Testing of Hypotheses Questions and Answers

Q: If, for a given \(\alpha\), \(0 \lt \alpha \lt 1\), non-randomized Neyman-Pearson and likelihood ratio tests of a simple hypothesis against a simple alternative exists, then :

The tests are equivalent.For a simple null against a simple alternative, the Neyman-Pearson most powerful test rejects when the likelihood ratio \(f_{1}(x)/f_{0}(x)\) exceeds a constant \(k\) chosen to give size \(\alpha\). The likelihood ratio test rejects when \(\lambda(x) = f_{0}(x)/\max(f_{0}(x), f_{1}(x))\) is small, which for \(k \gt 1\) is the same event. When a non-randomised test of exact size \(\alpha\) exists, both reduce to the same critical region.They are built differently (one from an optimality theorem, the other from a general principle), which is why the key prefers 'equivalent' to 'one and the same'. They are certainly not opposite.

Q: A sample of size \(n\) is obtained from a Poisson distribution with parameter \(m\). The most powerful (MP) test of less than size \(\alpha\) to test \(H_{0} : m \le m_{0}\) against \(H_{1} : m \gt m_{0}\) is given as :where \(T = \sum_{1}^{n} X_{i}\)

The keyed test is \(\phi(x) = 1\) if \(T \gt t_{0}\), 0 otherwise.The Poisson family has monotone likelihood ratio in \(T = \sum X_{i}\): for \(m_{1} \gt m_{0}\), \(f_{m_{1}}/f_{m_{0}} = e^{-n(m_{1} - m_{0})}(m_{1}/m_{0})^{T}\) increases with T. So the best test of \(m \le m_{0}\) against \(m \gt m_{0}\) rejects for large T, which rules out the two options rejecting for small T.T is discrete, so a test of exact size \(\alpha\) generally needs randomisation at \(T = t_{0}\) (option B). The question asks for the most powerful test of size less than \(\alpha\), that is the non-randomised one: reject when \(T \gt t_{0}\), with \(t_{0}\) the smallest value keeping \(P_{m_{0}}(T \gt t_{0}) \le \alpha\). That is option A.

Q: Let \(X_{1}\), \(X_{2}\), .......... \(X_{n}\) be iid random sample of size \(n\) from exponential distribution with mean \(\theta\). The MP test of size \(\alpha\) for testing \(H_{0} : \theta = \theta_{0}\) against \(H_{1} : \theta = \theta_{1} \lt \theta_{0}\), is :where \(T = \sum_{1}^{n} X_{i}\)

The MP test rejects when \(T \lt \theta_{0}\chi^{2}_{2n, 1-\alpha}/2\).For the exponential with mean \(\theta\), the likelihood ratio \(f_{\theta_{1}}/f_{\theta_{0}}\) is a decreasing function of \(T = \sum X_{i}\) when \(\theta_{1} \lt \theta_{0}\), so by Neyman-Pearson the MP test rejects for small T. Under \(H_{0}\), \(2T/\theta_{0}\) has a chi-square distribution with \(2n\) degrees of freedom, so\[ \begin{aligned} P\left(T \lt \tfrac{\theta_{0}}{2}c\right) &= P(\chi^{2}_{2n} \lt c) \\ &= \alpha \end{aligned} \]needs \(c\) to be the lower \(\alpha\) point, which in the upper-tail notation \(\chi^{2}_{\nu, p}\) (the point with area \(p\) to its right) is \(\chi^{2}_{2n, 1-\alpha}\). Options A and D put \(\theta_{0}\) in the denominator, and C uses the upper \(\alpha\) point, which would give size \(1 - \alpha\).

Q: A test \(\phi(x)\) is called an unbiased test if :

A test is unbiased if \(E_{H_{0}}\phi(X) \le \alpha\) and \(E_{H_{1}}\phi(X) \ge \alpha\).Unbiasedness means the test rejects at least as often when the alternative is true as when the null is: its size is at most \(\alpha\) and its power is at least \(\alpha\) at every alternative. Otherwise there would be an alternative under which the test is less likely to reject than under the null, which is perverse.Each of options A and B states only one half (A is just the size condition). Option D demands power equal to \(\alpha\), which describes a useless test that ignores the data, not an unbiased one.

