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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 ju. Then,
PIX ><
1
2. Chebyshev’s inequality: Let X be a discrete random variable with a finite mean ju
and a finite variance o”. Then,
PX-p [2 ho) <5
3, Weak Law of Large numbers: Let X;,Xo,...,Xn iid X with E[X] = p, Var(X) =
Define sample mean X = a ately Then,
n
a
P(X pl>8)<
=n
4, Using CLT to approximate probability: Let X1,Xo,...,Xq~ iid X with B[X] = 1, Var(X) =i
Define Y = X, + Xq+...+Xq. Then,
Y-mp
yno
5. Bias of an estimator: Bias(6,) = E(6) - 8
re Normal((), 1).
6. Method of moments: Sample moments, M,(X,, Xp,..., Xn) = Ue xe
Niel
Procedure: For one parameter 4
¢ Sample moment: my
¢ Distribution moment: E(X) = f(@)
¢ Solve for 6 from f(6) = my in terms of my.
eb: replace m; by M; in the above solution.
10.
dts
Likelihood of i.i.d. samples: Likelihood of a sampling xy, x9,..., In, denoted
n
L(a1,...,2n) =[] fx(ais1,,...)
i=l
Maximum likelihood (ML) estimation:
n
9},05,... = arg max fx (i591, 02, .-.)
5,85... rcs
Bayesian estimation: Let X;,..., X, ~ i.i.d.X, parameter 0.
Prior distribution of 8 : @ ~ fe(4).
Samples, S$ : (X, =21,..., X=.)
Posterior: 9 | (X; =21,..., X= 15)
Bayes’ rule: Posterior x Prior x Likelihood
Posterior density x fo(0) x P(X; = 1,...,Xn = 2, | O =)
Normal samples with unknown mean and known variance:
a Xn ~ iid. Normal(M,o?).
Prior M ~ Normal(, 08).
2 2
=[ ng o
Posterior mean: ji = X | ——2— } + —
ie (= aay, SHE
Hypothesis Testing
e Test for mean
Case (1): When population variance c” is known (2-test)
Test Ho Ha Test statistic | Rejection region
Lx
right-tailed | pp = po | u > po ie X- 0 X>c
~ hfe
SK
left-tailed | p= po | up < Ho X-—M x <i
_ yi
=
two-tailed |p=po} HAH) , X—-p [X — pol >
_ a
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
6406536030549.
Useful Data has been mentioned above.
6406536030550.
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 Statistics for Data Science II examination. Selecting **"Useful Data has been mentioned above."** acknowledges the reference sheet.
**Answer: Option A**
A: **Correct:** This option confirms and acknowledges the useful data and formulas provided above.
B: **Incorrect:** Standard exam instructions require selecting the first option to acknowledge the formula sheet.
A supermarket manager claims that the average billing time for customers is 12 minutes with a
known standard deviation of 3.5 minutes. To verify this claim, a random sample of 49 customers
was studied, showing an average billing time of 12.9 minutes. Let the null and alternative
hypothesis be:
;
.
Find the P-value. Enter the answer correct to two decimal places.
Published solution
**1. Identify the given values and hypotheses:**
- Null hypothesis: $H_0: \mu = 12$
- Alternative hypothesis: $H_A: \mu > 12$ (right-tailed test)
- Population standard deviation (known): $\sigma = 3.5\text{ minutes}$
- Sample size: $n = 49$
- Observed sample mean: $\bar{x} = 12.9\text{ minutes}$
**2. Compute the $z$-test statistic:**
$$z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} = \frac{12.9 - 12}{3.5 / \sqrt{49}} = \frac{0.9}{3.5 / 7} = \frac{0.9}{0.5} = 1.8$$
**3. Find the $P$-value for the right-tailed test:**
From the "Useful data" table in Question 2, $F_Z(1.8) = P(Z \le 1.8) = 0.9641$.
Therefore, the $P$-value is:
$$\text{P-value} = P(Z > 1.8) = 1 - F_Z(1.8) = 1 - 0.9641 = 0.0359 \approx 0.04$$
**Answer: 0.04**
**1. Identify the joint PDF and its support:**
The joint density of $X$ and $Y$ is:
$$f(x, y) = \begin{cases} 12y^2, & 0 \le y \le x \le 1, \\ 0, & \text{otherwise.} \end{cases}$$
**2. Integrate over $x$ to find the marginal density $f_Y(y)$:**
For a fixed $y \in [0, 1]$, the condition $0 \le y \le x \le 1$ means $x$ ranges from $y$ to $1$:
$$f_Y(y) = \int_{y}^{1} 12y^2 \, dx = 12y^2 [x]_{x=y}^{x=1} = 12y^2(1 - y), \quad 0 \le y \le 1$$
and $f_Y(y) = 0$ otherwise.
