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**About Statistical Inference for Estimation in Data Science Course**

This course covers statistical inference, sampling distributions, and confidence intervals. Students will learn how to define and make good estimators, the method of moments estimation, the maximum likelihood estimation, and ways to make confidence intervals that can be used in a wider range of situations.

This course is part of CU Boulder’s Master of Science in Data Science (MS-DS) degree, which is offered on the Coursera platform. It can be taken for academic credit. The MS-DS is a degree that brings together professors from Applied Mathematics, Computer Science, Information Science, and other departments at CU Boulder. The MS-DS is perfect for people who have a wide range of undergraduate education and/or work experience in computer science, information science, mathematics, and statistics. Admission is based on performance, and there is no application process. Visit https://www.coursera.org/degrees/master-of-science-data-science-boulder to find out more about the MS-DS programme.

Photo by Christopher Burns on Unsplash was used to make the logo.

WHAT YOU’LL FIND OUT

Find out what makes an estimator “good” and be able to compare different ones.

Use the methods of maximum likelihood and method of moments estimation to build good estimators.

Build confidence intervals for one and two population means, one and two population proportions, and a population variance, and explain what they mean.

**Course Apply Link – Statistical Inference for Estimation in Data Science **

**Statistical Inference for Estimation in Data Science Quiz Answers**

### Week 1 Quiz Answers

#### Quiz 1: Recognizing Discrete Distributions

Q1. Radiation from a Cesium-137 source is measured using a scintillation detector. The detector absorbs the radiation and re-admits the energy as visible photons at a certain rate commensurate with the half-life of Cesium-137.

Which of the following distributions may be used to best describe the number of photons emitted during a one hour period of time?

- Geometric
- Bernoulli
- Poisson
- Binomial

Q2. Suppose George is eating at his favorite restaurant one cold winter afternoon. He sits at his usual table, right by the window so he can entertain himself watching pedestrians stroll down the sidewalk of the moderately busy city street. In between sips of his bitter coffee, he notices that the occasional passerby slips on a small patch of ice right outside the restaurant. Now, as opposed to being a more caring sort and warning the waiter that a patch of ice needed salting, George continues to watch. He begins to think over time that there is a more or less constant probability of any one person slipping. He wants to know how many people might walk safely by before one slips. You can assume that each pedestrian is either well spaced enough to avoid seeing the person ahead of them fall or too engrossed in a cell phone to notice. Which below distribution could be used?

- Binomial
- Geometric
- Poisson
- Bernoulli

Q3. Jerry is a high school student who works at the local arcade as his summer job. This summer he’s been assigned to the prize booth, doling out stuffed animals, keychains, and multi-colored bouncy balls to kids with wads of tickets in their hands. Having been there for several summers, Jerry’s boss has let him in on some of the secrets of the business. For example, one of the games consists of a ring that can light up. A single dot of light speeds around the ring once tokens are put in the machine. If the player clicks a button at just the right time to stop the light at the marked point, they win. However, Jerry knows that even if the player pushes the button at just the right point, the machine is setup to only stop the light at the marked spot with a fixed probability. Jerry, standing there at the prize booth, wonders how many times out of ten tries a player might win the game assuming they pressed the button perfectly each and every time. Which distribution is this?

- Bernoulli
- Geometric
- Poisson
- Binomial

Q4. Jennifer is a construction worker at the site of a new skyscraper. The building-in-progress sits next to the bridge that spans the river flowing through the heart of downtown. After a hectic morning listening to the commands of her boss and the complaints of her coworkers, she likes to take her lunch up to a partially finished upper floor and watch cars pass on the bridge beneath her as she eats. After several days watching the traffic go by, and with an interest in all things statistics, she begins to wonder if she could model how many cars go by during her one hour lunch break. Which distribution might be used for this?

- Bernoulli
- Poisson
- Binomial
- Geometric

Q5. Candice works as a lifeguard at the local pool. In between shifts of scanning the water for trouble and whistling at kids to remind them of the no running rule, she and three fellow lifeguards relax in the break room. Most days they spend their fifteen minute breaks playing cards, specifically a game called euchre. However, one afternoon while Candice is on break a fellow lifeguard instead wants to play Texas hold’em poker. Candice searches in her pocket for a quarter to toss to settle the dispute, but can’t seem to find one. Instead she tosses him the unopened candy bar she was about to eat and tells him to throw it in the air and call out which of the two sides will be face up when it lands. He argues that it’s not a fair flip since the candy bar isn’t symmetrically shaped. Candice, who has always been carefree, shrugs and points out he is the one making the call so he has the advantage. What distribution might describe this toss?

