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**About Statistical Inference and Hypothesis Testing in Data Science Applications Course**

This course will focus on the theory and practise of testing hypotheses, especially as they relate to data science. Students will learn how to use hypothesis tests so that they can use data to make good decisions. The general logic of testing hypotheses, error and error rates, power, simulation, and the right way to calculate and understand p-values will be given extra attention. Also, the wrong use of testing ideas, especially p-values, and the ethical consequences of that use will be looked at.

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 program.

**WHAT YOU’LL FIND OUT**

- Explain what a composite hypothesis is and how important a test is when the null hypothesis is a composite one.
- Explain what a test statistic, significance level, and rejection region are for a hypothesis test. Give the shape of a no-go zone.
- Do tests to find out what the real population variance is.
- Find the sample mean and sample a minimum of the exponential distribution’s sampling distributions.

**Course Apply Link – Statistical Inference and Hypothesis Testing in Data Science Applications **

**Statistical Inference and Hypothesis Testing in Data Science Applications Quiz Answers**

### Week 1 Quiz Answers

#### Quiz 1: Introduction to Hypothesis Testing Quiz Answers

Q1. A manufacturer of 40-amp fuses wants to make sure that the mean amperage at which its fuses burn out is in fact 40.

If the mean amperage is higher than 40, the manufacturer might be liable for damage to an electrical system due to fuse malfunction.

To verify the amperage of the fuses, a sample of fuses is to be selected and inspected. A hypothesis test is performed on

H_{0}: \mu = 40 \,\,\, \text{versus} \,\,\, H_{1}: \mu > 40H

Which of the following statements are true for describing Type I and Type I errors. (Check all that apply.)

- Type I: The mean amperage is good and the manufacturer concludes that it is acceptable, which saves the manufacturer money.
- Type II: The mean amperage is greater than 40, but the manufacturer concludes that it is less than 40.
- Type II: The mean amperage is too high, but the manufacturer concludes that it is acceptable which opens them up to liability for damaged electrical systems.
- Type I: The mean amperage is good, but the manufacturer concludes that it is too high and they launch an expensive and unnecessary recall process.

Q2. Minor surgery on horses under field conditions requires a reliable short-term anesthetic producing good muscle relaxation, minimal cardiovascular and respiratory changes, and a quick, smooth recovery with minimal after effects so that horses can be left unattended.

A study reports that for a sample of n=73n=73 horses to which ketamine is administered under certain conditions, the sample average lateral recumbency (lying-down) time was 18.8618.86 min.

Lateral recumbency time is known to be normally distribited with a standard deviation of \sigma=8.6σ=8.6 min.

Do these data suggest that true average lateral recumbency time under these conditions is less than 20 min?

Test the hypotheses

H_{0} :\mu = 20 \,\,\, \text{versus} \,\,\, H_{1}: \mu< 20H

:μ=20versusH

:μ<20

at level of significance 0.100.10.

Give the appropriate critical value and conclusion.

- 1.28, Fail to reject H_{0}H

- 1.28, Reject H_{0}H

- -1.28, Reject H_{0}H

- -1.28, Fail to reject H_{0}H

Q3. The recommended daily dietary allowance for zinc among males older than age 50 years is 15 mg/day. A study reports the following summary data on intake for a sample of 12 males age 65-74 years: \overline{x} = 11.3

Assuming that the true distribution for zinc intake is normally distributed with variance 6.436.43, does the data indicate that the average daily zinc intake in the population of all males 65-74 falls below the recommended allowance?

Consider testing the hypotheses H_{0}: \mu=15H

:μ=15 versus H_{1}: \mu< 15H

:μ<15 at level of significance \alpha=0.05α=0.05.

Below what value does \overline{x}

need to be for us to reject H_{0}H

in favor of H_{1}H

? Give your answer to two decimal places.

Enter answer here

Q4. The director of manufacturing for ACME industries is interested in an online training program that can be used to train the firm’s maintenence employees for machine-repair operations.

- To evaluate the training method, the director of manufacturing has agreed to train 15 employees with the new approach. The director will approve the program as long as the mean training time stays below 50 days. However, the director firmly believes that it will take, on average, 50 or more days.
- Assuming that training times are normally distributed with a standard deviation of \sigma=3.2σ=3.2 days, what is the largest sample mean that could be observed that will make the company switch to the online program when testing the relevant hypotheses at level \alpha=0.01α=0.01?

Round your answer to 2 decimal places.