Q: Suppose X has a distribution with probability mass function that of discrete uniform on \(\{-1, 0, 1\}\) under \(H_{0}\) and \(U(-1, 1)\) under \(H_{1}\). Then the test function\[ \phi(x) = \begin{cases} 1 & \text{if } x \lt 0 \\ 0 & \text{if } x \ge 0 \end{cases} \]has size :

The size of the test is 1/3.The size is the probability of rejecting \(H_{0}\) when \(H_{0}\) is true. The test rejects when \(x \lt 0\), and under \(H_{0}\) the only negative value is \(-1\), with probability 1/3:\[ E_{H_{0}}[\phi(X)] = P_{H_{0}}(X = -1) = \tfrac{1}{3} \]Size is computed under \(H_{0}\) alone, so it makes no difference that \(H_{1}\) is continuous; the test is perfectly well defined. 1/2 is \(P_{H_{1}}(X \lt 0)\), which is the power, not the size.

Q: A frequency data is classified in 9 classes and Gamma distribution is fitted to it after estimating the parameters. If a \(\chi^{2}\) goodness of fit test is to be used without combining the classes, the degrees of freedom associated with \(\chi^{2}\) test are :

The degrees of freedom are 6.For a chi-square goodness-of-fit test with \(k\) classes and \(s\) parameters estimated from the data,\[ \nu = k - 1 - s = 9 - 1 - 2 = 6 \]One degree of freedom is lost because the expected frequencies are made to total the observed total, and one more for each estimated parameter; the gamma distribution has two (shape and scale).9 forgets every restriction, 8 subtracts only the total, and 7 counts only one estimated parameter.

Q: The distributions of X under \(H_{0}\) and \(H_{1}\) are given by :\(x\)1234\(H_{0}\)0.250.250.250.25\(H_{1}\)0.170.170.170.32The most powerful test at level \(\alpha = 0.05\) is given by :

The most powerful test rejects \(H_{0}\) with probability 0.2 if X = 4.By the Neyman-Pearson lemma, reject where \(P_{1}(x)/P_{0}(x)\) is largest. The ratio is \(0.17/0.25 = 0.68\) at \(x = 1, 2, 3\) and \(0.32/0.25 = 1.28\) at \(x = 4\), so \(x = 4\) comes first. Rejecting at \(x = 4\) always would give size 0.25, too large, so randomise there:\[ 0.25\gamma = 0.05 \Rightarrow \gamma = 0.2 \]The non-randomised test in A has size 0.25, not 0.05. Rejecting with probability 0.25 or 0.32 gives sizes 0.0625 and 0.08, also above the level.

Q: If one wishes to determine the variation in the body size of frogs from four different populations, the appropriate test statistics would be :

The appropriate test is the F-test.Body size is a measured variable and there are four groups to compare. Comparing the means and variation of more than two groups is done by analysis of variance (ANOVA), whose test statistic is F, the ratio of between-group to within-group variance.Student's t-test compares only two means, so it would need six separate tests here. Chi-square tests frequencies or counts in categories, not measurements, and regression analysis measures how one variable depends on another rather than comparing groups.

Q: To test \(H_{0} : p = p_{0}\) against the alternative \(H_{1} : p \ne p_{0}\), where \(p\) is the population proportion and \(p_{0}\) is a specified value of \(p\). Suppose \(P_{n}\) denotes the sample proportion based on a random sample of size \(n\) from the population. The test statistic for testing \(H_{0} : p = p_{0}\) against \(H_{1} : p \ne p_{0}\) is given by :

The test statistic is \(\dfrac{\sqrt{n}(P_{n} - p_{0})}{\sqrt{p_{0}(1 - p_{0})}}\).Under \(H_{0}\), \(P_{n}\) has mean \(p_{0}\) and variance \(p_{0}(1 - p_{0})/n\). Standardising,\[ Z = \frac{P_{n} - p_{0}}{\sqrt{p_{0}(1 - p_{0})/n}} \]which is approximately N(0, 1) for large \(n\) and is compared with \(\pm z_{\alpha/2}\). This is option A after multiplying top and bottom by \(\sqrt{n}\).Option B is not divided by the standard deviation, option C leaves out the \(\sqrt{n}\), and option D divides by the variance instead of its square root.

11.

