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Question 2 MCQ · 0.0 marks
Continuous random variables:
Distribution PDF (fx(k)) CDP (Fx(z)) E[X] Var(X)
0 a<a F
Uniform|a, ] te ae ore acar<d ath (i)
b-a b-a 2 12
1 z>b
0 r<0 1 1
Ar = ES bark
Exp(A) dee 0 _ joni T 7
= a:
1 exp ( iG -#) )
Normal(p,0?) | oV2a 20? No closed form a oe
—00 <2 < 00
f ee a, a.
Gamma(a, 3) Ta” e?r r>0 3 BP
Tat) a1 3-1
(1-2) a ap
Beta(a,A) | Tar) a+B | @+Ba+B+)
O<2<1
1. Markov’s inequality: Let X be a discrete random variable taking non-negative values
with a finite mean yt. Then,
P(X 20) <
ol
2. Chebyshev’s inequality: Let X be a discrete random variable with a finite mean ps
and a finite variance o?. Then,
3. Weak Law of Large numbers: Let X;, Xo,...,
XitXat.-+ Xm qhen
o.
Define sample mean X =
4. Using CLT to approximate probability: Let X\, X2
pt, Var(X) = 07.
P(|X—p|2 ko) <
n
1
RP
es
P(X - pl > 8) <5
2
Define Y = X, + X, +... +X, Then,
Y—nmp
no
= Normal(0, 1).
X, ~ iid X with E[X] = p, Var(X) =
on X, ~ iid X with E[X] =
5. Bias of an estimator: Bias(,0) = E[6] — 4.
6. Method of moments: Sample moments, Mj(X;, Xo,..., X)= Le Xt
i
Ni=t
Procedure: For one parameter @
e Sample moment: m,
¢ Distribution moment: E(X) = f(@)
¢ Solve for @ from f(@) = m, in terms of m.
ob: replace m; by M, in the above solution.
7. Likelihood of i.i.d. samples: Likelihood of a sampling x, 22,..., 2n, denoted
L(v1,---42n)=]] fx (i: 1,82, --)
i=1
8. Maximum likelihood (ML) estimation:
O;,63,... = arg max TT fx(2j;01,02,...)
PO ca
9. Bayesian estimation: Let X),...,X, ~ i.i.d.X, parameter 9.
Prior distribution of 9 : 0 ~ fo(4).
Samples, S : (X; = 21,...,X, = 2)
Posterior: 9 | (X, = 21,...,Xp = 2p)
Bayes’ rule: Posterior oc Prior x Likelihood
Posterior density « fa(#) x P(X; =a11,...,Xn =an | 9 =6)
10. Normal samples with unknown mean and known variance:
2 Sie X,, ~ iid. Normal(M, 0).
Prior M ~ Normal (tip, 03).
a ea
Posterior mean: ji = X a + Ho (a)
nop +0? no? +0?
11. Hypothesis Testing
e Test for mean
Case (1): When population variance o? is known (z-test)
Test Ho Ha Test statistic | Rejection region
T=xK
right-tailed | jo = pio | pp > flo pe Sa X>e
~_ gi
TX
left-tailed | pp = po | p< po ge X — po X<e
_ vi
Tae
two-tailed | = po | p # po z= X- [X — pig] >
__ ya
Case (2): When population variance o? is unknown (t-test)
Test yeh Hy Test statistic | Rejection region
T=X%
right-tailed | js = pig | @ > flo . X-p X>c
cot Sie
T=%
left-tailed | = po | 1 < po X-p X<e
&
T=
two-tailed | = uo | 1 # Ho ' X= | X-pml>e
aT i
¢ \*-test for variance:
Test Hy Ha Test statistic Rejection region
_1)92
right-tailed | o =o) | 0 > oy | T= fn a ea Soe
0
22
lefttailed | o =a) 0<o | T=" a nyt <2
fl
a
_1)g2 S? > c where — = P(S? > c*) or
two-tailed | ¢=09 | o #09 ri ) wx! 3
% S? < where a P(S? <¢’)
¢ Two samples 2-test for means:
Test ih Hy Test statistic Rejection region
T=X-Y ye
Tight-tlled | yx = po | pu > po X-Yan Norma. +) if Hy is true X-Y>e
1M
T=Y-X i _
Vet-talled | Ha | th < th Y—X~ Nona 2.2) if Hp is true Y-X>e
mm
T=X-Y sem _
twortalled yu = pa | pn # on X-Yx Namal +) if Ho is true X-Y|>e
1%
6406536925334.
Useful Data has been mentioned above.
6406536925335.
This data attachment is just for a reference & not for an evaluation.
