Statistical Inference MATH 1281. All Quizzes 1-9 Statistics Inference MATH 1281. All Graded 10.00 ou

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All quizzes statistics inference – 1281 Self-quiz unit one Started on Tuesday, 12 April 2022, 2:32 PM State Finished Completed on Tuesday, 12 April 2022, 2:35 PM Time taken 3 mins 1 sec Grade 10.00 out of 10.00 (100%) Information Flag question Information text The data set "cars.csv" was introduced in Chapter 9 of the book. This data set contains information on various models of cars. Download this file to your computer and store it in the working directory of R. The file can be downloaded from MATH 1281 Data Files in the course page for MATH 1281.

The variable "width" in the data set contains the width of the body of the car in inches. The variable "length" contains the length of the body of the car in inches. Produce a new variable by the name "area" that contains the area of the body of the car. This variable should be produced by multiplying the the width of the body by the length of the body. Answer the following questions with respect to the variable "area": Question 1 Correct Mark 1.00 out of 1.00 Flag question Question text The variable "area" is a: Select one: a.

Numeric b.

Factor Feedback Upload the file "cars.csv" to a data set by the name "cars" using the code: cars <read.csv("cars.csv"). Form the variable "area" using the code: area <- cars$length*cars$width.

The variable "area" is a product of two numeric variables and it obtains numeric values as well.

The correct answer is: Numeric Question 2 Correct Mark 1.00 out of 1.00 Flag question Question text The average area of the body of a car is: Answer: Feedback Apply the code: mean(area). The sample average turn out to be equal to 11493.36. Alternatively, you may apply the code: summary(area) and obtain the value of the mean from the produced report.

The correct answer is: 11493.36 Question 3 Correct Mark 1.00 out of 1.00 Flag question Question text The median area of the body of a car is: Answer: Feedback Apply the code: median(area). The sample median turn out to be equal to 11314.2. Alternatively, you may apply the code: summary(area) and obtain the value of the median from the produced report.

The correct answer is: 11314.2 Question 4 Correct Mark 1.00 out of 1.00 Flag question Question text The interquartile range of the area is: Answer: Feedback Apply the code: quantile(area,c(0.25,0.75)). The first quartile (Q1) is equal to 10709.72 and the third quartile (Q3) is equal to 12275.90. The difference between Q3 and Q1 is the interquartile range and is equal to IQR = Q3 - Q1 = 1566.18.

Alternatively, you may apply the code: summary(area) and obtain the values of Q1 and Q3 from the produced report.

The correct answer is: 1566.18 Question 5 Correct Mark 1.00 out of 1.00 Flag question Question text The number of outlier observations in the variable "area" is: Select one: a.

0 b.

1 c.

2 d.

3 e.

More than 3 Feedback

You may produce the box-plot of the variable using the code: boxplot(area). One point in the upper part of the plot represents the outliers. In order to make sure that this point represents a single data point you may produce the table with the code: table(area). The frequency of the largest value is 1.

An alternative method for counting the number of data points above the threshold Q3 + 1.5 (Q3Q1) may use the code: sum(12275.9 + 1.5*1566.18 < area). The count is 1.

The correct answer is: 1 Question 6 Correct Mark 1.00 out of 1.00 Flag question Question text The number of missing observations in the variable "area" is: Select one: a.

0 b.

1 c.

2 d.

3 Feedback There are no missing values in this variable.

The correct answer is: 0 Information Flag question Information text A rectangular body is square if the length is equal to the width. Consider, for each car, the difference between the length and the width of the body. Call this variable X. If cars tend to be square then we would expect the expectation of this variable to be equal to 0.

In Chapter 12 we will present a statistical test for testing that the expectation of a variable is equal to 0. The test statistic for this problem is given by: T = X/[S/ √n], where X is the sample average, S is the sample standard deviation, and n is the sample size.

In the following questions you are required to determine the region that contains 95% of the values of the test statistic if indeed cars tend to be square in their shape and to check whether or not the evaluation of the test statistic for the "cars" data set falls within this region.

Question 7 Correct Mark 1.00 out of 1.00 Flag question Question text Let X1, X2, ..., X205 be a sample of Normal random variables with expectation μ=0 and variance σ2=100. (Namely, standard deviation of σ=10.) The region that contains 95% of the distribution of the statistic T = X/[S/ √205] is best described by:

Select one: a.

[-19.7,19.7] b.

[88.3,127.7] c.

[-1.97,1.97] d.