Enter answer here

Q5. Which of the following statements is necessarily true in hypothesis testing.

- A Type I error is more serious/consequential than a Type II error.
- A Type II error is more serious/consequential than a Type I error.
- Both Type I and Type II errors are equally serious/consequential.
- The relative severity of one error over the other is problem dependent.

### Week 2 Quiz Answers

#### Quiz 1: Constructing Tests Quiz Answers

Q1. A commonly prescribed drug for relieving anxiety is believed to 60% effective. Experimental results with a new drug administered to a random sample of 100 adults who are suffering with anxiety showed that 68 experienced relief.

- Is this sufficient evidence to conclude that the new drug is superior to the one commonly prescribed?
- Carry out the relevant hypothesis test using a 0.03 level of significance. Give the appropriate critical value and decision.
- 1.88, Reject H_{0}H

0

. The new drug appears to be superior at 0.03 level of significance. - -0.97, Fail to reject H_{0}H

0

. The new drug does not appear to be superior at 0.03 level of significance. - 1.88, Fail to reject H_{0}H

0

. The new drug does not appear to be superior at 0.03 level of significance. - -1.88, Reject H_{0}H

0

. The new drug appears to be superior at 0.03 level of significance.

Q2. “Hardness” of a material in engineering is defined as the resistance to indentation. It is determined by measuring the permanent depth of the indentation.

- An engineer measured the “Brinell hardness” of 6 pieces of ductile iron

that were subcritically annealed. The resulting data were: - 179, 156, 167,183, 178, 165
- Assuming that Brinell hardness is normally distributed with standard deviation \sigma=4.2σ=4.2, consider testing
- H_{0}: \mu = 170H

0

:μ=170 versus H_{1}: \mu>170H

1

:μ>170.

What is the P-Value for this test and what is the conclusion based on this data set at 0.20 level of significance?

- 0.8416, Fail to Reject H_{0}H

0

. - 0.2184, Fail to Reject H_{0}H

0

. - 0.7776, Fail to Reject H_{0}H

0

. - 0.8416, Reject H_{0}H

0

Q3. Which option below best describes the following two plots?

- Plot A indicates a heavy-tailed distribution and Plot B indicates a light-tailed distribution.
- Both Plots A and B indicate heavy tailed distributions.
- Plot A indicates a light-tailed distribution and Plot B indicates a normal distribution.
- Plot A indicates a light-tailed distribution and Plot B indicates a heavy-tailed distribution.

Q4. A random sample of 10 chocolate energy bars from a certain company has, on average, 232 calories with a standard deviation of 15 calories.

Which of the following is a desirable shape of the power function for this test?

Q5. A manufacturer of ketchup uses a machine to automatically dispense

ingredients into bottles that move along an assembly line. The machine is working properly when 14 ounces are dispensed.

Suppose that the amount dispensed is normally distributed with a true standard deviation of 0.08 ounces. A random sample of 20 bottles had a sample mean of 13.8 ounces per bottle.

Is there evidence that the machine should be stopped and recalibrated? Test the relevant hypotheses at 0.05 level of significance.

Which of the following is true.

- Reject that the machine is properly calibrated if the sample mean is below 13.88 ounces or above 14.12 ounces.
- Reject that the machine is properly callibrated if the sample mean is below 13.76 ounces or above 14.84 ounces.
- Reject that the machine is properly calibrated if the sample mean is below 13.96 ounces or above 14.04 ounces.
- Reject that the machine is properly callibrated if the sample mean is below 13.6 ounces or above 14.4 ounces.

### Week 2 Quiz Answers

#### Quiz 1: More Hypothesis Tests!

Q1. A manufacturer of 40-amp fuses wants to make sure that the mean amperage at which its fuses burn out is in fact 40.

If the mean amperage is lower than 40, customers will complain because the fuses require replacement too often. If the mean amperage is higher than 40, the manufacturer might be liable for damage to an electrical system due to fuse malfunction.

To verify the amperage of the fuses, a sample of 10 fuses are selected and inspected. The average amperage at which fuses in the sample burned out was 38.8. The sample variance was 2.8.

Assuming burn out amperage is normally distributed, consider testing the hypotheses

H_{0}: \mu=40 \,\,\, \text{versus} \,\,\, H_{1}: \mu \ne 40*H*0:*μ*=40versus*H*1:*μ*=40.

Find the P-value for this test.