Published solution
This is a formula/reference card. It supplies probability distributions, estimation formulas and hypothesis-test statistics for later questions. Choose the form appropriate to the model: a sample mean has variance sigma squared divided by sample size, a two-sample difference includes both variances, and significance is calculated under the null hypothesis. The administrative card itself does not require a statistical computation.
**Answer: A.**
A: **Correct:** Acknowledges that the useful reference information is supplied.
B: **Incorrect:** This also describes its reference purpose, but the source selects the first acknowledgement.
The sum from i=11 through 100 contains 90 independent observations. Dividing by 90 makes the estimator their sample mean, so its expectation is mu and its bias is zero. Its variance is \((1/90^2)90\sigma^2=\sigma^2/90\). MSE=variance+bias² therefore equals sigma²/90. The first ten unused observations do not change its sampling variance.
**Answer: A.**
A: **Correct:** Correct variance of an unbiased mean of 90 observations.
B: **Incorrect:** Omits the variance contribution from the 90 summed observations.
C: **Incorrect:** Would apply to the mean of all 100 observations, which is not this estimator.
D: **Incorrect:** This is the variance of one observation, not the mean of 90.
A density must integrate to 1. The first interval contributes 0.2×1=0.2, the third contributes 0.4×1=0.4, and the middle contributes \(\int_1^2 cx\,dx=3c/2\). Hence 0.2+1.5c+0.4=1, so 1.5c=0.4 and c=4/15. With this positive constant the piecewise density is nonnegative and normalised.
**Answer: D.**
A: **Incorrect:** 3/15 makes total probability 0.9 rather than 1.
B: **Incorrect:** 1/10 makes total probability 0.75.
C: **Incorrect:** 3/10 makes total probability 1.05.
D: **Correct:** 4/15 supplies the missing probability mass 0.4 in the middle interval.
A pharmaceutical company produces vaccine vials, and historically, the probability that a vial is
defective is
. A new sterilization method is introduced, and the company wants to test
whether this method has changed the defect rate and, specifically, whether it has increased the
probability of defective vials. To test this, 250 vials are produced using the new process and
inspected. The company will reject the new sterilization method if the total number of defective
vials is greater than 7 out of 250. Using the normal approximation to the binomial distribution,
find the significance level of the test.
Enter the correct option number for
.
Published solution
The historical null is p=0.02. Because the suspected adverse change is an increase in defects, the alternative is p>0.02, labelled option 4. Enter the label 4. The upper-count rejection rule is consistent with this right-sided alternative rather than a lower-sided or unrestricted two-sided alternative.
**Answer: 4.**
A pharmaceutical company produces vaccine vials, and historically, the probability that a vial is
defective is
. A new sterilization method is introduced, and the company wants to test
whether this method has changed the defect rate and, specifically, whether it has increased the
probability of defective vials. To test this, 250 vials are produced using the new process and
inspected. The company will reject the new sterilization method if the total number of defective
vials is greater than 7 out of 250. Using the normal approximation to the binomial distribution,
find the significance level of the test.
Enter the correct option number for
.
Published solution
Under the null, the defect count is Binomial(250,0.02). Its mean is np=250×0.02=5, listed as option 3. This is the mean count, not the defect probability itself. Enter the option label 3.
**Answer: 3.**
A pharmaceutical company produces vaccine vials, and historically, the probability that a vial is
defective is
. A new sterilization method is introduced, and the company wants to test
whether this method has changed the defect rate and, specifically, whether it has increased the
probability of defective vials. To test this, 250 vials are produced using the new process and
inspected. The company will reject the new sterilization method if the total number of defective
vials is greater than 7 out of 250. Using the normal approximation to the binomial distribution,
find the significance level of the test.
Enter the correct option number for
.
Published solution
The null variance is np(1-p)=250×0.02×0.98=4.9. The standard deviation is sqrt(4.9)≈2.214, listed as option 1. Standardisation divides by this square root, rather than by variance 4.9. Enter label 1.
**Answer: 1.**
A pharmaceutical company produces vaccine vials, and historically, the probability that a vial is
defective is
. A new sterilization method is introduced, and the company wants to test
whether this method has changed the defect rate and, specifically, whether it has increased the
probability of defective vials. To test this, 250 vials are produced using the new process and
inspected. The company will reject the new sterilization method if the total number of defective
vials is greater than 7 out of 250. Using the normal approximation to the binomial distribution,
find the significance level of the test.
Enter the correct option number for
.
Published solution
The paper's displayed normal approximation uses threshold 7 directly, without a continuity correction: \(z=(7-5)/\sqrt{4.9}\approx0.9035\). The listed rounded z-score is 0.903, option 5. Enter the option label 5. A continuity-corrected treatment of the discrete event N>7 would use 7.5 and give a different approximation; the supplied blanks are explicitly built around the uncorrected calculation.