- Bernoulli
- Geometric
- Poisson
- Binomial

#### Quiz 2: Probability, Expectation, and Variance

Q1. A big box store sells three different models of refrigerators having 20, 21.5, and 26.8 cubic feet of usable storage space, respectively. Let X = the amount of storage space purchased by the next customer to buy a freezer. Suppose that X has probability mass function given by *x**p*(*x*)200.121.50.726.80.2

If the price of a refrigerator with X cubic feet of capacity is (20X+100)(20*X*+100), what is the expected price paid by the next customer to buy a freezer?

- $722.77
- $538.12
- $548.20
- $560.00

Q2. In proof testing of circuit boards, the probability that any particular diode will fail is 0.030.03. Suppose a circuit board contains 150 diodes and that each one performs or fails independently of all others. How many diodes would you expect to fail, and what is the standard deviation of the number that are expected to fail?

- 4 and 0.0014
- 1.5 and 0.0002
- 5 and 0.0020
- 4.5 and 2.0893

Q3. Let X denote the distance, in meters, that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose that for Solomon Island prehensile-tailed skinks, X has an exponential distribution with rate parameter \lambda = 0.014*λ*=0.014.

What is the probability that the distance exceeds the mean distance by more than 2 standard deviations?

- 0.18
- 0.05
- 0.10
- 0.95

Q4. The mode of a continuous distribution with probability density function f(x)*f*(*x*) is the value that maximizes f*f*.

What is the mode of the gamma distribution with parameters \alpha*α* and \beta*β* for \alpha>1*α*>1?

- \frac{\alpha}{\beta^{2}}
*β*2*α* - \frac{\alpha}{\beta}
*βα* - \frac{\alpha-1}{\Gamma(\alpha) \beta}Γ(
*α*)*βα*−1 - \frac{\alpha-1}{\beta}
*βα*−1

Q5. Suppose that X*X* is a Poisson random variable with parameter \lambda*λ*. What is E[X^{2}]*E*[*X*2]?

- \lambda
*λ* - \lambda^{2}
*λ*2 - \frac{\lambda}{1+\lambda}1+
*λλ* - \lambda(1+\lambda)
*λ*(1+*λ*)

#### Quiz 3: Method of Moments Estimation

Q1. Let X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* be a random sample from the Bernoulli distribution with parameter p*p*. Let \widehat{p}*p* be the method of moments estimator for p*p*.

Which of the following is true. (Check all that apply.)

- \widehat{p}
*p* is an unbiased estimator of p*p* - \widehat{p}
*p* is the sample mean. - \widehat{p}
*p* is the proportion of values in the sample that are equal to 1. - Var[\widehat{p}]=1
*Var*[*p*]=1

Q2. Let X_{1}, X_{2},\ldots, X_{n}*X*1,*X*2,…,*X**n* be a random sample from the geometric distribution with parameter p*p*, that takes on the values 0,1,2,\ldots0,1,2,….

Which of the following is the method of moments estimator of p*p*.

- \overline{X}
*X* - \frac{n}{2}2
*n* - \frac{1}{\overline{X}+1}
*X*+11 - \frac{1-\overline{X}}{\overline{X}}
*X*1−*X*

Q3. Let X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* be a random sample from the Beta distribution with parameters a*a* and b*b*.

Which of the following are the method of moments estimators for a*a* and b*b*?

- \widehat{a}=\frac{\overline{X}^{2}(1-\overline{X})}{\frac{1}{n} \sum_{i=1}^{n} X_{i}^{2}}-\overline{X}
*a*=*n*1∑*i*=1*n**Xi*2*X*2(1−*X*)−*X*and \widehat{b}=\widehat{a} \,\frac{1-\overline{X}}{\overline{X}}*b*=*aX*1−*X* - \widehat{a}=\frac{\overline{X}^{2}(1-\overline{X})}{\frac{1}{n} \sum_{i=1}^{n} X_{i}^{2}}-\overline{X}^{2}
*a*=*n*1∑*i*=1*n**Xi*2*X*2(1−*X*)−*X*2 and \widehat{b}=\widehat{a}(\overline{X}+1)*b*=*a*(*X*+1) - \widehat{a} = \frac{\overline{X}}{\overline{X}+1}
*a*=*X*+1*X* and \widehat{b}=\widehat{a}(\overline{X}+1)*b*=*a*(*X*+1) - \widehat{a} = \frac{\overline{X}}{\overline{X}+1}
*a*=*X*+1*X* and \widehat{b} = \left( \frac{\overline{X}}{\overline{X}+1}\right)^{2}*b*=(*X*+1*X*)2