- 0.0233
- 0.0495
- 0.0248
- 0.0467

Q2. The director of manufacturing for ACME industries is interested in a computer-assisted training program that can be used to train the firm’s maintenence employees for machine-repair operations. To evaluate the training method, the director of manufacturing has requested an estimate of the mean training time required with the computer assisted program. Suppose that management has agreed to train 15 employees with the new approach. The resulting sample mean training time is 53.87 days and the resulting sample standard deviation is 6.82 days.

Assuming that training times are normally distributed, is there evidence in the data to conclude that the true mean training time is greater than 50?

Use \alpha=0.05*α*=0.05.

Give the appropriate test statistic, critical value, and conclusion.

- The test statistic is approximately 2.22.2. The critical value is approximately 1.6451.645. The data does suggest that the true mean training time is greater than 50 when using \alpha=0.05
*α*=0.05. - The test statistic is approximately 2.22.2. The critical value is approximately 1.761.76. The data does suggest that the true mean training time is greater than 50 when using \alpha=0.05
*α*=0.05. - The test statistic is approximately 2.22.2. The critical value is approximately -1.76.−1.76. The data does not suggest that the true mean training time is greater than 50 when using \alpha=0.05
*α*=0.05. - The test statistic is approximately 2.22.2. The critical value is approximately 1.701.70. The data does suggest that the true mean training time is greater than 50 when using \alpha=0.05
*α*=0.05.

Q3. 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.

Let S^{2}*S*2 be the sample variance.

What is the distribution of S^{2}*S*2?

- \Gamma \left( \frac{n-1}{2}, \frac{n-1}{2 \sigma^{2}} \right)Γ(2
*n*−1,2*σ*2*n*−1) - \sigma^{2} \cdot \chi^{2}(2n)
*σ*2⋅*χ*2(2*n*) - \Gamma \left( \frac{n-1}{2}, \frac{1}{2} \right)Γ(2
*n*−1,21) - \chi^{2}(n-1)
*χ*2(*n*−1)

Q4. A national supermarket chain wants to redesign self checkout lanes throughout the country. Two designs have been suggested. Data on customer checkout times, in minutes, have been collected at stores where the systems have already been installed. The results were as follows.

System A | System B | |
---|---|---|

sample size | 120 | 100 |

sample mean | 4.1 | 3.3 |

sample standard deviation | 2.1 | 1.5 |

Assuming normality of customer checkout times for both systems, consider testing

H_{0}: \mu_{A}=\mu_{B} \,\,\, \text{versus} \,\,\, H_{1}: \mu_{A}>\mu_{B},*H*0:*μ**A*=*μ**B*versus*H*1:*μ**A*>*μ**B*,

where \mu_{A}*μ**A* is the true mean for System A and \mu_{B}*μ**B* is the true mean for System B.

Give the value of the appropriate test statistic to 2 decimal places.

Q5. When is Welch’s t-test used to compare two population means, and what distributional assumptions are needed?

- Welch’s t-test is used when at least one of the two sample sizes are small, the true variances for the populations are unknown, and we can’t assume they are equal. We need to assume that the two populations are normally distributed.
- Welch’s t-test is used when both sample sizes are small, the true variances for the populations are known. We need to assume that the two populations are normally distributed.
- Welch’s t-test is used when both sample sizes are large and the true variances are unknown so we must use sample variances. We don’t need to assume normality because of the Central Limit Theorem.
- Welch’s t-test is used when at least one of the two sample sizes are small, and the true variances for the populations are unknown but assumed to be equal. We need to assume that the two populations are normally distributed.

### Week 4 Quiz Answers

#### Quiz 1: Best Tests and Some General Skills Quiz Answers

Q1. Let X_{1}*X*1 be a sample of size 1 from the continuous uniform distribution over the interval from 00 to \theta*θ*. This means that X_{1}*X*1 has pdf

f(x) =

⎧⎩⎨1*θ*0,,0<*x*<*θ*otherwise

*f*(*x*)={*θ*10,,0<*x*<*θ*otherwise

Consider testing the hypotheses

H_{0}: \theta=1 \,\,\, \text{versus} \,\,\, H_{1}: \theta<1*H*0:*θ*=1versus*H*1:*θ*<1

at level of significance \alpha*α*, based on this single observation X_{1}*X*1.

Which of the following is a valid test.