If, for a given \(\alpha\), \(0 \lt \alpha \lt 1\), non-randomized Neyman-Pearson and likelihood ratio tests of a simple hypothesis against a simple alternative exists, then :

A.They are equivalent
B.They are one and the same
C.They are exactly opposite
D.One can't say anything about it

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The tests are equivalent.

For a simple null against a simple alternative, the Neyman-Pearson most powerful test rejects when the likelihood ratio \(f_{1}(x)/f_{0}(x)\) exceeds a constant \(k\) chosen to give size \(\alpha\). The likelihood ratio test rejects when \(\lambda(x) = f_{0}(x)/\max(f_{0}(x), f_{1}(x))\) is small, which for \(k \gt 1\) is the same event. When a non-randomised test of exact size \(\alpha\) exists, both reduce to the same critical region.

They are built differently (one from an optimality theorem, the other from a general principle), which is why the key prefers 'equivalent' to 'one and the same'. They are certainly not opposite.

Discuss in Forum

12.

A sample of size \(n\) is obtained from a Poisson distribution with parameter \(m\). The most powerful (MP) test of less than size \(\alpha\) to test \(H_{0} : m \le m_{0}\) against \(H_{1} : m \gt m_{0}\) is given as :

where \(T = \sum_{1}^{n} X_{i}\)

A.\(\phi(x) = 1\); \(T \gt t_{0}\) / \(0\); otherwise
B.\(\phi(x) = 1\); \(T \gt t_{0}\) / \(\gamma\); \(T = t_{0}\) / \(0\); otherwise
C.\(\phi(x) = 1\); \(T \lt t_{0}\) / \(0\); otherwise
D.\(\phi(x) = 1\); \(T \lt t_{0}\) / \(\gamma\); \(T = t_{0}\) / \(0\); otherwise

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The keyed test is \(\phi(x) = 1\) if \(T \gt t_{0}\), 0 otherwise.

The Poisson family has monotone likelihood ratio in \(T = \sum X_{i}\): for \(m_{1} \gt m_{0}\), \(f_{m_{1}}/f_{m_{0}} = e^{-n(m_{1} - m_{0})}(m_{1}/m_{0})^{T}\) increases with T. So the best test of \(m \le m_{0}\) against \(m \gt m_{0}\) rejects for large T, which rules out the two options rejecting for small T.

T is discrete, so a test of exact size \(\alpha\) generally needs randomisation at \(T = t_{0}\) (option B). The question asks for the most powerful test of size less than \(\alpha\), that is the non-randomised one: reject when \(T \gt t_{0}\), with \(t_{0}\) the smallest value keeping \(P_{m_{0}}(T \gt t_{0}) \le \alpha\). That is option A.

Discuss in Forum

13.

Let \(X_{1}\), \(X_{2}\), .......... \(X_{n}\) be iid random sample of size \(n\) from exponential distribution with mean \(\theta\). The MP test of size \(\alpha\) for testing \(H_{0} : \theta = \theta_{0}\) against \(H_{1} : \theta = \theta_{1} \lt \theta_{0}\), is :

where \(T = \sum_{1}^{n} X_{i}\)

A.\(\phi(x) = 1\); \(T \lt \dfrac{2\chi^{2}_{2n, 1-\alpha}}{\theta_{0}}\) / \(0\); otherwise
B.\(\phi(x) = 1\); \(T \lt \dfrac{\theta_{0}\chi^{2}_{2n, 1-\alpha}}{2}\) / \(0\); otherwise
C.\(\phi(x) = 1\); \(T \lt \dfrac{\theta_{0}\chi^{2}_{2n, \alpha}}{2}\) / \(0\); otherwise
D.\(\phi(x) = 1\); \(T \lt \dfrac{2\chi^{2}_{2n, \alpha}}{\theta_{0}}\) / \(0\); otherwise

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The MP test rejects when \(T \lt \theta_{0}\chi^{2}_{2n, 1-\alpha}/2\).