Q4. Let X_{1}, X_{2}, \ldots,X_{n}*X*1,*X*2,…,*X**n* be a random sample from the continuous Pareto distribution as given in the table of distributions for this course.

Which of the following is the method of moments estimator of the parameter \gamma*γ*?

- \frac{1}{\overline{X}}+1
*X*1+1 - \frac{1}{\overline{X}+1}
*X*+11 - \overline{X}
*X* - \frac{1}{\overline{X}-1}
*X*−11

Q5. Method of moments estimators are guaranteed to be unbiased estimators.

- True
- False

### Week 2 Quiz Answers

#### Quiz 1: Finding MLEs Quiz Answers

Q1. Let X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* be a random sample from the Poisson distribution with parameter \lambda*λ*.

Which of the following is the maximum likelihood estimator for \lambda*λ*?

- The sample standard deviation S
*S*. - The maximum value in the sample: \max(X_{1}, X_{2}, \ldots, X_{n})max(
*X*1,*X*2,…,*Xn*). - The sample mean \overline{X}
*X*. - The sample variance S^{2}
*S*2.

Q2. Suppose that X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* is a random sample from the \Gamma(\alpha,\beta)Γ(*α*,*β*) distribution **where \alpha α is known**.

Which of the following is the maximum likelihood estimator for \beta*β*?

- \alpha/ \,\overline{X}
*α*/*X* - \overline{X}^{\alpha-1}
*Xα*−1 - \sum_{i=1}^{n} (X_{i}/\alpha)∑
*i*=1*n*(*Xi*/*α*) - \sum_{i=1}^{n}X_{i}^{2}/\alpha∑
*i*=1*n**Xi*2/*α*

Q3. Consider a random sample X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* from the shifted exponential distribution with pdf

f(x;\theta) = *f*(*x*;*θ*)=

{*e*−(*x*−*θ*)0,,*x*≥*θ**x*<*θ*

{*e*−(*x*−*θ*)0,,*x*≥*θ**x*<*θ*

Which of the following is the maximum likelihood estimator for \theta*θ*?

- \max (X_{1}, X_{2}, \ldots, X_{n})max(
*X*1,*X*2,…,*Xn*) - \overline{X}
*X* - 1/ \, \overline{X}1/
*X* - \min (X_{1}, X_{2}, \ldots, X_{n})min(
*X*1,*X*2,…,*Xn*)

Q4. Consider a random sample X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* from the continuous Pareto distribution with pdf

f(x;\gamma) = *f*(*x*;*γ*)=

⎧⎩⎨*γ*(1+*x*)*γ*+10,,*x*≥0*x*<0

{(1+*x*)*γ*+1*γ*0,,*x*≥0*x*<0

Here \gamma>0*γ*>0 is a parameter to be estimated.

Which of the following is the maximum likelihood estimator for \gamma*γ*?

- n / \, \sum_{i=1}^{n} \ln (1+X_{i})
*n*/∑*i*=1*n*ln(1+*Xi*) - \prod_{i=1}^{n} \frac{1}{1+X_{i}}∏
*i*=1*n*1+*Xi*1 - \max(X_{1}, X_{2}, \ldots, X_{n})max(
*X*1,*X*2,…,*Xn*) - \overline{X}
*X*

Q5. Suppose that X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* is a random sample from the shifted exponential distribution with rate \lambda*λ* which has pdf

f(x;\lambda,\theta) = *f*(*x*;*λ*,*θ*)=

{*λ**e*−*λ*(*x*−*θ*)0,,*x*≥*θ**x*<*θ*

{*λ**e*−*λ*(*x*−*θ*)0,,*x*≥*θ**x*<*θ*

Which of the following are the MLEs for \lambda*λ* and \theta*θ*?