**Answer: 5.**
A pharmaceutical company produces vaccine vials, and historically, the probability that a vial is
defective is
. A new sterilization method is introduced, and the company wants to test
whether this method has changed the defect rate and, specifically, whether it has increased the
probability of defective vials. To test this, 250 vials are produced using the new process and
inspected. The company will reject the new sterilization method if the total number of defective
vials is greater than 7 out of 250. Using the normal approximation to the binomial distribution,
find the significance level of the test.
Calculate the value for
. Enter the answer correct to three decimal place.
Published solution
Following the paper's uncorrected approximation, the significance is P(Z>0.903). The supplied table gives Phi(0.903)=0.8166, so alpha≈1-0.8166=0.1834, rounding to 0.183. This is the null probability of rejecting for a high count under the approximation used in the preceding blanks. It is not the exact binomial tail or the continuity-corrected tail.
**Answer: 0.183.**
Under H0, the independent mean difference has variance 8/16+18/36=0.5+0.5=1, so its standard deviation is 1. The two-sided rule rejects when |Xbar-Ybar|>c. Thus \(c=\Phi^{-1}(1-\alpha/2)=-\Phi^{-1}(\alpha/2)\). The observed means 41 and 40 do not enter the null critical-value formula.
**Answer: B, D.**
A: **Incorrect:** Uses the lower alpha quantile and no two-tail adjustment.
B: **Correct:** Correct upper quantile for a two-sided significance alpha.
C: **Incorrect:** Has the wrong tail probability and sign for a positive two-sided threshold.
D: **Correct:** By normal symmetry, this equals the upper 1-alpha/2 quantile.
be a discrete random variable taking values {1, 2, 3} with respective probabilities
, where
is a parameter. Consider the samples {1, 2, 2, 1, 3, 2, 1, 3,
2, 3} from
. Using a Uniform[0, 1] prior, find the posterior mean of
. Enter the answer correct to
two decimal places.
Published solution
The outcomes 1,2,3 have probabilities p,(1-p)/2,(1-p)/2. In the sample, 1 occurs 3 times and the other values occur 7 times. Thus the likelihood is proportional to p³(1-p)^7; the factor 2^-7 does not depend on p. With a uniform Beta(1,1) prior, the posterior is Beta(4,8), whose mean is 4/(4+8)=1/3≈0.33. Counts of outcomes 2 and 3 combine because they share the same p-dependent factor.
**Answer: 0.33.**
For Poisson(lambda), each observation contributes e^-lambda lambda^x/x!. The sample 2,3,1 has total count 6 across 3 observations, so the exact likelihood is \(e^{-3\lambda}\lambda^6/(2!3!1!)=e^{-3\lambda}\lambda^6/12\). All listed likelihood choices omit constant factors independent of lambda. Up to that common proportionality convention, B is e^-3lambda lambda^6.
**Answer: B.**
A: **Incorrect:** Uses power 5 although the sample sum is 6.
B: **Correct:** Correct lambda-dependent likelihood, up to the omitted constant 1/12.
C: **Incorrect:** Uses only one observation's exponential factor instead of three.
D: **Incorrect:** The positive exponential sign and success power are wrong.
be a discrete random variable with probability mass function:
Consider the random sample {2,3,1}. Assume that the prior distribution of
is Gamma
.
Based on the above data, answer the given subquestions.
Find the posterior mean of
using the given prior distribution.
Published solution
The shape-rate Gamma(2,1) prior has density proportional to lambda^(2-1)e^-lambda. Multiplying by the sample likelihood lambda^6 e^-3lambda gives lambda^7 e^-4lambda, a Gamma(8,4) posterior. Its mean is shape/rate=8/4=2. The omitted factorial constant from the likelihood cancels during posterior normalisation.
**Answer: 2.**
Based on the above data, answer the given subquestions.
If
, then find the value of
.
Published solution
The PMF masses 2a,3a,a,4a sum to 10a=1, so a=0.1. Its expectation is 0(2a)+2(3a)+4a+x4(4a)=10a+4ax4. Given E[X]=3, substitute a=0.1 to get 1+0.4x4=3. Hence x4=2/0.4=5.
**Answer: 5.**
Based on the above data, answer the given subquestions.
Define a new random variable
. Find the value of
. Enter the answer
correct to one decimal place.
Published solution
Y=X+1, so 1<Y<4 means 0<X<3. The only support value in that interval is X=2; X=0 is excluded and the other values are 4 and 5. Normalisation gives a=0.1, so P(Y in (1,4))=P(X=2)=3a=0.3. Be careful with the strict lower endpoint.