For the exponential with mean \(\theta\), the likelihood ratio \(f_{\theta_{1}}/f_{\theta_{0}}\) is a decreasing function of \(T = \sum X_{i}\) when \(\theta_{1} \lt \theta_{0}\), so by Neyman-Pearson the MP test rejects for small T. Under \(H_{0}\), \(2T/\theta_{0}\) has a chi-square distribution with \(2n\) degrees of freedom, so

\[ \begin{aligned} P\left(T \lt \tfrac{\theta_{0}}{2}c\right) &= P(\chi^{2}_{2n} \lt c) \\ &= \alpha \end{aligned} \]

needs \(c\) to be the lower \(\alpha\) point, which in the upper-tail notation \(\chi^{2}_{\nu, p}\) (the point with area \(p\) to its right) is \(\chi^{2}_{2n, 1-\alpha}\). Options A and D put \(\theta_{0}\) in the denominator, and C uses the upper \(\alpha\) point, which would give size \(1 - \alpha\).

Discuss in Forum

14.

A test \(\phi(x)\) is called an unbiased test if :

A.\(E_{H_{0}}\phi(X) \le \alpha\)
B.\(E_{H_{1}}\phi(X) \ge \alpha\)
C.\(E_{H_{0}}\phi(X) \le \alpha\) and \(E_{H_{1}}\phi(X) \ge \alpha\)
D.\(E_{H_{0}}\phi(X) = \alpha = E_{H_{1}}\phi(X)\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

A test is unbiased if \(E_{H_{0}}\phi(X) \le \alpha\) and \(E_{H_{1}}\phi(X) \ge \alpha\).

Unbiasedness means the test rejects at least as often when the alternative is true as when the null is: its size is at most \(\alpha\) and its power is at least \(\alpha\) at every alternative. Otherwise there would be an alternative under which the test is less likely to reject than under the null, which is perverse.

Each of options A and B states only one half (A is just the size condition). Option D demands power equal to \(\alpha\), which describes a useless test that ignores the data, not an unbiased one.

Discuss in Forum

15.

Consider the problem of testing \(H_{0} : \theta = \theta_{0}\) against \(H_{1} : \theta \ne \theta_{0}\), where \(\theta_{0}\) is specified value of \(\theta\) and is the parameter of one parameter Cramer family. If \(\lambda(x)\) is the likelihood ratio statistic based on sample of size \(n\),

A.\(-2\log\lambda(x) \to \chi^{2}_{1}\ \forall\, \theta\) as \(n \to \infty\)
B.\(-2\log\lambda(x) \to \chi^{2}_{n}\ \forall\, \theta\) as \(n \to \infty\)
C.\(-2\log\lambda(x) \to \chi^{2}_{1}\), for \(\theta = \theta_{0}\), as \(n \to \infty\)
D.\(-2\log\lambda(x) \to N(0, 1)\), for \(\theta = \theta_{0}\) as \(n \to \infty\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

\(-2\log\lambda(x) \to \chi^{2}_{1}\) for \(\theta = \theta_{0}\), as \(n \to \infty\).

This is Wilks' theorem. Under regularity conditions (the Cramer conditions), when the null hypothesis is true the statistic \(-2\log\lambda\) converges in distribution to a chi-square with degrees of freedom equal to the number of parameters fixed by \(H_{0}\), here one.

The result holds only under \(H_{0}\): under a fixed alternative the statistic grows without bound, so 'for all \(\theta\)' in A and B is wrong, and B also has \(n\) degrees of freedom. The limit is a chi-square, not a standard normal (D); its square root, signed, would be normal.

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16.

Suppose X has a distribution with probability mass function that of discrete uniform on \(\{-1, 0, 1\}\) under \(H_{0}\) and \(U(-1, 1)\) under \(H_{1}\). Then the test function

\[ \phi(x) = \begin{cases} 1 & \text{if } x \lt 0 \\ 0 & \text{if } x \ge 0 \end{cases} \]

has size :

A.Not properly defined since \(H_{0}\) is discrete and \(H_{1}\) is continuous
B.1/3
C.0
D.1/2

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The size of the test is 1/3.

The size is the probability of rejecting \(H_{0}\) when \(H_{0}\) is true. The test rejects when \(x \lt 0\), and under \(H_{0}\) the only negative value is \(-1\), with probability 1/3:

\[ E_{H_{0}}[\phi(X)] = P_{H_{0}}(X = -1) = \tfrac{1}{3} \]

Size is computed under \(H_{0}\) alone, so it makes no difference that \(H_{1}\) is continuous; the test is perfectly well defined. 1/2 is \(P_{H_{1}}(X \lt 0)\), which is the power, not the size.