- \widehat{\lambda} = \frac{n}{\sum_{i=1}^{n} (X_{i}-X_{(n)})}\,\,
*λ*=∑*i*=1*n*(*Xi*−*X*(*n*))*n* and \,\, \widehat{\theta} = X_{(n)} \,\,*θ*=*X*(*n*) where \,\,X_{(n)}= \max (X_{1}, X_{2}, \ldots, X_{n})*X*(*n*)=max(*X*1,*X*2,…,*Xn*) - \widehat{\lambda} = \frac{n}{\sum_{i=1}^{n} (X_{i}-X_{(1)})}\,\,
*λ*=∑*i*=1*n*(*Xi*−*X*(1))*n* and \,\, \widehat{\theta} = X_{(1)} \,\,*θ*=*X*(1) where \,\,X_{(1)}= \min (X_{1}, X_{2}, \ldots, X_{n})*X*(1)=min(*X*1,*X*2,…,*Xn*) - \widehat{\lambda} = \overline{X}-X_{(1)}\,\,
*λ*=*X*−*X*(1) and \,\, \widehat{\theta} = X_{(1)} \,\,*θ*=*X*(1) where \,\,X_{(1)}= \min (X_{1}, X_{2}, \ldots, X_{n})*X*(1)=min(*X*1,*X*2,…,*Xn*) - \widehat{\lambda} = \overline{X}\,\,
*λ*=*X*and \,\, \widehat{\theta} = X_{(1)} \,\,*θ*=*X*(1) where \,\,X_{(1)}= \min (X_{1}, X_{2}, \ldots, X_{n})*X*(1)=min(*X*1,*X*2,…,*Xn*)

#### Quiz 2: Invariance, Mean-Squared Error, and Efficiency

Q1. Suppose that X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* is a random sample from the exponential distribution with rate \lambda>0*λ*>0.

Find the MLE for P(X>1)*P*(*X*>1) where X*X* is a future sampled value.

- \frac{1}{n} \sum_{i=1}^{n} \ln X_{i}
*n*1∑*i*=1*n*ln*Xi* - e^{-1/\overline{X}}
*e*−1/*X* - \sum_{i=1}^{n} \ln (X_{i}/\overline{X})∑
*i*=1*n*ln(*Xi*/*X*) - 1-e^{-\overline{X}}1−
*e*−*X*

Q2. Let X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* be a random sample from any distribution with mean \mu*μ* and variance \sigma^{2}*σ*2.

Consider the following two unbiased estimators of \mu*μ*.

\widehat{\mu}_{1} = \overline{X}*μ*1=*X*

and

\widehat{\mu}_{2} = \frac{X_{1}}{3} + \frac{2(X_{2}+ X_{3}+ \cdots +X_{n})}{3(n-1)}*μ*2=3*X*1+3(*n*−1)2(*X*2+*X*3+⋯+*X**n*)

Find the efficiency of \widehat{\mu}_{1}*μ*1 relative to \widehat{\mu}_{2}*μ*2.

- \frac{n^{2}+3n}{9(n-1)}9(
*n*−1)*n*2+3*n* - \frac{n-1}{3n}3
*nn*−1 - \frac{2n^{2}-3n}{8}82
*n*2−3*n* - \frac{2(n-1)}{3n^{2}}3
*n*22(*n*−1)

Q3. Suppose that X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* is a random sample from any distribution with mean \mu*μ* and variance \sigma^{2}*σ*2.

Consider the estimator of \sigma^{2}*σ*2 given by

\widehat{\sigma^{2}} = \frac{\sum_{i=1}^{n} (X_{i}-\overline{X})^2}{n}*σ*2=*n*∑*i*=1*n*(*X**i*−*X*)2.

What is the bias of this estimator?

- \frac{\mu^{2}}{\sigma^{2}}
*σ*2*μ*2 - S^{2}
*S*2 - -\frac{1}{n} \sigma^{2}−
*n*1*σ*2 - 00

Q4. The *median* of a continuous distribution corresponding to a rnadom variable X*X*, is a number \xi*ξ* such that P(X \leq \xi)=1/2*P*(*X*≤*ξ*)=1/2.

If X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* is a random sample from the distribution with pdf

f(x)=*f*(*x*)=

⎧⎩⎨3*x*2*θ*30,,0<*x*<*θ*otherwise,

{*θ*33*x*20,,0<*x*<*θ*otherwise,

find the MLE for the median of the distribution.