[106,110] Feedback The interval that contains 95% of the sampling distribution of the test statistic may be identified via simulations. Run the code: mu <- 0 sig <- 10 test.stat <- rep(0,10^5) for(i in 1:10^5) { X <- rnorm(205,mu,sig) test.stat[i] <- (mean(X))/sqrt(var(X)/205) } quantile(test.stat,c(0.025,0.975)) 2.5% 97.5% -1.958288 1.978535 The result is best described by [-1.97,1.97] The correct answer is: [-1.97,1.97] Question 8 Correct Mark 1.00 out of 1.00 Flag question Question text The standard deviation of the statistic T is Answer: Feedback Run the code: mu <- 0 sig <- 10 test.stat <- rep(0,10^5) for(i in 1:10^5) { X <- rnorm(205,mu,sig) test.stat[i] <- (mean(X))/sqrt(var(X)/205) } sd(test.stat) [1] 1.000478 The standard deviation is approximately equal to 1.

The correct answer is: 1 Question 9 Correct Mark 1.00 out of 1.00 Flag question

Question text Produce a variable by the name "x" that contains the difference between the length and the width of a car and compute the value of the test statistic for this variable. (Hint: Use the code "x <cars$length - cars$width" in order to produce "x" and apply the formula for the computation of the statistic to "x".) The value of the statistic for the observed data set "cars" is Answer: Feedback Produce the variable with the code: x <- cars$length - cars$width. The test statistic may be evaluated for the data using the code: mean(x)/sqrt(var(x)/205) The value of the test statistic is 146.1173.

The correct answer is: 146.1173 Question 10 Correct Mark 1.00 out of 1.00 Flag question Question text The value of the test statistic for the data set "cars" Select one: a. is inside the region that contains 95% of the sampling distribution of the statistics when cars are square b. is outside of the region that contains 95% of the sampling distribution of the statistics when cars are square Feedback The region is [-1.97,1.97]. The value 146.1173 is outside the region.

The correct answer is: is outside of the region that contains 95% of the sampling distribution of the statistics when cars are square Self-quiz unit two Started on Tuesday, 19 April 2022, 11:21 PM State Finished Completed on Tuesday, 19 April 2022, 11:24 PM Time taken 3 mins 49 secs Grade 10.00 out of 10.00 (100%) Information Flag question Information text In the book we compared two different estimators of the variance to each other. One estimator was the sample variance S2, an unbiased estimator of the variance. This estimator was computed using the function "var", which involves dividing the sum of squared deviations from the average by the number of observations minus 1 (n-1). The alternative estimator was obtained by dividing the sum of squares by the number of observations ( n ). This alternative estimator was equal to [(n-1)/n]S2.

In this exercise we compare two estimators of the standard deviation that are derived from the estimators of the variance. The first is S, the square root of S2, and the second is the square root

of the alternative estimator of the variance. The first estimator of the standard deviation may be computed with the function "sd". The alternative estimator may be computed by the multiplying the first estimator by √ [(n-1)/n].

We consider the bias, variance and mean square error of both estimators of the standard deviation. We assume a Normal sample of n=20 observations. The expectation of an observation is μ=5 and the variance is σ2 = 3. The next 5 questions refer to the following R code that simulates the sampling distribution of the two estimators:

Question 1 Correct Mark 1.00 out of 1.00 Flag question Question text In the context of the simulation, the value of the parameter that is being estimated is: (Choose the number closest to the answer) Select one: a.

1.71 b.

3.00 c.

1.73 d.

1.69 Feedback

The variance of the measurement is equal to 3. The standard deviation, the value that is being estimated, is equal to √ 3 = 1.732051. After rounding up we obtain 1.73 as the solution.

The correct answer is: 1.73 Question 2 Correct Mark 1.00 out of 1.00 Flag question Question text The bias of the first estimator of the standard deviation is: (Choose the number closest to the answer) Select one: a.

1.71 b.

-0.023 c.

1.73 d.

0.00 Feedback The bias of the first estimator is the difference between its expectation, 1.708774, and the value of the estimated parameter, √ 3. The difference is equal to -0.02327681 ≅ -0.023.

The correct answer is: -0.023 Question 3 Correct Mark 1.00 out of 1.00 Flag question Question text The bias of the alternative estimator of the standard deviation is: (Choose the number closest to the answer) Select one: a.

1.71 b.

-0.023 c.

1.73 d.

-0.067 Feedback The bias of the second estimator is the difference between its expectation, 1.665507, and the value of the estimated parameter, √ 3. The difference is equal to -0.06654381 ≅ -0.067.