**Answer: Option B**
A: **Incorrect:** The integrand is $12y^2$, not $12y$, so the factor $y^2$ remains after integrating with respect to $x$.
B: **Correct:** Integrating $12y^2$ with respect to $x$ from $x = y$ to $x = 1$ yields $f_Y(y) = 12y^2(1 - y)$ for $0 \le y \le 1$.
C: **Incorrect:** The marginal density of $Y$ cannot depend on the dummy variable $x$ in its support ($0 \le y \le 1$, not $0 \le y \le x$).
D: **Incorrect:** Both the expression $12y$ and the support $y \le x \le 1$ are incorrect.
be a discrete random variable with the probability mass function
where
is an unknown parameter. Let
be an estimator of
from
i.i.d
,
given as
Check whether
is an unbiased estimator of
, otherwise find an unbiased estimator of
.
**Solution:**
Since we know that
will be an unbiased estimator of
if
.
Now,
Now, the values of
and
can be calculated as
Now, substituting the values in equation(1), we get
Hence, from equation(2) we can conclude that
is an
estimator of
. Also,
will be
an unbiased estimator of
.
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 condition for an unbiased estimator (Blank A):**
By definition, an estimator $T$ is an unbiased estimator of a parameter $\theta$ if its expected value equals $\theta$:
$$E(T) = \theta$$
**2. Match with the numbered options list:**
In the given options list, **$E(T) = \theta$** is option **(5)**.
**Answer: 5**
be a discrete random variable with the probability mass function
where
is an unknown parameter. Let
be an estimator of
from
i.i.d
,
given as
Check whether
is an unbiased estimator of
, otherwise find an unbiased estimator of
.
**Solution:**
Since we know that
will be an unbiased estimator of
if
.
Now,
Now, the values of
and
can be calculated as
Now, substituting the values in equation(1), we get
Hence, from equation(2) we can conclude that
is an
estimator of
. Also,
will be
an unbiased estimator of
.
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 whether $T$ is biased or unbiased (Blank E):**
From equation (2), $E(T) = \theta + 1 \ne \theta$. Since the expected value of $T$ does not equal $\theta$, $T$ is a **biased** estimator of $\theta$.
**2. Match with the numbered options list:**
In the options list, **biased** is option **(10)**.
**Answer: 10**
be a discrete random variable with the probability mass function
where
is an unknown parameter. Let
be an estimator of
from
i.i.d
,
given as
Check whether
is an unbiased estimator of
, otherwise find an unbiased estimator of
.
**Solution:**
Since we know that
will be an unbiased estimator of
if
.
Now,
Now, the values of
and
can be calculated as
Now, substituting the values in equation(1), we get
Hence, from equation(2) we can conclude that
is an
estimator of
. Also,
will be
an unbiased estimator of
.
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. Construct an unbiased estimator from $T$ (Blank F):**
Since $E(T) = \theta + 1$, by linearity of expectation:
$$E(T - 1) = E(T) - 1 = (\theta + 1) - 1 = \theta$$
Therefore, **$T - 1$** is an unbiased estimator of $\theta$.
**2. Match with the numbered options list:**
In the options list, **$T - 1$** is option **(12)**.
**Answer: 12**
A nutrition company claims that the average protein content of its energy bars is 20 grams, with a
known standard deviation of 4 grams. A sports academy suspects that the actual mean protein
content may be less than 20 grams. To test the claim, the academy collects a random sample of 64
bars. The academy decides to reject the company’s claim if the mean protein content of the
sample
and accepts it otherwise. Assuming that the company’s claim is true, what is the
probability that the academy rejects the claim?
Published solution
**1. Set up the sampling distribution under the company's claim ($H_0$):**
- True mean under the claim: $\mu = 20\text{ grams}$
- Known population standard deviation: $\sigma = 4\text{ grams}$
- Sample size: $n = 64$
- Standard error of the sample mean $\bar{X}$:
$$\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{4}{\sqrt{64}} = \frac{4}{8} = 0.5$$
**2. Calculate the probability of rejecting the claim ($\bar{X} < 19.2$):**
Standardize $\bar{X} = 19.2$:
$$Z = \frac{19.2 - 20}{0.5} = \frac{-0.8}{0.5} = -1.6$$
Using the symmetry of the standard normal distribution and $F_Z(1.6) = 0.9452$ from the Useful Data table:
$$P(\bar{X} < 19.2) = P(Z < -1.6) = 1 - F_Z(1.6) = 1 - 0.9452 = 0.0548 \approx 0.05$$
**Answer: 0.05**
A laboratory is testing a new temperature sensor to check whether it provides more consistent
readings than the old model. The lab records the temperature (in \(^{o}\)C) from 5 repeated
measurements using the new sensor:
10, 8, 2, 5, 15
The old temperature sensor has a known standard deviation of
.