- \sqrt[3]{1/2} \cdot \max(X_{1}, X_{2}, \ldots, X_{n})31/2⋅max(
*X*1,*X*2,…,*Xn*) - \xi^{3}/\theta^{3}
*ξ*3/*θ*3 - \overline{X}^{1/3}/\theta^{3}
*X*1/3/*θ*3 - \sqrt{2/3} \cdot \min(X_{1}, X_{2}, \ldots, X_{n})2/3⋅min(
*X*1,*X*2,…,*Xn*)

Q5. Let X_{1},X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* be a random sample from the Poisson distribution with parameter \lambda>0*λ*>0. Let \widehat{\theta}*θ* be the MLE for \lambda*λ*.

What is the mean-squared error of \widehat{\lambda}*λ* as an estimator of \lambda*λ*?

- 00
- \lambda/n
*λ*/*n* - \lambda^{2}/\sqrt{n}
*λ*2/*n* - \lambda^{2}/n
*λ*2/*n*

### Week 3 Quiz Answers

#### Quiz 1: The Cramer-Rao Lower Bound

Q1. Let X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* be a random sample from the normal distribution with mean \mu*μ* and variance \sigma^{2}*σ*2. Assume that \sigma^{2}*σ*2 is known.

Find the Cramer-Rao lower bound for the variance of all unbiased estimators of \mu*μ*.

- \frac{\sigma^{2}}{n}
*nσ*2 - \frac{\mu^{2}}{n}
*nμ*2 - \frac{\mu}{\sigma \sqrt{n}}
*σn**μ* - \frac{\mu}{n}
*nμ*

Q2. Let X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* be a random sample from the normal distribution with mean \mu*μ* and variance \sigma^{2}*σ*2. Assume that \mu*μ* is known.

Find the Cramer-Rao lower bound for the variance of all unbiased estimators of \sigma^{2}*σ*2.

- \frac{2 (\sigma^{2})^{2}}{n}
*n*2(*σ*2)2 - \frac{\mu}{\sigma \sqrt{n}}
*σn**μ* - \frac{\sigma^{2}}{n}
*nσ*2 - \frac{\mu^{2}}{n}
*nμ*2

Q3. Suppose that X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* be a random sample from the Poisson distribution with parameter \lambda>0*λ*>0.

Find the Cramer-Rao lower bound for the variance of all unbiased estimators of P(X_{i}=0)*P*(*Xi*=0).

- \frac{\lambda^{2}}{n}
*nλ*2 - \frac{\lambda e^{-2 \lambda}}{n}
*nλe*−2*λ* - \frac{\lambda^{2}e^{-\lambda}}{n}
*nλ*2*e*−*λ* - \frac{\lambda}{n}
*nλ*

#### Quiz 2: Further Computations with MLEs Quiz Answers

Q1. Let X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* be a random sample from the exponential distribution with rate \lambda>0*λ*>0.

Let \widehat{\lambda}*λ*, be the MLE for \lambda*λ*.

Compute the efficiency of \widehat{\lambda}*λ*.

- \frac{n \lambda^{2}}{(n-1)^{2}(n-2)}(
*n*−1)2(*n*−2)*nλ*2 - \frac{(n-1)^{2}(n-2)}{n^{3}}
*n*3(*n*−1)2(*n*−2) - \frac{n}{(n-1)^{2}}(
*n*−1)2*n* - \frac{\lambda^{2}}{(n-1)^{2}(n-2)}(
*n*−1)2(*n*−2)*λ*2

Q2. Let X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* be a random sample from the exponential distribution with rate \lambda*λ*.

Find the asymptotic distribution of the maximum likelihood estimator for the “tail probability” P(X>1)*P*(*X*>1) for some fixed x>0*x*>0.

(Note: You do not actually need to find the MLE itself!)

- N \left( e^{-\lambda} , \lambda^{2} e^{-2 \lambda}/n \right)
*N*(*e*−*λ*,*λ*2*e*−2*λ*/*n*) - N(\lambda,e^{-2\lambda}/n)
*N*(*λ*,*e*−2*λ*/*n*) - N(e^{-\lambda},1/(n \lambda^{2}))
*N*(*e*−*λ*,1/(*nλ*2)) - N(\lambda,\lambda^{2}/n)
*N*(*λ*,*λ*2/*n*)

Q3. Let \widehat{\gamma}_{n}*γ**n* be the MLE for the parameter of the Pareto distribution with pdf

f(x;\gamma) = *f*(*x*;*γ*)=

⎧⎩⎨*γ*(1+*x*)*γ*+10,,*x*>0otherwise

.{(1+*x*)*γ*+1*γ*0,,*x*>0otherwise.

