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Question 2 MCQ · 0.0 marks
1. Markov’s inequality: Let X be a discrete random variable taking non-negative values
with a finite mean pz. Then,
P(X 20 <
ele
2. Chebyshev’s inequality: Let X be a discrete random variable with a finite mean ju
and a finite variance 0”. Then,
1
PX a2 ko) <5
3. Weak Law of Large numbers; Let X;,Xo,...,Xn iid X with E[X] = p, Var(X) = v.
Define sample mean X = sRitorh Then,
n
PKs ><
bs reals
(X-al> 8) <5
4, Using CLT to approximate probability: Let X1,Xo,...,Xn~ iid X with B[X] = 1, Var(X) =e
Define Y = X, + X+...+Xq. Then,
Y-mp
yno
5. Bias of an estimator: Bias({,0) = E[6] - 4.
wv Normal(0,1).
6. Method of moments: Sample moments, Mj(X;,X9,...,X,) =-)2 X*
Procedure:_ For one parameter
¢ Sample moment: my
¢ Distribution moment: E(X) = f(#)
¢ Solve for 6 from f(0) = my in terms of my.
od: replace m by M; in the above solution.
7. Likelihood of i.i.d. samples: Likelihood of a sampling 71, 29,..., In, denoted
n
DGS cao ea) =[] fx(ais1,,...)
i=1
8. Maximum likelihood (ML) estimation:
n
9;,05,... = arg max fx (aa; 1,2,.-.)
Hh Ot
9. Bayesian estimation: Let Xj,..., X, ~ i.i.d.X, parameter 0.
Prior distribution of 8 : 0 ~ fe(4).
Samples, $ : (X, =21,..., X= 25)
Posterior: 9 | (X; =21,...,Xn =n)
Bayes’ rule: Posterior x Prior x Likelihood
Posterior density x fo(0) x P(X; = 1,...,Xn = 2, | O =)
10. Normal samples with unknown mean and known variance:
a Xn ~ iid. Normal(M,o?).
Prior M ~ Normal (jo, 08).
2 2
(19 o
Posterior mean: ji = X | ——2— } + —
ss (= Raa, ee
11. Hypothesis Testing
e Test for mean
Case (1): When population variance c” is known (z-test)
Test Ho Ha Test statistic | Rejection region
1? G
right-tailed | z= po | pp > po gut X>c
_ fe
(ee,
left-tailed | p= po | up < bo . X — b a3
__ ova
F=X
two-tailed | = po | 1 4 Ho X-pw | [X-pol>e
aha
Case (2): When population variance o
2
is unknown (t-test)
Test Ho Hy Test statistic | Rejection region
T=¥
right-tailed | ju = pig | pe > plo X-m X>c
tra = SG
T=X
left-tailed | j= yo | pe < joo ; X-w X<c
eae 7
ie
two-tailed | p= po | pF po X-w| X-ml>e
thi = Ta
\?-test for variance:
Test Ho Hy Test statistic Rejection region
. . —1)s?
right-tailed | ¢ = 09 | 7 > 09 ro x wey Soe
%
_1)82
ummm \oeas exes le oad ney Sed
%
_1)92 8? > 2 where = = P(S? > 2) or
twotailed |o=o| oo |7=2—F a |" NB ee
% S? < c where z= P(S* < c’)
¢ Two samples z-test for means:
Test Hy Hy Test statistic Rejection region
T=X-Y 2 2 A= a
Tight-tailed | jax = pa | o > Ha X-Yo Nomal(0 eet, 2) if Hp is true MNS
mn
T=¥-X _ _
leftctailed J j= ya] 14 <4 7X Norma a.) if Ho is true Y-X>e
nm
T=X-Y S. wh _
twortailed | un = pa | on # Ha X-Ywo Nonmal(0 ons 2) if Ho is true [X-¥|>e
mm
6406536003408.
Useful Data has been mentioned above.
6406536003409.
This data attachment is just for a reference & not for an evaluation.
