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Question 3 MSQ · 3.0 marks
A company manufactures batteries and claims that the standard deviation of the operating life of
a battery is 4 hours. The quality control manager suspects that the actual variability is higher than
the claimed value. A random sample of 17 batteries gave a sample standard deviation of 6 hours.
Choose the correct set of option(s).
6406537406648.6406537406649.6406537406650.
We accept company's claim at 5% level of significance.
6406537406651.
We reject company's claim at 5% level of significance.
Published solution
The suspicion is higher variability, so test H0:sigma=4 against HA:sigma>4. For a normal population, \(T=(n-1)S^2/\sigma_0^2=16(36)/16=36\) follows chi-square with 16 degrees of freedom under H0. The supplied upper 5% critical value is 26.296. Since 36>26.296, reject the company's claimed standard deviation at 5%. The variance test needs the normality assumption for this exact chi-square calibration.
**Answer: A, D.**
A: **Correct:** Uses the correct claimed standard deviation and right-tailed alternative.
B: **Incorrect:** Tests a decrease, contrary to the manager's suspicion.
C: **Incorrect:** The statistic lies in the rejection region.
D: **Correct:** 36 exceeds the supplied right-tail critical value.
Significance is the probability of rejecting a true null. The rule accepts H0 when X<c, so it rejects when X>=c. Under H0, X has mean 20 and variance 16, hence standard deviation 4. Therefore \[\alpha=P_{20}(X\ge c)=1-F_Z\left(\frac{c-20}{4}\right).\] Endpoints do not matter for a normal variable. Evaluate this probability under mean 20, not under the alternative mean 24.
**Answer: B.**
A: **Incorrect:** This is the acceptance probability, not the rejection probability.
B: **Correct:** Correctly standardises under the null and takes the upper tail.
C: **Incorrect:** Uses the alternative distribution rather than the null.
D: **Incorrect:** Uses the variance 16 in place of the standard deviation 4 and takes the wrong tail.
Find the value of Enter your answer correct to three decimal places.
Published solution
Integrate the joint density over 0<x<1/2 and 0<y<1: \[P(X<1/2,Y<1)=\int_0^{1/2}\int_0^1 3xy(1-x)\,dy\,dx=\frac32\int_0^{1/2}(x-x^2)\,dx.\] The antiderivative is (3/2)(x²/2-x³/3). At 1/2 it equals (3/2)(1/8-1/24)=1/8=0.125. The integration region lies entirely inside the stated support.
**Answer: 0.125.**
A botanist is studying a rare plant species and trying to estimate the probability
that a given
seed will germinate under controlled conditions. Historical data
indicates that the prior distribution of
follows a
distribution. After planting 15 seeds,
where each seed's germination is independent of the others, the botanist observes that 6 out of
the 15 seeds germinated. Find the posterior mean of
. Enter your answer correct to two decimal
places.
Published solution
A Beta(alpha,beta) prior with s successes and f failures gives Beta(alpha+s,beta+f) posterior for a Bernoulli success probability. Here alpha=4,beta=5,s=6,f=15-6=9, so the posterior is Beta(10,14). Its mean is 10/(10+14)=10/24≈0.416667. Rounded to two decimals, enter 0.42. The second beta parameter counts failures, not the total number of trials.
**Answer: 0.42.**
be a random sample from a population having probability mass
function
Find the maximum likelihood estimator of
. Also, find the maximum likelihood estimate of
for
the sample {2, 1, 3, 2, 2}
**Solution:**
Since
are independent, the likelihood function is
Therefore,
The log-likelihood is
Hence,
Differentiating
with respect to
, and set the derivative equal to zero, we obtain
Now, maximum likelihood estimate for the sample {2, 1, 3, 2, 2} will be
.
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
Each observation is Binomial(3,theta). Independence gives a product likelihood. Collecting powers yields \[L(\theta)=\left[\prod_{i=1}^n\binom3{x_i}\right]\theta^{\sum_i x_i}(1-\theta)^{3n-\sum_i x_i}.\] This is option 2. Binomial coefficients multiply, and the total number of Bernoulli trials across n observations is 3n. Enter the option number 2.
**Answer: 2.**
be a random sample from a population having probability mass
function
Find the maximum likelihood estimator of
. Also, find the maximum likelihood estimate of
for
the sample {2, 1, 3, 2, 2}
**Solution:**
Since
are independent, the likelihood function is
Therefore,
The log-likelihood is
Hence,
Differentiating
with respect to
, and set the derivative equal to zero, we obtain
Now, maximum likelihood estimate for the sample {2, 1, 3, 2, 2} will be
.