The correct answer is: -0.067 Question 4 Correct Mark 1.00 out of 1.00

Flag question Question text The mean squared error (MSE) of the first estimator of the standard deviation is: (Choose the number closest to the answer) Select one: a.

0.0792 b.

0.0780 c.

-0.023 d.

0.00 Feedback The mean square error (MSE) of the first estimator is equal to the sum of the variance of the estimator and the square of the bias. Run the code: > 0.07870757 + (1.708774-sqrt(3))^2 [1] 0.07924938 After rounding up we get 0.0792.

The correct answer is: 0.0792 Question 5 Correct Mark 1.00 out of 1.00 Flag question Question text The mean squared error (MSE) of the second estimator of the standard deviation is: (Choose the number closest to the answer) Select one: a.

0.0792 b.

0.0780 c.

-0.023 d.

0.00 Feedback The mean square error (MSE) of the first estimator is equal to the sum of the variance of the estimator and the square of the bias. Run the code: > 0.07477219 + (1.665507-sqrt(3))^2 [1] 0.07920027 After rounding up we get 0.0792.

The correct answer is: 0.0792 Information Flag question Information text In the book it was proposed to use 1/X (one over the sample average) as an estimator of the rate λ of the Exponential distribution. In the following 3 questions you are requested to compute the bias, the variance and the mean square error (MSE) of the given estimator for an Exponential

sample of size n=25 observations and when λ=3. You may carry out these computations by conducting a simulation in R: Question 6 Correct Mark 1.00 out of 1.00 Flag question Question text The expectation of the estimator is: (Give an answer of the form x.xxx, i.e. with 3 significance digits) Answer: Feedback Run the simulation: > lam <- 3 > lam.hat <- rep(0,10^5) > for(i in 1:10^5) +{ + X <- rexp(25,lam) + lam.hat[i] <- 1/mean(X) +} > mean(lam.hat) [1] 3.126287 . The expectation of the estimator is approximately equal to 3.126.

The correct answer is: 3.126 Question 7 Correct Mark 1.00 out of 1.00 Flag question Question text The mean square error (MSE) of the estimator is: (Give an answer of the form x.xxx, i.e. with 3 significance digits) Answer: Feedback Run the simulation: > lam <- 3 > lam.hat <- rep(0,10^5) > for(i in 1:10^5) +{ + X <- rexp(25,lam) + lam.hat[i] <- 1/mean(X) +} > var(lam.hat)+(mean(lam.hat)-3)^2 [1] 0.4386786 . The MSE of the estimator is approximately equal to 0.439.

The correct answer is: 0.442 Question 8 Correct Mark 1.00 out of 1.00 Flag question Question text The estimator is an unbiased estimator:

Select one: a.

TRUE b.

FALSE Feedback The estimator is biased since its expectation, 3.126, is not equal to the value of the parameter, 3.000, that it tries to estimate.

The correct answer is: FALSE Information Flag question Information text The next 2 questions refer to the following information: The variable "num.of.doors" in the data set "cars" counts the number of cars in the sample that have four doors and the number that have two doors. In the given data set the number of cars with four doors is 114 and the number of cars with two doors is 89. Two observations have a missing value for this variable. Denote by p the probability that a car has two doors.

Question 9 Correct Mark 1.00 out of 1.00 Flag question Question text The estimated value of p, ignoring missing values, is: (Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback The estimator of a probability of an event is the relative frequency of the event in the sample.

The event is "having 2 doors". This occurs in the sample 89, out of 89+114=203 times. The relative frequency is 89/203=0.4384236 The correct answer is: 0.438 Question 10 Correct Mark 1.00 out of 1.00 Flag question Question text If the value of the parameter p is 0.45 then the standard deviation of the estimator of the parameter is: (Consider, as before, the sample size without the missing observations and give an answer of the form x.xxx with 3 significance digits) Answer: Feedback The variance of the estimator is p(1-p)/n, where n=203 is the sample size. The standard deviation is the square root of the variance. If p=0.45 then the variance is 0.45*0.55/203 = 0.001219212 and the standard deviation is 0.03491721.

The correct answer is: 0.0349 Self-quiz unit three Started on Saturday, 23 April 2022, 9:17 PM

State Finished Completed on Saturday, 23 April 2022, 9:54 PM Time taken 36 mins 18 secs Marks 9.00/9.00 Grade 10.00 out of 10.00 (100%) Information Flag question Information text A case study that deals with the effect of prior reputation on the evaluation of an instructor is presented in Question 11.1 in the book (Section 11.5). Students were exposed to either one of two conditions, C and P, and were asked to rate a lecture given by the instructor. It was found that the average rating among the 25 students that were exposed to condition C was 2.613 and the average rating among the 24 students that were exposed to condition P was 2.236. The sample standard deviations were 0.533 and 0.543, respectively.