- Reject H_{0}
*H*0, in favor of H_{1}*H*1 if X_{1} < \alpha*X*1<*α*. - Reject H_{0}
*H*0, in favor of H_{1}*H*1 if X_{1} > \alpha*X*1>*α*. - Reject H_{0}
*H*0, in favor of H_{1}*H*1 if X_{1} < 1- \alpha*X*1<1−*α*. - Reject H_{0}
*H*0, in favor of H_{1}*H*1 if X_{1} >1- \alpha*X*1>1−*α*.

Q2. Compute the following quantity.

\chi^{2}_{0.01,9}+t_{0.90,6}*χ*0.01,92+*t*0.90,6

Give your answer to two decimal places.

Q3. Let X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* be a random sample from the \Gamma(2,\beta)Γ(2,*β*) distribution.

Give the Neyman-Pearson ratio for finding the best test of

H_{0}: \beta=2 \,\,\, \text{versus} \,\,\, H_{1}: \beta=3.*H*0:*β*=2versus*H*1:*β*=3.

- \left( \frac{2}{3}\right)^{n} \displaystyle e^{\sum_{i=1}^{n} X_{i}}(32)
*ne*∑*i*=1*n**Xi* - \left( \frac{2}{3}\right)^{n} \left( \prod_{i=1}^{n} X_{i}\right) \displaystyle e^{\sum_{i=1}^{n} X_{i}}(32)
*n*(∏*i*=1*n**Xi*)*e*∑*i*=1*n**Xi* - \left( \frac{2}{3}\right)^{2n} \displaystyle e^{\sum_{i=1}^{n} X_{i}}(32)2
*ne*∑*i*=1*n**Xi* - \left( \frac{2}{3}\right)^{2n} \displaystyle e^{-\sum_{i=1}^{n} X_{i}}(32)2
*ne*−∑*i*=1*n**Xi*

Q4. Suppose that X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* is a random sample from the continuous uniform distribution over the interval from 00 to 11, which has pdf

- f(x) =
- ⎧⎩⎨1
*θ*0,,0<*x*<*θ*otherwise -
*f*(*x*)={*θ*10,,0<*x*<*θ*otherwise - Find the pdf for the maximum value in the sample.
- f(x) =
- ⎧⎩⎨1
*θn*0,,0<*x*<*θ*otherwise *f*(*x*)={*θn*10,,0<*x*<*θ*otherwise- f(x) =
- ⎧⎩⎨1−(1−
*xθ*)*n*0,,0<*x*<*θ*otherwise *f*(*x*)={1−(1−*θx*)*n*0,,0<*x*<*θ*otherwise- f(x) =
- ⎧⎩⎨(
*xθ*)*n*0,,0<*x*<*θ*otherwise *f*(*x*)={(*θx*)*n*0,,0<*x*<*θ*otherwise- f(x) =
- {
*nθnxn*−10,,0<*x*<*θ*otherwise *f*(*x*)={*θnn**xn*−10,,0<*x*<*θ*otherwise

Q5. Let X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* be a random sample from the N(\mu,\sigma^{2})*N*(*μ*,*σ*2) where **\mu μ is known**.

On which statistic is the best test of

H_{0}: \sigma^{2} = \sigma_{0}^{2} \,\,\, \text{versus} \,\,\, H_{1}: \sigma^{2}>\sigma_{0}^{2}*H*0:*σ*2=*σ*02versus*H*1:*σ*2>*σ*02

based?

- \overline{X}
*X* - \sum_{i=1}^{n} (X_{i}-\mu)^{2}∑
*i*=1*n*(*Xi*−*μ*)2 - \sum_{i=1}^{n} X_{i}∑
*i*=1*n**Xi* - \sum_{i=1}^{n} X_{i}^{2}∑
*i*=1*n**Xi*2

#### Quiz 2: Uniformly Most Powerful Tests and F-Tests

Q1. A teacher believes that the standard deviation of scores for a particular Midterm that he gives every semester is 4 points. His current students claim that the standard deviation is more than 4 points.

Let \sigma^{2}*σ*2 be the true variance for the Midterm scores.

A random sample of 10 Midterm scores from the current semester has an observed standard deviation of s=5.2*s*=5.2 points.

Assuming that test scores are normally distributed, consider testing the hypotheses

H_{0}: \sigma^{2}=16 \,\,\, \text{versus} \,\,\, H_{1}: \sigma^{2}>16*H*0:*σ*2=16versus*H*1:*σ*2>16.

Give the P-value for the appropriate test, rounded to 3 decimal places.