Published solution
**1. Review the reference sheet:**
Question 2 provides the official formula and table reference sheet ("Useful data") for the examination. Selecting **"Useful Data has been mentioned above."** acknowledges the reference sheet.
**Answer: Option A**
A: **Correct:** This option acknowledges the formulas and statistical tables provided in the question.
B: **Incorrect:** Standard exam instructions require selecting the first option to confirm the reference data.
**1. Use the independence of $X$ and $Y$:**
Since $X$ and $Y$ are i.i.d. $\text{Exp}(1)$ random variables, they are independent, so:
$$P(0 < X < 1,\; 0 < Y < 1) = P(0 < X < 1) \cdot P(0 < Y < 1)$$
**2. Compute $P(0 < X < 1)$ for $\text{Exp}(1)$:**
From the Useful Data table, the CDF of $\text{Exp}(\lambda)$ with $\lambda = 1$ is $F_X(x) = 1 - e^{-x}$ for $x > 0$, and $e^{-1} = 0.37$ is given in Useful Data:
$$P(0 < X < 1) = 1 - e^{-1} = 1 - 0.37 = 0.63$$
(Using exact $e^{-1} \approx 0.36788$, $1 - e^{-1} \approx 0.63212$.)
**3. Multiply the probabilities:**
$$P(0 < X < 1,\; 0 < Y < 1) = (1 - e^{-1})^2 \approx (0.6321)^2 \approx 0.40$$
(With $e^{-1} = 0.37$, $(0.63)^2 = 0.3969 \approx 0.40$.)
**Answer: 0.40**
**1. Find $c$ such that $\hat{\mu}$ is an unbiased estimator of $\mu$:**
Given $X_1, \dots, X_n \sim \text{i.i.d. } N(\mu, \sigma^2)$, we have $E[\bar{X}] = \mu$ and $\text{Var}(\bar{X}) = \frac{\sigma^2}{n}$.
For $\hat{\mu} = \left(\frac{n+1}{n}\right)\bar{X} + c$ to be unbiased, we require $E[\hat{\mu}] = \mu$:
$$E[\hat{\mu}] = \left(\frac{n+1}{n}\right)E[\bar{X}] + c = \left(1 + \frac{1}{n}\right)\mu + c = \mu \implies c = -\frac{\mu}{n}$$
**2. Compute $\text{Risk}(\hat{\mu}) = \text{MSE}(\hat{\mu})$:**
Since $\hat{\mu}$ is unbiased, its risk (mean squared error) equals its variance:
$$\text{Risk}(\hat{\mu}) = \text{Var}(\hat{\mu}) = \text{Var}\left(\frac{n+1}{n}\bar{X} + c\right) = \left(\frac{n+1}{n}\right)^2 \text{Var}(\bar{X})$$
Substitute $\text{Var}(\bar{X}) = \frac{\sigma^2}{n}$:
$$\text{Risk}(\hat{\mu}) = \frac{(n+1)^2}{n^2} \cdot \frac{\sigma^2}{n} = \frac{(n+1)^2}{n^3}\sigma^2$$
**Answer: Option A**
A: **Correct:** Unbiasedness gives $c = -\mu/n$, and $\text{Var}(\hat{\mu}) = ((n+1)/n)^2 (\sigma^2/n) = \frac{(n+1)^2}{n^3}\sigma^2$.
B: **Incorrect:** $c$ must be negative ($-\mu/n$), and the variance of $\bar{X}$ is $\sigma^2/n$, not $\sigma^2$.
C: **Incorrect:** Both $c = -n/\mu$ and the unsquared factor $(n+1)/n$ are incorrect.
D: **Incorrect:** $c = -\mu/n$ has denominator $n$, not $n+1$, and the coefficient $(n+1)/n$ must be squared when taking variance.