Question 1 Correct Mark 1.00 out of 1.00 Flag question Question text You want to construct a confidence interval for the expected rating of students that were exposed to condition P. The confidence level of the interval should be 85%. The quantile of the tdistribution that should be used for the construction is: (Assume a Normal distribution for the rating) Select one: a. qt(0.85,24) b. qt(0.925,24) c. qt(0.85,23) d. qt(0.925,23) Feedback The central part should include 85% of the distribution. The tails should include 15%. Half of this probability, or 7.5%, goes to the upper tail and half to the lower tail. The probability above the quantile is 0.075. Therefore, the probability below the quantile should be 0.925.

The number of degrees of freedom is the number of observations minus 1. The number of observations is 24. Hence, the number of degrees of freedom is 23.

Combining the two arguments we get that quantile should be qt(0.925,23).

The correct answer is: qt(0.925,23) Question 2 Correct Mark 1.00 out of 1.00 Flag question Question text

The confidence interval with confidence level of 85% for the expected rating of students that were exposed to condition P is: (Assume a Normal distribution for the rating. Choose the interval closest to the answer) Select one: a.

[2.07, 2.40] b. [ 2.12, 2.35] c.

[2.50, 2.73] d.

[2.59, 2.64] Feedback The sample average is 2.236 and the sample standard deviation is 0.543. The estimated standard deviation of the sample average is 0.543/√24, where 24 is the sample size. We apply the formula for computing a confidence interval for the expectation of a Normal measurement: 2.236 + qt(c(0.075,0.925),23)*0.543/sqrt(24) . The output of the computation is: [2.070929, 2.401071], which after rounding up gives [2.07,2.40].

The correct answer is: [2.07, 2.40] Question 3 Correct Mark 1.00 out of 1.00 Flag question Question text You want to construct a confidence interval for the variance of the rating of students that were exposed to condition P. The confidence level of the interval should be 80%. A quantile of the chisquare distribution that should be used for the construction is: (Assume a Normal distribution for the rating) Select one: a. qchisq(0.85,23) b. qchisq(0.10,23) c. qchisq(0.90,24) d. qchisq(0.95,23) Feedback The central part should contain 80% of the distribution. The tails should contain 20%. Half of this probability, or 10%, goes to the upper tail and other 10% goes to the lower tail. It follows that the quantiles of interest are those associated with 10% and with 90%.

The number of degrees of freedom is the number of observations minus 1. The number of observations is 24. Hence, the number of degrees of freedom is 23.

Combining the two arguments we get that quantile should be considered are qchisq(0.10,23) and qchisq(0.90,23). Only the former appears among the options.

The correct answer is: qchisq(0.10,23)

Question 4 Correct Mark 1.00 out of 1.00 Flag question Question text The confidence interval with confidence level of 80% for the variance of the rating of students that were exposed to condition P is: (Assume a Normal distribution for the rating. Choose the interval closest to the answer) Select one: a.

[0.212, 0.457] b.

[0.390, 0.841] c.

[0.239, 0.395] d.

[0.439, 0.727] Feedback The sample standard deviation is 0.543 and the sample variance is the square of this number. We apply the formula for computing a confidence interval for the variance of a Normal measurement: 23/qchisq(c(0.90,0.10),23)*0.543^2. The output of the computation is: [0.2118770, 0.4567314], which after rounding up gives [0.212, 0.457].

The correct answer is: [0.212, 0.457] Information Flag question Information text Confidence intervals are computed for the expectation and the variance of a measurement on the basis of 20 observations. The confidence interval are constructed under the assumption that the measurement has a Normal distribution. In reality, the distribution of the observations is the Uniform(0.5,5.5) distribution. Recall that the expectation of the measurement is 3 and the variance is 2.083333. Use simulations to answer the following questions: Question 5 Correct Mark 1.00 out of 1.00 Flag question Question text Consider a confidence interval for the expectation with a nominal confidence level of 95%. The actual confidence level is: (Select the closest number.)

Select one: a.

0.93 b.

0.95 c.

0.97 d.