Based on the above data, answer the given subquestions.
Compute the sample variance of the 5 temperature readings from the new sensor. Enter the
answer correct to one decimal place.
A laboratory is testing a new temperature sensor to check whether it provides more consistent
readings than the old model. The lab records the temperature (in \(^{o}\)C) from 5 repeated
measurements using the new sensor:
10, 8, 2, 5, 15
The old temperature sensor has a known standard deviation of
.
Based on the above data, answer the given subquestions.
At
significance level, determine whether the new sensor has less variability than the old
sensor.
6406536030568.
The new sensor does not show less variability than the old sensor.
6406536030569.
The new sensor has less variability than the old sensor.
Published solution
**1. Set up the hypotheses for the $\chi^2$-test for variance:**
- Old sensor standard deviation: $\sigma_0 = 1^\circ\text{C} \implies \sigma_0^2 = 1$
- $H_0: \sigma^2 = 1$ vs. $H_A: \sigma^2 < 1$ (testing if the new sensor has *less* variability, left-tailed test)
- Sample size: $n = 5$ (degrees of freedom $df = n - 1 = 4$)
- Sample variance: $s^2 = 24.5$
**2. Compute the test statistic and compare with the critical value:**
$$T = \frac{(n - 1)s^2}{\sigma_0^2} = \frac{4 \times 24.5}{1} = 98$$
For a left-tailed test at $\alpha = 0.05$ with $df = 4$, the rejection region is $T < F_{\chi_4^2}^{-1}(0.05) = 0.711$ (from Useful Data).
Since $T = 98 \not< 0.711$ (in fact, the sample variance $24.5$ is much larger than $1$), we fail to reject $H_0$.
**3. Conclude:**
**The new sensor does not show less variability than the old sensor.**
**Answer: Option A**
A: **Correct:** Since the sample variance $s^2 = 24.5$ is much larger than $\sigma_0^2 = 1$ (test statistic $T = 98 > 0.711$), we do not reject $H_0$.
B: **Incorrect:** To show less variability, the sample variance would need to be significantly smaller than $1$, whereas $s^2 = 24.5$.
A factory manufactures light bulbs, and each bulb has a probability
of being defective. To
estimate this probability using Bayesian methods, assume the prior distribution of
to be
Beta(3, 3). A quality inspector tests 12 independent bulbs. If 5 bulbs are found to be defective,
answer the given subquestions.
Determine the posterior distribution of
.
6406536030570.
Beta
6406536030571.
Beta
6406536030572.
Beta
6406536030573.
Beta
Published solution
**1. Recall the Beta-Binomial conjugate update:**
If the prior distribution for the success/defective probability $p$ is $\text{Beta}(\alpha, \beta)$ and we observe $x$ defectives in $n$ independent Bernoulli trials, the posterior distribution of $p$ is:
$$p \mid x \sim \text{Beta}(\alpha + x,\; \beta + n - x)$$
**2. Substitute the given values:**
- Prior: $\text{Beta}(3, 3)$, so $\alpha = 3$ and $\beta = 3$.
- Sample: $n = 12$ bulbs tested, $x = 5$ defective bulbs, and $n - x = 12 - 5 = 7$ non-defective bulbs.
- Posterior parameters:
$$\alpha_{\text{post}} = 3 + 5 = 8, \qquad \beta_{\text{post}} = 3 + 7 = 10$$
Thus, the posterior distribution is **$\text{Beta}(8, 10)$**.
**Answer: Option D**
A: **Incorrect:** This adds $4$ and $6$ instead of $x = 5$ defectives and $n - x = 7$ non-defectives.
B: **Incorrect:** This swaps the number of defectives ($5$) and non-defectives ($7$).
C: **Incorrect:** The prior is $\text{Beta}(3, 3)$, not $\text{Beta}(4, 0)$.
D: **Correct:** Updating $\text{Beta}(3, 3)$ with $5$ defectives and $7$ non-defectives yields $\text{Beta}(3 + 5, 3 + 7) = \text{Beta}(8, 10)$.
A factory manufactures light bulbs, and each bulb has a probability
of being defective. To
estimate this probability using Bayesian methods, assume the prior distribution of
to be
Beta(3, 3). A quality inspector tests 12 independent bulbs. If 5 bulbs are found to be defective,
answer the given subquestions.