**Setup :** A machine is set to fill packets with an average of 500g of flour with a known standard
deviation of 20g. Let the null and alternative hypotheses be
and
. To
verify this, a quality team selects a random sample of 64 packets and defined the critical region as
. Find the power of the test against the alternative
. (See the "Useful data"
given in question number 2 for
values. )
**Solution:** Since the population standard deviation is known, the sampling distribution of
under
the null hypothesis will be
distribution with mean
and variance
.
To find the power of the test, we will solve:
Consider the following options for the blanks.
Answer the given sub-questions.
**Note:** Enter only the correct option number (e.g. 8) without any brackets or punctuation.
Enter the correct option number for
.
Published solution
**1. Determine the sampling distribution of $\bar{X}$ (Blank A):**
By the Central Limit Theorem (or for normal populations with known population standard deviation $\sigma = 20\text{ g}$ and large sample size $n = 64$), the sampling distribution of $\bar{X}$ under the null hypothesis is a **Normal** distribution.
**2. Match with the numbered options list:**
In the given options list, **Normal** is option **(12)**.
**Answer: 12**
**Setup :** A machine is set to fill packets with an average of 500g of flour with a known standard
deviation of 20g. Let the null and alternative hypotheses be
and
. To
verify this, a quality team selects a random sample of 64 packets and defined the critical region as
. Find the power of the test against the alternative
. (See the "Useful data"
given in question number 2 for
values. )
**Solution:** Since the population standard deviation is known, the sampling distribution of
under
the null hypothesis will be
distribution with mean
and variance
.
To find the power of the test, we will solve:
Consider the following options for the blanks.
Answer the given sub-questions.
**Note:** Enter only the correct option number (e.g. 8) without any brackets or punctuation.
Enter the correct option number for
.
Published solution
**1. Find the mean of $\bar{X}$ under the null hypothesis $H_0: \mu = 500$ (Blank B):**
Under $H_0$, the mean of the sample mean $\bar{X}$ is:
$$E[\bar{X}] = \mu_0 = 500$$
**2. Match with the numbered options list:**
In the options list, **$500$** is option **(2)**.
**Answer: 2**
**Setup :** A machine is set to fill packets with an average of 500g of flour with a known standard
deviation of 20g. Let the null and alternative hypotheses be
and
. To
verify this, a quality team selects a random sample of 64 packets and defined the critical region as
. Find the power of the test against the alternative
. (See the "Useful data"
given in question number 2 for
values. )
**Solution:** Since the population standard deviation is known, the sampling distribution of
under
the null hypothesis will be
distribution with mean
and variance
.
To find the power of the test, we will solve:
Consider the following options for the blanks.
Answer the given sub-questions.
**Note:** Enter only the correct option number (e.g. 8) without any brackets or punctuation.
Enter the correct option number for
.
Published solution
**1. Find the variance of $\bar{X}$ (Blank C):**
Given population standard deviation $\sigma = 20$ and sample size $n = 64$, the variance of the sample mean $\bar{X}$ is:
$$\text{Var}(\bar{X}) = \frac{\sigma^2}{n} = \frac{20^2}{64} = \frac{400}{64} = 6.25$$
**2. Match with the numbered options list:**
In the options list, **$6.25$** is option **(3)**.
**Answer: 3**
**Setup :** A machine is set to fill packets with an average of 500g of flour with a known standard
deviation of 20g. Let the null and alternative hypotheses be
and
. To
verify this, a quality team selects a random sample of 64 packets and defined the critical region as
. Find the power of the test against the alternative
. (See the "Useful data"
given in question number 2 for
values. )
**Solution:** Since the population standard deviation is known, the sampling distribution of
under
the null hypothesis will be
distribution with mean
and variance
.
To find the power of the test, we will solve:
Consider the following options for the blanks.
Answer the given sub-questions.
**Note:** Enter only the correct option number (e.g. 8) without any brackets or punctuation.
Enter the correct option number for
.