0.99 Feedback Run the simulation: > a <- 0.5 > b <- 5.5 > n <- 20 > X.bar <- rep(0,10^5) > S <- rep(0,10^5) > for(i in 1:10^5) +{ + X <- runif(n,a,b) + X.bar[i] <- mean(X) + S[i] <- sd(X) +} > mu.LCL <- X.bar - qt(0.975,n-1)*S/sqrt( n ) > mu.UCL <- X.bar + qt(0.975,n-1)*S/sqrt( n ) > mean((3 >= mu.LCL) & (3 <= mu.UCL)) [1] 0.94833 The closest answer is 0.95.

The correct answer is: 0.95 Question 6 Correct Mark 1.00 out of 1.00 Flag question Question text Consider a confidence interval for the variance with a nominal confidence level of 95%. The actual confidence level is: (Select the closest number.)

Select one: a.

0.93 b.

0.95 c.

0.97 d.

0.99 Feedback Run the simulation: > a <- 0.5 > b <- 5.5 > n <- 20 > X.bar <- rep(0,10^5) > S <- rep(0,10^5) > for(i in 1:10^5) +{ + X <- runif(n,a,b) + X.bar[i] <- mean(X) + S[i] <- sd(X) +}

> var.LCL <- (n-1)*S^2/qchisq(0.975,n-1) > var.UCL <- (n-1)*S^2/qchisq(0.025,n-1) > mean((2.083333 >= var.LCL) & (2.083333 <= var.UCL)) [1] 0.99603 The closest answer is 0.99.

The correct answer is: 0.99 Information Flag question Information text A poll to determine the popularity of a candidate for election was conducted among 500 people.

It was found that 25 of those people are not eligible for voting, and among those that are eligible to vote, 207 support the candidate. Let p be the probability that an eligible voter supports the candidate.

Question 7 Correct Mark 1.00 out of 1.00 Flag question Question text How many people eligible for voting should have been included in the survey in order to assure that the radius of a 95% confidence interval for p is no more than 0.01: (Choose the smallest sufficient number among the given choices.)

Select one: a.

9,100 b.

9,300 c.

9,500 d.

9,700 Feedback The radius of the confidence interval is bounded by qnorm(0.975)/(2√n). Therefore, , in order to assure a radius of 0.01, the sample size should be at least (qnorm(0.975)/0.02)^2 This expression produces 9603.647. We get that 9,700 is the smallest of the options still larger than the minimal required sample size.

The correct answer is: 9,700 Question 8 Correct Mark 1.00 out of 1.00 Flag question Question text A 95% confidence interval for p is: (Choose the interval closest to the answer) Select one: a.

[0.391, 0.480] b.

[0.520, 0.609]

c.

[0.527, 0.602] d.

[0.400, 0.473] Feedback We use the formula for the computation of a confidence interval for a proportion and run the code: p.hat+qnorm(c(0.025,0.975))*sqrt(p.hat*(1-p.hat)/475). The resulting confidence interval is [0.3911971, 0.4803818]. After rounding up we get the interval [0.391, 0.480].

The correct answer is: [0.391, 0.480] Question 9 Correct Mark 1.00 out of 1.00 Flag question Question text The estimated value of p is: (Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback The number of eligible voters is 500-25=475. Of them, 207 support the candidate. Therefore, the sample proportion is 207/475=0.4357895 The correct answer is: 0.4357895 Graded-quiz unit three Started on Sunday, 24 April 2022, 6:30 PM State Finished Completed on Sunday, 24 April 2022, 7:44 PM Time taken 1 hour 13 mins Marks 20.00/20.00 Grade 100.00 out of 100.00 Information Flag question Information text A case study that deals with the effect of prior reputation on the evaluation of an instructor is presented in Question 11.1 in the book (Section 11.5). Students were exposed to either one of two conditions, C and P, and were asked to rate a lecture given by the instructor. It was found that the average rating among the 25 students that were exposed to condition C was 2.613 and the average rating among the 24 students that were exposed to condition P was 2.236. The sample standard deviations were 0.533 and 0.543, respectively.

Question 1 Correct Mark 1.00 out of 1.00 Flag question Question text You want to construct a confidence interval for the expected rating of students that were exposed to condition P. The confidence level of the interval should be 80%. The quantile of the tdistribution that should be used for the construction is: (Assume a Normal distribution for the rating)

Select one: a. qt(0.85,23) b. qt(0.925,24) c. qt(0.90,23) d. qt(0.925,23) Feedback The correct answer is: qt(0.90,23) Question 2 Correct Mark 1.00 out of 1.00 Flag question Question text The confidence interval with confidence level of 80% for the expected rating of students that were exposed to condition P is: (Assume a Normal distribution for the rating. Choose the interval closest to the answer) Select one: a.