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
Take the log of the likelihood: \[\ell(\theta)=\sum_i\log\binom3{x_i}+\left(\sum_ix_i\right)\log\theta+\left(3n-\sum_ix_i\right)\log(1-\theta).\] Products become sums, while the powers become coefficients. This expression is option 6. The combinatorial term is constant in theta but is still part of the complete log-likelihood.
**Answer: 6.**
be a random sample from a population having probability mass
function
Find the maximum likelihood estimator of
. Also, find the maximum likelihood estimate of
for
the sample {2, 1, 3, 2, 2}
**Solution:**
Since
are independent, the likelihood function is
Therefore,
The log-likelihood is
Hence,
Differentiating
with respect to
, and set the derivative equal to zero, we obtain
Now, maximum likelihood estimate for the sample {2, 1, 3, 2, 2} will be
.
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
Let S=sum(xi). Differentiating the nonconstant log-likelihood gives S/theta-(3n-S)/(1-theta)=0. Rearranging yields S=3n theta, so \(\hat\theta=S/(3n)=\bar X/3\). The second derivative is negative for interior samples, confirming a maximum. This estimator is option 8. The factor 3 is needed because each observation counts successes in three trials.
**Answer: 8.**
be a random sample from a population having probability mass
function
Find the maximum likelihood estimator of
. Also, find the maximum likelihood estimate of
for
the sample {2, 1, 3, 2, 2}
**Solution:**
Since
are independent, the likelihood function is
Therefore,
The log-likelihood is
Hence,
Differentiating
with respect to
, and set the derivative equal to zero, we obtain
Now, maximum likelihood estimate for the sample {2, 1, 3, 2, 2} will be
.
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.
Calculate the value for
. Enter the answer correct to two decimal places.
Published solution
The sample values 2,1,3,2,2 sum to 10 across n=5 observations. The MLE for a Binomial(3,theta) parameter is total successes divided by total trials: \(\hat\theta=10/(3\cdot5)=2/3\approx0.666667\). Correct to two decimals, enter 0.67. The raw sample mean is 2 successes per observation, not a probability.
**Answer: 0.67.**
be a discrete random variable with probability mass function
Based on the above data, answer the given subquestions.
Find the value of
. Enter the answer correct to one decimal place.
Published solution
The strict event 0<X<3 contains only X=1 and X=2. Their PMF values are (1+1)/10=0.2 and (2+1)/10=0.3. Adding gives 0.5. Do not include either endpoint: X=0 has mass 0.1 and X=3 has mass 0.4, and they are excluded by the strict inequalities.
**Answer: 0.5.**
be a discrete random variable with probability mass function
Based on the above data, answer the given subquestions.
Find the value of
.
Published solution
First compute E[X]=0(0.1)+1(0.2)+2(0.3)+3(0.4)=2. By linearity, E[X+2]=E[X]+2=4. This does not require transforming the entire PMF, although doing so would give the same shifted expectation.
**Answer: 4.**
The lifetime (in hours) of a certain electronic component is modeled by a continuous random
variable
with PDF
Based on the above data, answer the given subquestions.
Given that a randomly selected component has a lifetime of more than 2 hours, what is the
probability that its lifetime is less than 3 hours? Enter the answer correct to two decimal places.
Published solution
Use the conditional probability ratio: \[P(X<3\mid X>2)=\frac{P(2<X<3)}{P(X>2)}.\] For density x/8, the numerator is [x²/16] from 2 to 3=(9-4)/16=5/16. The denominator is [x²/16] from 2 to 4=(16-4)/16=12/16. Thus the ratio is 5/12≈0.416667, rounding to 0.42. The observation X>2 restricts the reference population for the denominator.
**Answer: 0.42.**
Y=X² is increasing for 0<X<4. Its inverse is x=sqrt(y) with derivative 1/(2sqrt(y)). The support becomes 0<y<16. Therefore fY(y)=fX(sqrt(y))|dx/dy|=(sqrt(y)/8)/(2sqrt(y))=1/16 on (0,16), zero elsewhere. The Jacobian factor is essential; simply substituting sqrt(y) into the original density is insufficient.
**Answer: A.**
A: **Correct:** The constant 1/16 on an interval of length 16 is the correct transformed density.
B: **Incorrect:** Omits the inverse-derivative factor and does not have the correct normalisation.
C: **Incorrect:** Retains a spurious y-dependent inverse-square-root factor.
D: **Incorrect:** Treats the transformed value as though it still had the original x/8 density.