Published solution
**1. Identify the definition of the power of a hypothesis test (Blank D):**
The power of a test $(1 - \beta)$ against an alternative $H_A$ is the probability of **rejecting $H_0$** when $H_A$ is true:
$$1 - \beta = P(\text{Reject } H_0 \mid H_A)$$
**2. Match with the numbered options list:**
In the options list, **$\text{Reject } H_0$** is option **(6)**.
**Answer: 6**
**Setup :** A machine is set to fill packets with an average of 500g of flour with a known standard
deviation of 20g. Let the null and alternative hypotheses be
and
. To
verify this, a quality team selects a random sample of 64 packets and defined the critical region as
. Find the power of the test against the alternative
. (See the "Useful data"
given in question number 2 for
values. )
**Solution:** Since the population standard deviation is known, the sampling distribution of
under
the null hypothesis will be
distribution with mean
and variance
.
To find the power of the test, we will solve:
Consider the following options for the blanks.
Answer the given sub-questions.
**Note:** Enter only the correct option number (e.g. 8) without any brackets or punctuation.
Enter the correct option number for
.
Published solution
**1. Standardize $\bar{X} = 505.83$ under $\mu = 507$ (Blank E):**
With $\mu = 507$ and standard error $\sigma_{\bar{X}} = \sqrt{6.25} = 2.5$:
$$Z = \frac{505.83 - 507}{2.5} = \frac{-1.17}{2.5} = -0.468$$
**2. Match with the numbered options list:**
In the options list, **$-0.468$** is option **(9)**.
**Answer: 9**
**Setup :** A machine is set to fill packets with an average of 500g of flour with a known standard
deviation of 20g. Let the null and alternative hypotheses be
and
. To
verify this, a quality team selects a random sample of 64 packets and defined the critical region as
. Find the power of the test against the alternative
. (See the "Useful data"
given in question number 2 for
values. )
**Solution:** Since the population standard deviation is known, the sampling distribution of
under
the null hypothesis will be
distribution with mean
and variance
.
To find the power of the test, we will solve:
Consider the following options for the blanks.
Answer the given sub-questions.
**Note:** Enter only the correct option number (e.g. 8) without any brackets or punctuation.
Enter the correct option number for
.
Published solution
**1. Evaluate $1 - F_Z(-0.468)$ using Useful Data (Blank F):**
From Question 2's Useful Data, $F_Z(-0.468) = 0.32$.
Therefore, the power of the test is:
$$1 - F_Z(-0.468) = 1 - 0.32 = 0.68$$
**2. Match with the numbered options list:**
In the options list, **$0.68$** is option **(15)**.
**Answer: 15**
, the
observed sample mean is 52.4, and the sample standard deviation is 4. Let the null and alternative
hypotheses be
and
. Consider a test that rejects
if
for some
constant
at 5% level of significance. Find the value of
. Enter the answer correct to two decimal
places.
Published solution
**1. Identify the test distribution and parameters:**
- Since the population variance $\sigma^2$ is unknown and $X \sim \text{Normal}(\mu, \sigma^2)$, we use a one-sample right-tailed **$t$-test** with degrees of freedom $df = n - 1 = 25 - 1 = 24$.
- Null hypothesis: $H_0: \mu_0 = 50$
- Sample size: $n = 25$, sample standard deviation: $s = 4$, so standard error $s / \sqrt{n} = 4 / 5 = 0.8$.
- At the $5\%$ level of significance ($\alpha = 0.05$), from the Useful Data table in Question 2:
$$F_{t_{24}}(1.711) = 0.95 \implies t_{24,\,0.05} = 1.711$$
**2. Find the critical value $c$:**
The rejection region $\bar{X} > c$ corresponds to $\frac{\bar{X} - \mu_0}{s / \sqrt{n}} > 1.711$:
$$c = \mu_0 + t_{24,\,0.05} \cdot \frac{s}{\sqrt{n}} = 50 + 1.711 \times 0.8 = 50 + 1.3688 = 51.3688 \approx 51.37$$
**Answer: 51.37**
be the number of questions asked by a randomly selected student during a live session. The
probability mass function of
is given by
Based on the above data, answer the given subquestions.