**Answer: 0.3.**
represent the number of trials required to achieve a certain outcome
in an experiment, where
{1, 2, 3, 4, 5, 6}. The probability that the outcome occurs in 1 to 5
trials is
each, while the probability that it takes 6 trials is
. A random sample of five
independent observations of
is recorded as:
4, 2, 5, 3, 1
Based on the above data, answer the given subquestions.
Find the method of moments estimate of
for the given sample. Enter the answer correct to one
decimal place.
Published solution
The distribution has probability theta at each of 1,...,5 and 1-5theta at 6. Its mean is theta(1+2+3+4+5)+6(1-5theta)=6-15theta. The sample mean is (4+2+5+3+1)/5=3. Equating moments gives 6-15theta=3, so theta=0.2. This lies at the valid boundary 0<=theta<=1/5, where P(X=6)=0.
**Answer: 0.2.**
represent the number of trials required to achieve a certain outcome
in an experiment, where
{1, 2, 3, 4, 5, 6}. The probability that the outcome occurs in 1 to 5
trials is
each, while the probability that it takes 6 trials is
. A random sample of five
independent observations of
is recorded as:
4, 2, 5, 3, 1
Based on the above data, answer the given subquestions.
The experiment is considered efficient if the expected number of trials required to achieve the
outcome is at most 4. Using the estimated value of
, determine whether this requirement is
satisfied.
6406536925362.
The requirement is satisfied.
6406536925363.
The requirement is not satisfied.
Published solution
With the method-of-moments estimate theta=0.2, the estimated population mean is 6-15(0.2)=3 trials. This is at most 4, so the stated efficiency requirement is satisfied under that fitted model. This conclusion is based on the estimate, not a certainty about the unknown population mean.
**Answer: A.**
A: **Correct:** The estimated mean 3 satisfies the upper bound 4.
B: **Incorrect:** The estimated mean does not exceed 4.
A box contains some green balls and some yellow balls. Meera claims that if a ball is picked
randomly from the box, then the probability
that the ball will be green is 0.25. Arjun suspects
that this probability is too high. To test the claim, he randomly picks 8 balls from the box. If two or
more green balls are obtained among the 8 balls, then he accepts Meera's claim; otherwise, he
rejects it.
Based on the above data, answer the given subquestions.
Define the null hypothesis and the alternative hypothesis.
The claimed probability is 0.25, while Arjun suspects that claim is too high, meaning the actual probability is smaller. Thus H0:p=0.25 and HA:p<0.25. Rejecting for fewer than two green balls is consistent with this lower-sided alternative.
**Answer: C.**
A: **Incorrect:** Tests either direction of change rather than specifically a lower green probability.
B: **Incorrect:** Tests a higher probability, contrary to the suspicion.
C: **Correct:** Uses the stated null value and the correct lower-sided alternative.
D: **Incorrect:** Reverses the claimed-value null and the alternative.
A box contains some green balls and some yellow balls. Meera claims that if a ball is picked
randomly from the box, then the probability
that the ball will be green is 0.25. Arjun suspects
that this probability is too high. To test the claim, he randomly picks 8 balls from the box. If two or
more green balls are obtained among the 8 balls, then he accepts Meera's claim; otherwise, he
rejects it.
Based on the above data, answer the given subquestions.
Find the significance level of the test. Enter the answer correct to three decimal places.
Published solution
Using the intended independent Bernoulli-draw model, G~Binomial(8,0.25) under H0. The rule rejects when G<2, so \[\alpha=P(G=0)+P(G=1)=0.75^8+8(0.25)0.75^7\approx0.367081.\] Round to 0.367. If sampling without replacement from a small finite box were intended, composition information and a hypergeometric model would be needed; the source calculation uses the binomial assumption.
**Answer: 0.367.**
.
Based on the above data, answer the given subquestions.
Find the value of
for which
. Enter the answer correct to one decimal
place.
Published solution
For Poisson(lambda), P(X=2)=e^-lambda lambda²/2 and P(X=1)=e^-lambda lambda. The supplied equality gives 4lambda²/2=5lambda after cancelling e^-lambda. For positive lambda, divide by lambda: 2lambda=5, hence lambda=2.5. The positive-rate convention excludes the degenerate lambda=0 solution.
**Answer: 2.5.**
.
Based on the above data, answer the given subquestions.
Find
. Enter the answer correct to two decimal places.
Published solution
For a Poisson variable, E[X]=lambda and Var(X)=lambda. Therefore E[X²]=Var(X)+(E[X])²=lambda+lambda². Using the shared value lambda=2.5 gives 2.5+6.25=8.75. The second raw moment includes the variance term; it is not simply the squared mean.
**Answer: 8.75.**