[2.09, 2.38] b.

[2.47, 2.76] c.

[2.14, 2.33] d.

[2.52, 2.71] Feedback The correct answer is: [2.09, 2.38] Question 3 Correct Mark 1.00 out of 1.00 Flag question Question text The confidence interval with confidence level of 75% for the variance of the rating of students that were exposed to condition P is: (Assume a Normal distribution for the rating. Choose the interval closest to the answer) Select one: a.

[0.404, 0.805] b.

[0.250, 0.374] c.

[0.219, 0.437] d.

[0.460, 0.689] Feedback The correct answer is: [0.219, 0.437] Information Flag question Information text Confidence intervals are computed for the expectation and the variance of a measurement on the basis of 15 observations. The confidence interval are constructed under the assumption that the measurement has a Normal distribution. In reality, the distribution of the observations is the Exponential(1/3) distribution. Recall that the expectation of the measurement is 3 and the variance is 9. Use simulations to answer the following questions: Question 4 Correct Mark 1.00 out of 1.00 Flag question Question text Consider a confidence interval for the expectation with a nominal confidence level of 95%. The actual confidence level is: (Select the closest number.)

Select one: a.

0.75 b.

0.85 c.

0.90 d.

0.95 Feedback The correct answer is: 0.90 Question 5 Correct Mark 1.00 out of 1.00 Flag question Question text Consider a confidence interval for the variance with a nominal confidence level of 95%. The actual confidence level is: (Select the closest number.)

Select one: a.

0.75 b.

0.85 c.

0.90 d.

0.95 Feedback

The correct answer is: 0.75 Information Flag question Information text A poll to determine the popularity of a candidate for election was conducted among 600 people.

It was found that 57 of those people are not eligible for voting, and among those that are eligible to vote, 223 support the candidate. Let p be the probability that an eligible voter supports the candidate.

Question 6 Correct Mark 1.00 out of 1.00 Flag question Question text How many people eligible for voting should be included in the survey in order to assure that the radius of a 90% confidence interval for p is no more than 0.025: (Choose the smallest sufficient number among the given choices.)

Select one: a.

500 b.

700 c.

900 d.

1,100 Feedback The correct answer is: 1,100 Question 7 Correct Mark 1.00 out of 1.00 Flag question Question text A 90% confidence interval for p is: (Choose the interval closest to the answer) Select one: a.

[0.384, 0.438] b.

[0.376, 0.445] c.

[0.562, 0.616] d.

[0.555, 0.624] Feedback The correct answer is: [0.376, 0.445] Information Flag question

Information text The data set "cars.csv" was introduced in Chapter 9 of the book. This data set contains information on various models of cars. Download this file to your computer and store it in the working directory of R. The file can be downloaded from MATH 1281 Data Files in the course page for MATH 1281.

The variable "engine.size" in the data set contains the volume swept by all the pistons inside the cylinders in cubic inches. The variable "horsepower" contains the power produced by the engine in horsepowers. Produce a new variable by the name "ratio" that contains the ratio between the power of the engine and the size of the engine. This variable describes the number of horsepowers produced for each cubic inch of engine size. Answer the following questions with respect to the variable "ratio": Question 8 Correct Mark 1.00 out of 1.00 Flag question Question text The number of outlier observations in the variable "ratio" is: Select one: a.

0 b.

1 c.

2 d.

3 e.

More than 3 Feedback The correct answer is: More than 3 Information Flag question Information text A possible measure of efficiency of an engine is the power it produces for each unit of volume.

The efficiency may be measured for each car using the ratio between the power and the volume of the engine. Consider, for each car, the ratio between the power of the engine measured in horsepower and the volume of the engine measured by the volume swept by all the pistons inside the cylinders in cubic inches. Call this variable Y. In Chapter 12 we will present a statistical test for testing that the expectation of a variable is equal to a given value, say 1. The test statistic for this problem is given by: T = [Y-1]/[S/ √n], where Y is the sample average, S2 is the sample variance and n is the sample size.

In the following questions you are required to determine the region that contains 95% of the values of the test statistic if indeed the ratio between power and size of engine is equal to 1 and to check whether or not the evaluation of the test statistic for the "cars" data set falls within this region.

Question 9

Correct Mark 1.00 out of 1.00 Flag question Question text Let Y1, Y2, ..., Y203 be a sample of Normal random variables with expectation μ=1 and variance σ2=0.04. (Namely, standard deviation of σ=0.2.) The region that contains 95% of the distribution of the statistic T = [Y-1]/[S/ √203] is best described by: Select one: a.