Given that a student had asked at least one question, what is the probability that the student
asked exactly 2 questions? Enter the answer correct to two decimal places.
is an unknown parameter. Consider a random sample (0, 1, 1, 0, 1) from
.
Based on the above data, answer the given subquestions.
Find the difference between the value of the method of moment estimate and the maximum
likelihood estimate of
.
Published solution
**1. Find the Maximum Likelihood Estimate (MLE) of $\lambda$:**
For a Bernoulli sample $(0, 1, 1, 0, 1)$ with $3$ ones and $2$ zeros where $p = 1 - e^{-\lambda}$, the likelihood function is:
$$L(\lambda) = (1 - e^{-\lambda})^3 (e^{-\lambda})^2 = (1 - e^{-\lambda})^3 e^{-2\lambda}$$
By the invariance property of MLE, since the MLE of the Bernoulli success probability $p$ is the sample mean $\hat{p}_{\text{MLE}} = \bar{x} = \frac{3}{5}$, the MLE of $\lambda$ satisfies:
$$1 - e^{-\hat{\lambda}_{\text{MLE}}} = \frac{3}{5} \implies \hat{\lambda}_{\text{MLE}} = -\ln\left(\frac{2}{5}\right)$$
**2. Compute the difference between the MM estimate and the ML estimate:**
Since $\hat{\lambda}_{\text{MM}} = \hat{\lambda}_{\text{MLE}} = -\ln(2/5)$, their difference is:
$$\hat{\lambda}_{\text{MM}} - \hat{\lambda}_{\text{MLE}} = 0$$
**Answer: 0**
A student is attempting to solve logic puzzles on an online platform. The number of attempts
required to solve a puzzle for the first time is modeled as a Geometric
distribution. The
student's attempts observed for solving 4 different puzzles are 2, 3, 1, and 2. Assume prior
distribution of
to be Beta
.
Based on the above data, answer the given subquestions.
Find the posterior distribution of
.
6406536003438.
Gamma
6406536003439.
Beta
6406536003440.
Beta
6406536003441.
Beta
Published solution
**1. Write the likelihood for the $\text{Geometric}(\theta)$ samples:**
From the Useful Data table in Question 2, the PMF of $\text{Geometric}(\theta)$ is $P(X = k) = (1 - \theta)^{k - 1}\theta$ for $k = 1, 2, \dots$.
For $n = 4$ independent observations $x_1 = 2, x_2 = 3, x_3 = 1, x_4 = 2$:
$$L(\theta) = \prod_{i=1}^4 \theta (1 - \theta)^{x_i - 1} = \theta^4 (1 - \theta)^{\sum (x_i - 1)} = \theta^4 (1 - \theta)^{(2+3+1+2) - 4} = \theta^4 (1 - \theta)^4$$
**2. Multiply by the $\text{Beta}(3, 4)$ prior:**
The prior density is $f(\theta) \propto \theta^{3 - 1}(1 - \theta)^{4 - 1} = \theta^2 (1 - \theta)^3$.
Therefore, the posterior density is proportional to:
$$f(\theta \mid x) \propto \theta^2 (1 - \theta)^3 \cdot \theta^4 (1 - \theta)^4 = \theta^{6}(1 - \theta)^{7} = \theta^{7 - 1}(1 - \theta)^{8 - 1}$$
which is the **$\text{Beta}(7, 8)$** distribution.
**Answer: Option B**
A: **Incorrect:** The posterior for a probability parameter $\theta \in (0, 1)$ with a Beta prior is a Beta distribution, not Gamma.
B: **Correct:** Adding $n = 4$ successes to $\alpha = 3$ gives $7$, and adding $\sum(x_i - 1) = 4$ failures to $\beta = 4$ gives $8$, yielding $\text{Beta}(7, 8)$.
