C.i-True, ii-False, iii-True
D.i-False, ii-False, iii-True

Exam Year(s)

DSSSB TGT - 2023,

Answer: A

All three statements are true. In probability theory an outcome is the result of a random experiment. Statistical inference aims to estimate the uncertainty, or the sample to sample variation, in what is learnt from a sample. A sampling error is the difference between the true population values (parameters) and the measured sample values (statistics). The other options mark at least one of these as false, which is wrong. An outcome such as getting a 4 on one roll of a die is a result of a random experiment, which is the idea in statement i.

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Let \(X\) and \(Y\) be independently distributed central chi-squared random variables with degrees of freedom \(m\ (\ge 3)\) and \(n\ (\ge 3)\), respectively. If \(E\left(\frac{X}{Y}\right) = 3\) and \(m + n = 14\), then \(E\left(\frac{Y}{X}\right)\) is equal to

A.\(\frac{2}{7}\)
B.\(\frac{3}{7}\)
C.\(\frac{4}{7}\)
D.\(\frac{5}{7}\)

Exam Year(s)

GATE Mathematics - 2015,

Answer: D

\(E\left(\frac{Y}{X}\right)\) is equal to \(\frac{5}{7}\).

For a chi-squared variable with \(k\) degrees of freedom, \(E(\chi_{k}^{2}) = k\) and, for \(k \gt 2\), \(E\left(\frac{1}{\chi_{k}^{2}}\right) = \frac{1}{k - 2}\). By independence:

\[ \begin{aligned} E\left(\tfrac{X}{Y}\right) &= \tfrac{m}{n - 2} = 3, \quad m + n = 14 \\ 14 - n &= 3n - 6 \Rightarrow n = 5,\ m = 9 \\ E\left(\tfrac{Y}{X}\right) &= \tfrac{n}{m - 2} = \tfrac{5}{7} \end{aligned} \]

Taking \(E(Y/X)\) as the reciprocal \(\frac{1}{3}\) is the common mistake: the expectation of a reciprocal is not the reciprocal of the expectation, and \(\frac{1}{3}\) is not offered.

Discuss in Forum

Let \(\{X_{i}\}\) be a sequence of independent Poisson(\(\lambda\)) variables and let \(W_{n} = \frac{1}{n}\sum_{i=1}^{n} X_{i}\). Then the limiting distribution of \(\sqrt{n}(W_{n} - \lambda)\) is the normal distribution with zero mean and variance given by

A.1
B.\(\sqrt{\lambda}\)
C.\(\lambda\)
D.\(\lambda^{2}\)

Exam Year(s)

GATE Mathematics - 2018,

Answer: C

The variance is \(\lambda\).

A Poisson(\(\lambda\)) variable has mean \(\lambda\) and variance \(\lambda\). \(W_{n}\) is the sample mean of independent, identically distributed variables with finite variance, so by the central limit theorem

\[ \begin{aligned} \sqrt{n}(W_{n} - \lambda) &\to N(0, \sigma^{2}) \\ &= N(0, \lambda) \end{aligned} \]

The scaling by \(\sqrt{n}\) is exactly what keeps the variance at \(\sigma^{2}\) rather than shrinking it to \(\sigma^{2}/n\). \(\sqrt{\lambda}\) is the standard deviation, not the variance; \(\lambda^{2}\) squares the variance again; and 1 is the variance only after dividing by \(\sqrt{\lambda}\) as well.

Discuss in Forum

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be independent and identically distributed random variables with probability density function given by

\[ f_{X}(x; \theta) = \begin{cases} \theta e^{-\theta(x-1)}, & x \ge 1, \\ 0 & \text{otherwise.} \end{cases} \]

Also, let \(\overline{X} = \frac{1}{n}\sum_{i=1}^{n} X_{i}\). Then the maximum likelihood estimator of \(\theta\) is

A.\(1/\overline{X}\)
B.\((1/\overline{X}) - 1\)
C.\(1/(\overline{X} - 1)\)
D.\(\overline{X}\)

Exam Year(s)

GATE Mathematics - 2018,

Answer: C

The maximum likelihood estimator is \(1/(\overline{X} - 1)\).

The density is an exponential shifted to start at 1. For a sample with every \(x_{i} \ge 1\) the log-likelihood is

\[ \begin{aligned} \ell(\theta) &= n\ln\theta - \theta\sum_{i=1}^{n}(x_{i} - 1) \\ \ell'(\theta) &= \frac{n}{\theta} - \sum_{i=1}^{n}(x_{i} - 1) = 0 \\ \hat{\theta} &= \frac{n}{\sum x_{i} - n} = \frac{1}{\overline{X} - 1} \end{aligned} \]

\(\ell''(\theta) = -n/\theta^{2} \lt 0\), so this is a maximum. \(1/\overline{X}\) is the estimator for the unshifted exponential, which ignores the shift by 1, and \((1/\overline{X}) - 1\) applies the shift in the wrong place. \(\overline{X}\) estimates a mean, not a rate.