Using the posterior distribution, calculate the posterior mean of
. Enter the answer correct to two
decimal places.
Published solution
**1. Use the mean formula for a $\text{Beta}(\alpha, \beta)$ distribution:**
From the Useful Data table, the mean of a $\text{Beta}(\alpha, \beta)$ random variable is:
$$E[p] = \frac{\alpha}{\alpha + \beta}$$
**2. Substitute the posterior parameters $\alpha = 8$ and $\beta = 10$:**
$$\text{Posterior mean} = \frac{8}{8 + 10} = \frac{8}{18} = \frac{4}{9} \approx 0.4444 \approx 0.44$$
**Answer: 0.44**
**1. Use the PDF $f_X(x) = \frac{1}{2}(2 - x)$ on $0 < x < 2$ to compute $E[X]$:**
$$E[X] = \int_0^2 x \cdot \frac{1}{2}(2 - x)\,dx = \frac{1}{2} \int_0^2 (2x - x^2)\,dx$$
**2. Evaluate the definite integral:**
$$E[X] = \frac{1}{2} \left[x^2 - \frac{x^3}{3}\right]_0^2 = \frac{1}{2} \left(4 - \frac{8}{3}\right) = \frac{1}{2} \cdot \frac{4}{3} = \frac{2}{3}$$
**Answer: Option D**
A: **Incorrect:** $\frac{1}{2}$ is not the expected value; the density is higher near $x = 0$, pulling the mean to $\frac{2}{3}$.
B: **Incorrect:** $\frac{4}{3}$ is the value of the integral before multiplying by $k = \frac{1}{2}$.
C: **Incorrect:** $\frac{3}{4} = 0.75$ is the value of the CDF $F_X(1)$, not the expected value $E[X]$.
D: **Correct:** Integrating $\frac{1}{2}x(2 - x)$ from $0$ to $2$ gives $E[X] = \frac{2}{3}$.
Based on the above data, answer the given subquestions.
What is the value of
?
6406536030580.
0.01
6406536030581.
0.10
6406536030582.
0.70
6406536030583.
0.30
Published solution
**1. Sum all probabilities in the joint PMF table and set equal to $1$:**
The entries in the table are:
- Row $X = 0$: $a + 0.05 + 0.00 = a + 0.05$
- Row $X = 1$: $2a + 0.15 + 0.05 = 2a + 0.20$
- Row $X = 2$: $0.05 + a + 3a = 4a + 0.05$
Total probability:
$$(a + 0.05) + (2a + 0.20) + (4a + 0.05) = 7a + 0.30 = 1$$
**2. Solve for $a$:**
$$7a = 0.70 \implies a = 0.10$$
**Answer: Option B**
A: **Incorrect:** If $a = 0.01$, the total probability would be $0.37 \ne 1$.
B: **Correct:** Summing all cells gives $7a + 0.30 = 1$, so $a = 0.10$.
C: **Incorrect:** $0.70$ is the value of $7a$, not $a$.
D: **Incorrect:** $0.30$ is the sum of the constant entries in the table, not $a$.
represent the number of issues resolved during a customer support
call. The probability mass function of
is given by
where
. A random sample of five customer ratings is {1, 2, 2, 3, 2}.
Based on the above data, answer the given subquestions.
The service policy requires that the average number of issues resolved per call must be at least
1.5. Based on the estimated value of
, determine whether this requirement is satisfied.
6406536030586.
The requirement is satisfied.
6406536030587.
The requirement is not satisfied.
Published solution
**1. Evaluate the estimated average number of issues resolved per call:**
From the previous question, the expected number of issues resolved per call is:
$$E[X] = \frac{5}{3} + \theta$$
Substituting the estimated value $\hat{\theta} = \frac{1}{3}$ (or noting directly that under the method of moments, the estimated mean equals the sample mean $\bar{x}$):
$$\widehat{E[X]} = \frac{5}{3} + \frac{1}{3} = 2$$
**2. Compare with the service policy requirement:**
Since $2 \ge 1.5$, the requirement that the average number of issues resolved per call be at least $1.5$ **is satisfied**.
**Answer: Option A**
A: **Correct:** With $\hat{\theta} = 1/3$, the estimated average number of issues resolved is $E[X] = 5/3 + 1/3 = 2.0$, which is greater than $1.5$.
B: **Incorrect:** Even for any $\theta \ge 0$, $E[X] = 5/3 + \theta \ge 1.67 > 1.5$, so the requirement is clearly satisfied.