C: **Incorrect:** Adds $3$ instead of $4$ to $\alpha$ and $\beta$.
D: **Incorrect:** Adds $1$ instead of $4$ to $\alpha$.
A student is attempting to solve logic puzzles on an online platform. The number of attempts
required to solve a puzzle for the first time is modeled as a Geometric
distribution. The
student's attempts observed for solving 4 different puzzles are 2, 3, 1, and 2. Assume prior
distribution of
to be Beta
.
Based on the above data, answer the given subquestions.
Find the posterior mean. Enter the answer correct to two decimal places.
Published solution
**1. Use the mean formula for the $\text{Beta}(7, 8)$ posterior distribution:**
For $\theta \mid x \sim \text{Beta}(\alpha, \beta)$ with $\alpha = 7$ and $\beta = 8$, the posterior mean is:
$$E[\theta \mid x] = \frac{\alpha}{\alpha + \beta} = \frac{7}{7 + 8} = \frac{7}{15} \approx 0.4667 \approx 0.47$$
**Answer: 0.47**
A pharmaceutical company claims that its new drug reduces systolic blood pressure by an average
of 10 mmHg, with a known standard deviation of 2 mmHg. A medical researcher believes that the
drug may be less effective than claimed. To test this claim, she conducts a study on a random
sample of 64 patients and measures the reduction in systolic blood pressure after using the drug
for 2 weeks. The researcher decides to reject the company's claim if sample mean
and
accept it otherwise.
Based on the above data, answer the given subquestions.
**1. Formulate the null and alternative hypotheses:**
- Hypotheses are always stated in terms of the **population parameter** $\mu$ (never the sample statistic $\bar{X}$).
- The company's claim is that the average reduction is $10\text{ mmHg}$, which gives the null hypothesis:
$$H_0: \mu = 10$$
- The researcher believes the drug may be **less effective** than claimed (i.e., the true mean reduction is less than $10\text{ mmHg}$), which gives the left-tailed alternative hypothesis:
$$H_A: \mu < 10$$
**Answer: Option B**
A: **Incorrect:** The researcher suspects the drug is *less* effective ($\mu < 10$), not more effective ($\mu > 10$).
B: **Correct:** The null hypothesis is $H_0: \mu = 10$ and the researcher's alternative is $H_A: \mu < 10$.
C: **Incorrect:** Hypotheses must be stated about the population mean $\mu$, not the sample mean $\bar{X}$.
D: **Incorrect:** Hypotheses must be stated about $\mu$, not $\bar{X}$.
A pharmaceutical company claims that its new drug reduces systolic blood pressure by an average
of 10 mmHg, with a known standard deviation of 2 mmHg. A medical researcher believes that the
drug may be less effective than claimed. To test this claim, she conducts a study on a random
sample of 64 patients and measures the reduction in systolic blood pressure after using the drug
for 2 weeks. The researcher decides to reject the company's claim if sample mean
and
accept it otherwise.
Based on the above data, answer the given subquestions.
Find the power of the test for
. Enter the answer correct to two decimal places.
Published solution
**1. Identify the rejection region and standard error:**
- Rejection region: $\bar{X} < 9.42$
- Known population standard deviation: $\sigma = 2\text{ mmHg}$
- Sample size: $n = 64$
- Standard error of $\bar{X}$:
$$\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{2}{\sqrt{64}} = \frac{2}{8} = 0.25$$
**2. Compute the power of the test when the true mean is $\mu = 9$:**
$$\text{Power} = P(\text{Reject } H_0 \mid \mu = 9) = P(\bar{X} < 9.42 \mid \mu = 9)$$
Standardize using $\mu = 9$ and $\sigma_{\bar{X}} = 0.25$:
$$Z = \frac{9.42 - 9}{0.25} = \frac{0.42}{0.25} = 1.68$$
From the Useful Data table in Question 2, $F_Z(1.68) = 0.9535$.
Rounding to two decimal places gives **$0.95$**.
**Answer: 0.95**