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A combination of numbers has geometric mean as 6.0 and arithmetic mean as 6.3. These numbers are :

A.6, 6, 6
B.5, 6, 8
C.4, 6, 9
D.None of the above

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The numbers are 4, 6, 9.

Test them against both means:

\[ \begin{aligned} GM &= \sqrt[3]{4 \times 6 \times 9} = \sqrt[3]{216} = 6.0 \\ AM &= \frac{4 + 6 + 9}{3} = \frac{19}{3} \approx 6.3 \end{aligned} \]

For 6, 6, 6 both means are exactly 6, since the AM equals the GM only when all values are equal. For 5, 6, 8 the AM is 6.33 but the GM is \(\sqrt[3]{240} \approx 6.21\), not 6.0. As 4, 6, 9 fits both, 'None of the above' is wrong.

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The sampling procedure in which an interviewer is asked to interview 25 teachers, 50 public servants and 25 farmers is called :

A.Stratified sampling
B.Accidental sampling
C.Spatial sampling
D.Quota sampling

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

This is quota sampling.

In quota sampling the interviewer is given a fixed number of respondents to interview from each category, here 25 teachers, 50 public servants and 25 farmers, but chooses the actual individuals himself, usually as convenient. It is a non-probability method.

Stratified sampling also divides the population into groups, but selects units within each stratum at random, not by the interviewer's choice. Accidental (convenience) sampling sets no quotas, and spatial sampling selects locations in an area.

Discuss in Forum

Which of the following is not an example of non-sampling risk ?

A.Failing to evaluate results properly
B.Use of an audit procedure inappropriate to achieve to a given audit objective
C.Obtaining an unrepresentative sample
D.Failure to recognize an error

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The one that is not non-sampling risk is obtaining an unrepresentative sample.

In auditing, sampling risk is the risk that the sample examined does not represent the population, so the auditor's conclusion would differ from what testing every item would show. An unrepresentative sample is the very definition of sampling risk.

Non-sampling risk comes from everything else: using an audit procedure that does not suit the objective, failing to recognise an error in an item examined, or misjudging the results. These could happen even if every item were tested.

Discuss in Forum

When an investigator wants a sample containing in units which possess a rare attribute. The appropriate sampling procedure is :

A.Simple random sampling without replacement
B.Stratified sampling
C.Inverse sampling
D.Systematic sampling

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The appropriate procedure is inverse sampling.

When an attribute is rare, a fixed-size sample may contain few or none of the units of interest. In inverse sampling (introduced by J.B.S. Haldane) units are drawn one at a time until a fixed, predetermined number of units with the rare attribute has been found. The sample size is therefore random, and the proportion is estimated from it.

Simple random, stratified and systematic sampling all fix the sample size in advance, so they can easily miss a rare attribute altogether.

Discuss in Forum

The variance value for an F-distribution with parameters V₁ = 5 and V₂ = 9 is :

A.1.587
B.1.378
C.2.578
D.1.498

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The variance is 1.587.

For an F-distribution with \(\nu_{1}\) and \(\nu_{2}\) degrees of freedom (here 5 and 9; the variance exists only when \(\nu_{2}\) is greater than 4):

\[ \begin{aligned} V &= \frac{N}{D} \\ N &= 2\nu_{2}^{2}(\nu_{1} + \nu_{2} - 2) \\ D &= \nu_{1}(\nu_{2} - 2)^{2}(\nu_{2} - 4) \end{aligned} \]

\[ \begin{aligned} V &= \frac{2(81)(12)}{5(49)(5)} \\ &= \frac{1944}{1225} = 1.587 \end{aligned} \]

The mean of this distribution is \(\nu_{2}/(\nu_{2} - 2) = 9/7 = 1.286\), which none of the options gives; 1.378, 2.578 and 1.498 do not follow from the formula.

Discuss in Forum

10.

The first group sample has 100 items with mean 15 and standard deviation 3.0. If the whole group has 250 items with mean 15.6 and standard deviation \(\sqrt{13.44}\). What is the standard deviation of the second group ?

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The standard deviation of the second group is 4.

The second group has 250 - 100 = 150 items. Its mean, from the combined total:

\[ \begin{aligned} \bar{x}_{2} &= \frac{250(15.6) - 100(15)}{150} \\ &= \frac{2400}{150} = 16 \end{aligned} \]

So \(d_{1} = 15 - 15.6 = -0.6\) and \(d_{2} = 16 - 15.6 = 0.4\). The combined variance is 13.44, since the combined SD is \(\sqrt{13.44}\):

\[ \begin{aligned} 250(13.44) &= 100(9 + 0.36) \\ &\quad + 150(\sigma_{2}^{2} + 0.16) \\ 3360 &= 936 + 24 + 150\sigma_{2}^{2} \\ \sigma_{2}^{2} &= 16,\quad \sigma_{2} = 4 \end{aligned} \]

Check the others: with a second-group SD of 2, 6 or 8 the combined variance would come out as 6.24, 25.44 or 42.24 instead of 13.44.

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