A company claims that the mean lifetime of a particular type of battery is
hours. A researcher
suspects that the actual mean lifetime is less than
hours. The hypotheses are
A test is performed at a significance level of
. The resulting p-value is 0.03. It is also
known that the probability of failing to reject
when the true mean lifetime is
hours is
.
Based on the above data, answer the given subquestions.
Which of the following statements is correct?
6406537406669.
The probability of type I error is
.
6406537406670.
Since the p-value is less than
,
is rejected, and the probability of a Type
I error is
.
6406537406671.
The probability of a Type II error is
, and the power of the test is
,
against the alternative
6406537406672.
The power of the test is
, against the alternative
Published solution
The given probability of failing to reject when mu=95 is the Type II error beta=0.20 at that alternative. Power there is 1-beta=0.80. The significance level is alpha=0.05, controlling rejection under the null. The observed p-value 0.03 is below alpha and leads to rejection, but it is not itself the probability that the null is true or the test's Type I error rate.
**Answer: C.**
A: **Incorrect:** 0.20 is the specified Type II error at mu=95, not the Type I level.
B: **Incorrect:** Rejection is correct, but assigning the observed p-value as the Type I error probability is incorrect.
C: **Correct:** Correctly identifies beta=0.20 and power=0.80 at mu=95.
D: **Incorrect:** 0.20 is the miss probability, not the power.
A company claims that the mean lifetime of a particular type of battery is
hours. A researcher
suspects that the actual mean lifetime is less than
hours. The hypotheses are
A test is performed at a significance level of
. The resulting p-value is 0.03. It is also
known that the probability of failing to reject
when the true mean lifetime is
hours is
.
Based on the above data, answer the given subquestions.
For the hypothesis test described here, the population standard deviation is known to be
.
The test rejects
when
If the test is conducted at a significance level
,
determine the sample size
required for the given rejection rule.
Published solution
Under H0, the sample mean has mean 100 and standard deviation 10/sqrt(n). The left-tail rejection threshold is 96.71. For alpha=0.05 the supplied z critical point is -1.645. Set \[(96.71-100)/(10/\sqrt n)=-1.645.\] This becomes -0.329sqrt(n)=-1.645, giving sqrt(n)=5 and n=25. A normal model or suitable CLT approximation underlies this calibration.
**Answer: 25.**
that a randomly selected student passes a
particular examination. Before any students are observed, the prior distribution of
is given
below:
Five students are selected independently, and their examination outcomes
are observed to be:
Based on the above data, answer the given subquestions.
Find the value of
. Enter the answer correct to three decimal places.
Published solution
The exact ordered outcome sequence has three passes and two failures. Its probability conditional on p is p³(1-p)². Mix over the prior: \[P(S)=0.6(0.8^3\,0.2^2)+0.4(0.6^3\,0.4^2)=0.012288+0.013824=0.026112.\] Rounded to three decimals this is 0.026. No choose(5,3) factor appears because the question specifies the exact sequence S.
**Answer: 0.026.**
that a randomly selected student passes a
particular examination. Before any students are observed, the prior distribution of
is given
below:
Five students are selected independently, and their examination outcomes
are observed to be:
Based on the above data, answer the given subquestions.
Using the posterior mode, estimate the value of
.
Published solution
The posterior candidate weights are prior times likelihood. At p=0.8, the weight is 0.6×0.8³×0.2²=0.012288. At p=0.6, it is 0.4×0.6³×0.4²=0.013824. The larger weight is at p=0.6, so the posterior mode is 0.6, despite its smaller prior mass. A common normalising denominator does not alter this comparison.
**Answer: 0.6.**
For the estimator mu-hat=aXbar+b, linearity gives E(mu-hat)=a mu+b. Bias is estimator expectation minus the true target, so Bias(mu-hat)=a mu+b-mu=(a-1)mu+b. The constant b can alter bias without affecting the estimator's variance.
**Answer: B.**
A: **Incorrect:** Adds the target mean rather than subtracting it.
B: **Correct:** Correctly subtracts mu from a mu+b.
C: **Incorrect:** Omits subtraction of the true mean and uses the wrong sign on b.
D: **Incorrect:** Reverses the coefficient of mu; the stated bias convention is E(estimator)-parameter.
.
Based on the above data, answer the given subquestions.
Suppose
,
,
,
, and
. find the value of the risk
.
Published solution
Under squared-error loss, risk is variance plus squared bias. With a=2,b=1,mu=5,sigma²=4,n=8, the bias is (2-1)5+1=6. Variance of the estimator is a²sigma²/n=4×4/8=2. Therefore \[E[(\mu-\hat\mu)^2]=2+6^2=38.\] The direction inside the square does not affect this squared-error risk.
**Answer: 38.**