Which of the following are true about \widehat{\gamma}_{n}*γ**n*?

(Check all that apply.)

- \widehat{\gamma}_{n}
*γ**n* does not depend on n*n* - \displaystyle\lim_{n \rightarrow \infty} E[\widehat{\gamma_{n}}]= \gamma
*n*→∞lim*E*[*γn*]=*γ* - \widehat{\gamma}_{n}
*γ**n* is an unbiased estimator of \gamma*γ* - \widehat{\gamma}_{n}
*γ**n* has an asymptotically normal distribution

Q4. Let X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* be a random sample from any distribution with mean \mu*μ* and finite variance \sigma^{2}*σ*2.

What is the asymptotic distribution of \sum_{i=1}^{n}X_{i}∑*i*=1*n**Xi*?

- N(\mu,\sigma^{2})
*N*(*μ*,*σ*2) - N(\mu,\sigma^{2}/n)
*N*(*μ*,*σ*2/*n*) - \Gamma(\mu,\sigma^{2}/n)Γ(
*μ*,*σ*2/*n*) - N(n \mu, n \sigma^{2})
*N*(*nμ*,*nσ*2)

Q5. Let X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* e a random sample from the distribution with pdf

f(x) = \theta x^{\theta-1} \, I_{(0,1)}(x)*f*(*x*)=*θ**x**θ*−1*I*(0,1)(*x*).

To what does the sample mean \overline{X}*X* converge in probability?

- \frac{\theta}{\theta+1}
*θ*+1*θ* - 1+\frac{1}{\theta}1+
*θ*1 - 00
- \theta
*θ*

#### Quiz 3: Confidence Intervals Involving the Normal Distribution

Q1. Observations on “stabilized viscosity of asphalt specimens” are 27812900301328562888 Preliminary investigation of the data support the assumption that viscosity is at least approximately normally distributed. Give a 95% confidence interval for true average viscosity.

- (2852.6,2905.5)(2852.6,2905.5)
- (2783.27,2991.93)(2783.27,2991.93)
- (2901,4,2907.2)(2901,4,2907.2)
- (2856.2,2919.0)(2856.2,2919.0)

Q2. The NIST Standard Reference Database gave the following summary informationfor fracture strengths (MPa) of 138 ceramic bars fired in a particular kiln: \overline{x}=83.14*x*=83.14, s=2.73*s*=2.73.

Calculate an approximate 90\% confidence interval for true average fracture strength.

- (82.758, 83.522)(82.758,83.522)
- (80.683, 85.597)(80.683,85.597)
- (82.909, 83.371)(82.909,83.371)
- (81.931, 84.349)(81.931,84.349)

Q3. Suppose that a random sample of 42 bottles of a particular brand of wine is selected and the alcohol content of each bottle is determined. Let \mu*μ* denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting 95% confidence interval is (10.8,12.4)(10.8,12.4).

Which of the following is true when using the same data set. (Select all that apply.)

- A 90% confidence interval for \mu
*μ*will be an interval that is contained in (10.8,12.4)(10.8,12.4). - A 90% confidence interval will end up being narrower.
- A 90% confidence interval for \mu
*μ*will be an interval that contains the interval (10.8,12.4)(10.8,12.4). - The true mean \mu
*μ*is between 10.810.8 and 12.412.4 90% of the time.

Q4. The lengths, in inches for a particular population of sockeye salmon (*Oncorhynchus nerka*) are known to be normally distributed with variance \sigma^{2}=25.1*σ*2=25.1. A random sample of 12 sockeye salmon were pulled from the region and measured, producing a sample mean of \overline{x}= 23.7*x*=23.7 inches.

Derive an 80% confidence interval for the true population mean length \mu*μ*.

- (21.85, 25.55)(21.85,25.55)
- (21.73, 25.67)(21.73,25.67)
- (14.57,32.83)(14.57,32.83)
- (13.82,33.58)(13.82,33.58)

Q5. You are planning on taking a random sample from a normal distribution with variance 99 and making a 95% confidence interval for its mean \mu*μ*.