Q3. Consider a random sample X_{1}*X*1, of size 1, from the continuous distribution with pdf

f(x;\theta)=

⎧⎩⎨1−*θ*2(*x*−12),0,0<*x*<1otherwise

*f*(*x*;*θ*)={1−*θ*2(*x*−21),0,0<*x*<1otherwise

for some parameter -1<\theta< 1−1<*θ*<1.

Does a uniformly most powerful (UMP) test for

H_{0}: \theta = 0 \,\,\, \text{versus} \,\,\, H_{1}: \theta \ne 0*H*0:*θ*=0versus*H*1:*θ*=0

exist?

- Yes, and the form of the test will be to reject H_{0}
*H*0, in favor of H_{1}*H*1, if X_{1} \geq c*X*1≥*c*OR X_{1} \leq -c*X*1≤−*c*for some constant c>0*c*>0 to be determined. - Yes, and the form of the test will be to reject H_{0}
*H*0, in favor of H_{1}*H*1, if X_{1} \geq c*X*1≥*c*for some constant c*c*to be determined. - Yes, and the form of the test will be to reject H_{0}
*H*0, in favor of H_{1}*H*1, if X_{1} \leq c*X*1≤*c*for some constant c*c*to be determined. - No. UMP tests never exist for two-sided alternative hypotheses.

Q3. A random sample of size 88 from the N(\mu_{1},\sigma_{1}^{2})*N*(*μ*1,*σ*12) distribution has a sample variance of s_{1}^{2}=13.2*s*12=13.2.

- An independent random sample of size 66 from a N(\mu_{1},\sigma_{2}^{2})
*N*(*μ*1,*σ*22) has a sample variance of s_{2}^{2}=15.1*s*22=15.1. - Is there evidence to suggest that \sigma_{1}^{2} \ne \sigma_{2}^{2}
*σ*12=*σ*22? - Use \alpha=0.03
*α*=0.03 level of significance. - No. The test statistic is approximately 1.14391.1439 and we would reject the null hypothesis that the variances are equal, in favor of the stated alternative, if it is greater than 6.28006.2800.
- No. The test statistic is approximately 1.14391.1439 and we would reject the null hypothesis that the variances are equal, in favor of the stated alternative, if it is greater than 6.42876.4287.
- No. The test statistic is approximately 1.14391.1439 and we would reject the null hypothesis that the variances are equal, in favor of the stated alternative, if it is greater than 4.91504.9150.
- No. The test statistic is approximately 1.14391.1439 and we would reject the null hypothesis that the variances are equal, in favor of the stated alternative, if it is greater than 4.35304.3530.

Q4. Let X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* be a random sample from the N(\mu,1)*N*(*μ*,1).

Consider deriving the uniformly most powerful (UMP) test for

H_{0}: \mu=\mu_{0} \,\,\, \text{versus} \,\,\, H_{1}: \mu>\mu_{0}.*H*0:*μ*=*μ*0versus*H*1:*μ*>*μ*0.

We begin by finding the best test for

H_{0}: \mu=\mu_{0} \,\,\, \text{versus} \,\,\, H_{1}: \mu = \mu_{1},*H*0:*μ*=*μ*0versus*H*1:*μ*=*μ*1,

where \mu_{1}*μ*1 is some fixed value that is greater than \mu_{0}*μ*0.

Which rejection rules will give you the UMP test?

- Test A: Reject H_{0}
*H*0, n favor of H_{1}*H*1 if - \exp \left[ -\frac{1}{2} \left( \sum_{i=1}^{n} (X_{i}-\mu_{0})^{2} – \sum_{i=1}^{n} (X_{i}-\mu_{1})^{2} \right) \right] \leq kexp[−21(∑
*i*=1*n*(*Xi*−*μ*0)2−∑*i*=1*n*(*Xi*−*μ*1)2)]≤*k* - for some k
*k*to be determined. - Test B: Reject H_{0}
*H*0, n favor of H_{1}*H*1 if - \sum_{i=1}^{n} (X_{i}-\mu_{0})^{2} – \sum_{i=1}^{n} (X_{i}-\mu_{1})^{2} \geq k∑
*i*=1*n*(*Xi*−*μ*0)2−∑*i*=1*n*(*Xi*−*μ*1)2≥*k* - for some k
*k*to be determined. - Both tests A and B above.
- None of the above.

Q5. True or False. For a simple H_{0}*H*0 and simple H_{1}*H*1, the uniformly most powerful test of size \alpha*α* is the same as the best test.