Discuss in Forum

17.

A frequency data is classified in 9 classes and Gamma distribution is fitted to it after estimating the parameters. If a \(\chi^{2}\) goodness of fit test is to be used without combining the classes, the degrees of freedom associated with \(\chi^{2}\) test are :

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The degrees of freedom are 6.

For a chi-square goodness-of-fit test with \(k\) classes and \(s\) parameters estimated from the data,

\[ \nu = k - 1 - s = 9 - 1 - 2 = 6 \]

One degree of freedom is lost because the expected frequencies are made to total the observed total, and one more for each estimated parameter; the gamma distribution has two (shape and scale).

9 forgets every restriction, 8 subtracts only the total, and 7 counts only one estimated parameter.

Discuss in Forum

18.

The distributions of X under \(H_{0}\) and \(H_{1}\) are given by :

\(x\)	1	2	3	4
\(H_{0}\)	0.25	0.25	0.25	0.25
\(H_{1}\)	0.17	0.17	0.17	0.32

The most powerful test at level \(\alpha = 0.05\) is given by :

A.reject \(H_{0}\) if X = 4
B.reject \(H_{0}\) with probability 0.25 if X = 4
C.reject \(H_{0}\) with probability 0.2 if X = 4
D.reject \(H_{0}\) with probability 0.32 if X = 4

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The most powerful test rejects \(H_{0}\) with probability 0.2 if X = 4.

By the Neyman-Pearson lemma, reject where \(P_{1}(x)/P_{0}(x)\) is largest. The ratio is \(0.17/0.25 = 0.68\) at \(x = 1, 2, 3\) and \(0.32/0.25 = 1.28\) at \(x = 4\), so \(x = 4\) comes first. Rejecting at \(x = 4\) always would give size 0.25, too large, so randomise there:

\[ 0.25\gamma = 0.05 \Rightarrow \gamma = 0.2 \]

The non-randomised test in A has size 0.25, not 0.05. Rejecting with probability 0.25 or 0.32 gives sizes 0.0625 and 0.08, also above the level.

Discuss in Forum

19.

If one wishes to determine the variation in the body size of frogs from four different populations, the appropriate test statistics would be :

A.Student t-test
B.Chi-square
C.Regression analysis
D.F-test

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

The appropriate test is the F-test.

Body size is a measured variable and there are four groups to compare. Comparing the means and variation of more than two groups is done by analysis of variance (ANOVA), whose test statistic is F, the ratio of between-group to within-group variance.

Student's t-test compares only two means, so it would need six separate tests here. Chi-square tests frequencies or counts in categories, not measurements, and regression analysis measures how one variable depends on another rather than comparing groups.

Discuss in Forum

20.

To test \(H_{0} : p = p_{0}\) against the alternative \(H_{1} : p \ne p_{0}\), where \(p\) is the population proportion and \(p_{0}\) is a specified value of \(p\). Suppose \(P_{n}\) denotes the sample proportion based on a random sample of size \(n\) from the population. The test statistic for testing \(H_{0} : p = p_{0}\) against \(H_{1} : p \ne p_{0}\) is given by :

A.\(\dfrac{\sqrt{n}(P_{n} - p_{0})}{\sqrt{p_{0}(1 - p_{0})}}\)
B.\(\sqrt{n}(P_{n} - p_{0})\)
C.\(\dfrac{P_{n} - p_{0}}{\sqrt{p_{0}(1 - p_{0})}}\)
D.\(\dfrac{P_{n} - p_{0}}{p_{0}(1 - p_{0})}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The test statistic is \(\dfrac{\sqrt{n}(P_{n} - p_{0})}{\sqrt{p_{0}(1 - p_{0})}}\).

Under \(H_{0}\), \(P_{n}\) has mean \(p_{0}\) and variance \(p_{0}(1 - p_{0})/n\). Standardising,

\[ Z = \frac{P_{n} - p_{0}}{\sqrt{p_{0}(1 - p_{0})/n}} \]

which is approximately N(0, 1) for large \(n\) and is compared with \(\pm z_{\alpha/2}\). This is option A after multiplying top and bottom by \(\sqrt{n}\).

Option B is not divided by the standard deviation, option C leaves out the \(\sqrt{n}\), and option D divides by the variance instead of its square root.

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