What is the minimum sample size necessary to ensure that the length of this confidence interval is less than 2.72.7?

- 20
- 19
- 18
- 57

#### Quiz 4: Confidence Intervals for Differences Between Means

Q1. A manufacturer of a premium laminate and wood compostive flooring system would like to be able to say that their product will experience less warpage than hardwood flooring under certain fixed temperature and humidity conditions.

A random sample of size 1010 from the laminate system produced a sample mean warpage of 2.7 degrees and a random sample of size 1212 from the hardwood flooring gave a sample mean warpage of 3.2 degrees. It is believed that both systems have a true warpage variance of 0.80.8 degrees squared.

Assuming warpage for both products are normally distributed, find a 90% confidence interval for \mu_{1}-\mu_{2}*μ*1−*μ*2 where \mu_{1}*μ*1 is the true mean warpage for laminate and \mu_{2}*μ*2 is the true mean warpage for hardwood.

- (-1.251,0.251)(−1.251,0.251)
- (-1.130, 0.130)(−1.130,0.130)
- (-1.161,0.161)(−1.161,0.161)
- (-1.008,0.008)(−1.008,0.008)

Q2. A manufacturer of a premium laminate and wood compostive flooring system would like to be able to say that their product will experience less warpage than hardwood flooring under certain fixed temperature and humidity conditions.

A random sample of size 1010 from the laminate system produced a sample mean warpage of 2.7 degrees and a sample variance of 0.730.73. A random sample of size 1212 from the hardwood flooring gave a sample mean warpage of 3.2 degrees and a *sample* variance of 0.800.80. It is believed that the true variance for the laminate system is the same as the true variance for the hardwood system.

Assuming warpage for both products are normally distributed, find a 90% confidence interval for \mu_{1}-\mu_{2}*μ*1−*μ*2 where \mu_{1}*μ*1 is the true mean warpage for laminate and \mu_{2}*μ*2 is the true mean warpage for hardwood.

- (-1.147,0.147)(−1.147,0.147)
- (-0.007,0.003)(−0.007,0.003)
- (-0.981,0.019)(−0.981,0.019)
- (-1.117,0.117)(−1.117,0.117)

Q3. The Slant Shear Test (SST) is a widely used test for evaluating the bond of resinous repair materials to concrete. It was applied to two random samples of concrete slabs which were finshed in different ways. Population 1 has a smooth finish while Population 2 has a more textured finish.

The sample mean shear strength ( in N/mm^{2}2) and sample variance for the smooth finish sample of size 138 from were 18.1718.17 and 1.781.78, respectively. The sample mean and variance in shear strength for a random sample of size 110 hand-chiseled specimens, were were 21.6621.66 and 3.213.21.

Find a 95% confidence interval \mu_{1}-\mu_{2}*μ*1−*μ*2, the true mean difference in shear strengths.

- (-3.879,-3.101)(−3.879,−3.101)
- (-3.892,-3.088)(−3.892,−3.088)
- (-3.827,-3.153)(−3.827,−3.153)
- I can’t answer this question without knowing the distribution of shear strengths for the two populations.

### Week 5 Quiz Answers

#### Quiz 1: Confidence Intervals for Proportions and Variances

Q1. Suppose 250 randomly selected people in a community are surveyed to determine whether or not they support a tax increase to fund community infrastructure. Of the 250 surveyed, 112 were in favor of the tax increase.

Give a 95% confidence interval for the true proportion of people in the community who support the tax increase.

- (0.3867,0.5096)(0.3867,0.5096)
- (0.3969, 0.4991)(0.3969,0.4991)
- The sample size is not large enough for me to be able to answer this question.
- (0.4380, 0.4580)(0.4380,0.4580)

Q2. A new shopping mall is proposed on unincorporated land on the boundary between two cities. A random sample of 150 residents from City A who live within within 1 mile of the boundary was taken. Of this sample, 78 were in favor of the mall. Similarly, a random sample of 150 residents from City B who live within 1 mile of the boundary was taken. Of this sample, 92 were in favor of the mall.

Find a 95% confidence interval for p_{B}-p_{A}*pB*−*pA* where p_{A}*pA* and p_{B}*pB* are the true proportion of residents from each area who are in favor of the mall.

- (0.0034, 0.1833)(0.0034,0.1833)
- (-0.0183,0.2050)(−0.0183,0.2050)
- (-0.0004,0.1870)(−0.0004,0.1870)
- (0.5094, 0.6239)(0.5094,0.6239)

Q3. Consider the following data set.