[-19.7,19.7] b.

[88.3,127.7] c.

[-1.97,1.97] d.

[106,110] Feedback The correct answer is: [-1.97,1.97] Question 10 Correct Mark 1.00 out of 1.00 Flag question Question text Consider a variable by the name "y" that contains the ratio between the power of the engine and the size of the engine in the data set "cars". Use the fact that the sample average of the variable is y=0.8237 and and the sample standard deviation is s= 0.1833 in order to compute the value of the test statistic for this variable. The value of the statistic for the observed data set "cars" is: Answer: Feedback The correct answer is: -13.7032 Question 11 Correct Mark 1.00 out of 1.00 Flag question Question text The value of the test statistic for the data set "cars" Select one: a. is inside the region that contains 95% of the sampling distribution of the statistics when the expected ratio is 1 b. is outside of the region that contains 95% of the sampling distribution of the statistics when the expected ratio is 1 Feedback The correct answer is: is outside of the region that contains 95% of the sampling distribution of the statistics when the expected ratio is 1

Information Flag question Information text In the book we compared two different estimators of the variance to each other. One estimator was the sample variance S2, and unbiased estimator of the variance. This estimator was computed using the function "var", which involves dividing the sum of squared deviations from the average by the number of observations minus 1 (n-1). The alternative estimator was obtained by dividing the sum of squares by the number of observations ( n ). This alternative estimator was equal to [(n-1)/n]S2.

In this exercise we compare two estimators of the standard deviation that are derived from the estimators of the variance. The first is S, the square root of S2, and the second is the square root of the alternative estimator of the variance. The first estimator of the standard deviation may be computed with the "sd". The alternative estimator may be computed by the multiplying the first estimator by √ [(n-1)/n].

We consider the bias, variance and mean square error of both estimators of the standard deviation. We assume a Normal sample of n=15 observations. The expectation of an observation is μ=7 and the variance is σ2 = 5. The next 5 questions refer to the following R code that simulates the sampling distribution of the two estimators:

Question 12 Correct Mark 1.00 out of 1.00 Flag question Question text In the context of the simulation, the value of the parameter that is being estimated is: (Choose the number closest to the answer) Select one:

a.

2.24 b.

5.00 c.

2.16 d.

2.20 Feedback The correct answer is: 2.24 Question 13 Correct Mark 1.00 out of 1.00 Flag question Question text The bias of the first estimator of the standard deviation is: (Choose the number closest to the answer) Select one: a.

2.76 b.

0.00 c.

2.19 d.

-0.04 Feedback The correct answer is: -0.04 Question 14 Correct Mark 1.00 out of 1.00 Flag question Question text The bias of the alternative estimator of the standard deviation is: (Choose the number closest to the answer) Select one: a.

2.16 b.

-0.11 c.

0.00 d.

-0.08 Feedback The correct answer is: -0.11

Question 15 Correct Mark 1.00 out of 1.00 Flag question Question text The mean squared error (MSE) of the first estimator of the standard deviation is: (Choose the number closest to the answer) Select one: a.

0.177 b.

0.0780 c.

0.04 d.

0.00 Feedback The correct answer is: 0.177 Question 16 Correct Mark 1.00 out of 1.00 Flag question Question text The mean squared error (MSE) of the second estimator of the standard deviation is: (Choose the number closest to the answer) Select one: a.

0.177 b.

0.0780 c.

0.04 d.

0.00 Feedback The correct answer is: 0.177 Information Flag question Information text In the book it was proposed to use 1/X (one over the sample average) as an estimator of the rate λ of the Exponential distribution. In the following 3 questions you are requested to compute the bias, the variance and the mean square error (MSE) of the given estimator for an Exponential sample of size n=35 observations and when λ=7. You may carry out these computations by conducting a simulation in R: Question 17 Correct

Mark 1.00 out of 1.00 Flag question Question text The expectation of the estimator is: (Give an answer of the form x.xxx, i.e. with 3 significance digits) Answer: Feedback The correct answer is: 7.207907 Question 18 Correct Mark 1.00 out of 1.00 Flag question Question text The mean square error (MSE) of the estimator is: (Give an answer of the form x.xxx, i.e. with 3 significance digits) Answer: Feedback The correct answer is: 1.615481 Information Flag question Information text The next 2 questions refer to the following information: The variable "num.of.doors" in the data set "cars" counts the number of cars in the sample that have four doors and the number that have two doors. In the given data set the number of cars with four doors is 114 and the number of cars with two doors is 89. Two observations have a missing value for this variable. Denote by p the probability that a car has four doors.