- True
- False

### Week 5 Quiz Answers

#### Quiz 1: Adventures in GLRTs

Q1. Let X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* be a random sample of size n*n* from the continous distribution with pdf

f(x;\theta) =

{*θ*(1−*x*)*θ*−1,0,0<*x*<1otherwise

*f*(*x*;*θ*)={*θ*(1−*x*)*θ*−1,0,0<*x*<1otherwise

for some parameter \theta>0*θ*>0.

Consider testing

H_{0}: \theta=1 \,\,\, \text{versus} \,\,\, H_{1}: \theta \ne 1.*H*0:*θ*=1versus*H*1:*θ*=1.

Which of the following statements are true? (Check all that apply.)

- The uniformly most powerful (UMP) test is the same as the generalized likelihood ratio test (GLRT).
- A uniformly most powerful (UMP) test does not exist.
- Under H_{0}
*H*0, X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*Xn* are iid from the uniform distribution over the interval (0,1)(0,1). - For a large sample size, the generalized likelihood ratio \lambda(\vec{X})
*λ*(*X*) has an approximate \chi^{2}(1)*χ*2(1) distribution.

Q2. Let X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* be a random sample of size n*n* from the continous distribution with pdf

f(x;\theta) =

{*θ*(1−*x*)*θ*−1,0,0<*x*<1otherwise

*f*(*x*;*θ*)={*θ*(1−*x*)*θ*−1,0,0<*x*<1otherwise

for some parameter \theta>0*θ*>0.

Consider testing

H_{0}: \theta=1 \,\,\, \text{versus} \,\,\, H_{1}: \theta \ne 1*H*0:*θ*=1versus*H*1:*θ*=1.

What is the maximum likelihood estimator needed for computing the denominator of the generalized likelihood ratio (GLR)?

- \widehat{\theta} = \frac{n}{\sum_{i=1}^{n} (1-X_{i})}
*θ*=∑*i*=1*n*(1−*Xi*)*n* - \widehat{\theta} = \frac{\prod_{i=1}^{n} (1-X_{i})}{n}
*θ*=*n*∏*i*=1*n*(1−*Xi*) - \widehat{\theta} = \frac{-n}{\sum_{i=1}^{n} \ln(1-X_{i})}
*θ*=∑*i*=1*n*ln(1−*Xi*)−*n* - \widehat{\theta} = \overline{X}
*θ*=*X*

Q3. True or False.

A generalized likelihood ratio test will always be less powerful than a UMP test.

- True
- False

Q4. True or False.

When computing a generalized likelihood ratio (GLR) for some H_{0}*H*0 versus some H_{1}*H*1, the restricted MLE is always less than or equal to the unrestricted MLE.

- True
- False

Q5. Suppose that X_{1}, X_{2}, \ldots, X_{n}*X*1,*X*2,…,*X**n* is a random sample from the \Gamma(2,\beta)Γ(2,*β*) distribution . Suppose that n*n* is “large”.

Consider testing

H_{0}: \beta = 1 \,\,\, \text{versus} \,\,\, H_{1}: \beta \ne 1*H*0:*β*=1versus*H*1:*β*=1

using an approximate large sample generalized likelihood ratio test.

Suppose that the generalized likelihood ratio \lambda(\vec{X})*λ*(*X*) is observed to be \lambda(\vec{x})=0.67*λ*(*x*)=0.67.

Which of the following is an approximate large sample test using \alpha=0.05*α*=0.05?

- I can not answer this question without knowing the value of n
*n*. - The rule is to reject H_{0}
*H*0, in favor of H_{1}*H*1 if -2 \ln \lambda(\vec{X})>0.0039−2ln*λ*(*X*)>0.0039. For the given \lambda(\vec{x})*λ*(*x*), we reject H_{0}*H*0. - The rule is to reject H_{0}
*H*0, in favor of H_{1}*H*1 if -2 \ln \lambda(\vec{X})>5.0239−2ln*λ*(*X*)>5.0239. For the given \lambda(\vec{x})*λ*(*x*), we fail to reject H_{0}*H*0. - The rule is to reject H_{0}
*H*0, in favor of H_{1}*H*1 if -2 \ln \lambda(\vec{X})>3.841−2ln*λ*(*X*)>3.841. For the given \lambda(\vec{x})*λ*(*x*), we fail to reject H_{0}*H*0.

**Conclusion**

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