9.2229, \,\, 8.3665, \,\, 8.797058, \,\, 10.2195, \,\, 6.56299.2229,8.3665,8.797058,10.2195,6.5629

Assuming that these data come from a normal distribution, find a 92% confidence interval for the true variance of that distribution.

- (0.7230,11.5577)(0.7230,11.5577)
- (0.1423, 1.6065)(0.1423,1.6065)
- (0.6225,7,0282)(0.6225,7,0282)
- (0.0865,1.3831)(0.0865,1.3831)

Q4. Which of the following are true about a confidence interval for a normal population variance?

(Select all that apply.)

- The confidence interval must include the sample variance S^{2}
*S*2. - A 95% confidence interval for the variance must fully contain any 90% confidence interval for the variance.
- The confidence interval may be infinite.
- A 90% confidence interval will be wider than a 95% confidence interval.

#### Quiz 2: Build Your Own Confidence Intervals Quiz Answers

Q1. Let X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* be a random sample from the continuous uniform distribution on the interval from 00 to \theta*θ* with \theta>0*θ*>0.

Using the “cdf method” from this lesson, we can show that Y_{n} = \max (X_{1}, X_{2}, \ldots, X_{n})*Y**n*=max(*X*1,*X*2,…,*X**n*) has pdf given by

f(y;\theta) = *f*(*y*;*θ*)=

{*n**θ**n**y**n*−10,,0<*y*<*θ*otherwise

{*θ**n**n**y**n*−10,,0<*y*<*θ*otherwise

Which of the following is a valid 100(1-\alpha)100(1−*α*)% confidence interval for \theta*θ*.

*(Hint: You might want to look at Lesson 3 of Module 1 to figure out how to form a pivotal quantity.*

- \left( \frac{Y_{n}}{\alpha^{1/n}},\infty\right)(
*α*1/*nYn*,∞) - \left(0, \frac{Y_{n}}{(1-\alpha)^{1/n}}\right)(0,(1−
*α*)1/*nYn*) - \left( \frac{Y_{n}}{(1-\alpha)^{1/n}},\infty\right)((1−
*α*)1/*nYn*,∞) - \left( 0,\frac{Y_{n}}{\alpha^{n}}\right)(0,
*αnYn*)

Q2. Let X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* be a random sample from the shifted continuous exponential rate 11 distribution with pdf

f(x;\theta) = *f*(*x*;*θ*)=

{*e*−(*x*−*θ*)0,,*x*>*θ*otherwise

{*e*−(*x*−*θ*)0,,*x*>*θ*otherwise

Which of the following represents a 95% confidence interval for \theta*θ* based on the minimum value Y_{n} = \min (X_{1}, X_{2}, \ldots, X_{n})*Yn*=min(*X*1,*X*2,…,*Xn*).

- (0.05^{1/n}Y_{n}, \infty)(0.051/
*nYn*,∞) - (0,0.95^{1/n}Y_{n})(0,0.951/
*nYn*) - \left( Y_{n}+\frac{\ln 0.05}{n},Y_{n} \right)(
*Yn*+*n*ln0.05,*Yn*) - (nY_{n} -\ln 0.05, nY_{n}+\ln 0.05)(
*nYn*−ln0.05,*nYn*+ln0.05)

Q3. Let X_{1}*X*1 be a random sample of size 11 from the \Gamma(3,\beta)Γ(3,*β*) distribution with parameter \beta>0*β*>0.

Derive a 90% confidence interval for \beta*β* based on the single observation X_{1}*X*1.

(Yes, a sample of size 11 is silly but it makes for an easier computation and the point of this problem is to test the techniques for developing confidence intervals!)

- \left( 0, \frac{2\chi^{2}_{0.10,5}}{X_{1}} \right)(0,
*X*12*χ*0.10,52) - \left( \frac{2X_{1}}{\chi^{2}_{0.10,5}}, \infty\right)(
*χ*0.10,522*X*1,∞) - \left( 0, \frac{\chi^{2}_{0.10,6}}{2X_{1}} \right)(0,2
*X*1*χ*0.10,62) - \left( 0, \frac{\chi^{2}_{0.10,5}}{2X_{1}} \right)(0,2
*X*1*χ*0.10,52)

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