Question 19 Correct Mark 1.00 out of 1.00 Flag question Question text The estimated value of p, ignoring missing values, is: (Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback The correct answer is: 0.562 Question 20 Correct Mark 1.00 out of 1.00 Flag question Question text If the value of the parameter p is 0.55 then the standard deviation of the estimator of the parameter is: (Consider, as before, the sample size without the missing observations and give an answer of the form x.xxx with 3 significance digits) Answer: Feedback

The correct answer is: 0.0349 Self-quiz unit four Started on Monday, 2 May 2022, 9:08 PM State Finished Completed on Monday, 2 May 2022, 9:51 PM Time taken 43 mins 26 secs Grade 10.00 out of 10.00 (100%) Information Flag question Information text A rectangular body is square if the length is equal to the width. In the quizzes of Unit 1 the possibility of testing that cars are square on the basis of the information in the data set "cars.csv" was explored. In this exercise we re-do the test using the tools that were introduced in this unit.

Specifically, you are asked to apply the function "t.test" to the difference between the length and the width of cars. (Hint: Use the code "x <- cars$length - cars$width" in order to produce "x" and apply the function to "x".)

Question 1 Correct Mark 1.00 out of 1.00 Flag question Question text The (default) null hypothesis in the application of the function "t.test" to the variable "x" corresponds to the statement: Select one: a.

The length of the car is equal to the width of the car. b.

The length of the car is not equal to the width of the car. c.

The expected difference between the length and the width of the car is equal to 0. d.

The expected difference between the length and the width of the car is not equal to 0.

Feedback Hypothesis are formulated in terms of parameters of the distribution and not in terms of the observations themselves. The default hypothesis of the function "t.test" is "mu = 0", which corresponds to the expectation of the measurement being equal to zero. In the current setting the measurement is the difference between the length and the width. The default null is that the expectation of this difference is equal to zero.

The correct answer is: The expected difference between the length and the width of the car is equal to 0.

Question 2 Correct Mark 1.00 out of 1.00 Flag question Question text The null hypothesis is:

Select one: a.

Not rejected with a significance level of 5%. b.

Rejected with a significance level of 5%, but not with a significance level of 1%. c.

Rejected with a significance level of 1%, but not with a significance level of 0.1%. d.

Rejected with a significance level of 0.1% Feedback When we run the code: x <- cars$length - cars$width we get that the p-value is smaller than 2.2 x 10-16. In particular, the p-value is less than 0.1%. Therefore, the null hypothesis is rejected at the the 0.1% significance level.

The correct answer is: Rejected with a significance level of 0.1% Information Flag question Information text It is assumed, when constructing the t-test, that the measurements are Normally distributed. In this exercise we examine the robustness of the test to divergence from the assumption. You are required to compute the significance level of a one-sided t-test of H0: E(X) ≤ 5 versus H1: E(X) > 5. Assume there are n = 15 observations and use a t-test with a nominal 5% significance level.

Question 3 Correct Mark 1.00 out of 1.00 Flag question Question text Consider the test statistic T = [X-5]/[S/ √n], where X is the sample average, S2 is the sample variance and n is the sample size. The structure of the rejection region of the test is: Select one: a. {|T| > c} b. {T > c} c. {T < c} d. {T ≠ c} Feedback The alternative hypothesis corresponds to expectations larger than 5. The T statistic measured the difference between the sample average and the value 5. Larger values of T are an indication for the validity of the alternative hypothesis and an evidence against the validity of the null hypothesis. Therefore, we reject the null hypothesis in favor of the alternative hypothesis for values of the T statistic that are larger than a threshold.

The correct answer is: {T > c} Question 4 Correct

Mark 1.00 out of 1.00 Flag question Question text In order to obtain a significance level of 5% one should set the threshold c to be equal to: (Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback If the measurement has a Normal distribution then the distribution of the T statistic is the tdistribution on 15-1=14 degrees of freedom. The rejection region involves the upper right tail of this distribution. Consequently, the threshold is the 95-percentile of the t-distribution, which can be computed using the expression: qt(0.95,14). It turns out that the threshold is equal to 1.76131.

The correct answer is: 1.76131 Question 5 Correct Mark 1.00 out of 1.00 Flag question Question text Assume that the distribution of the observations is Poisson(5). Then the actual significance level of the test is (Use simulations to compute the significance level. Choose the closest answer.): Select one: a.

0.02 b.

0.04 c.

0.06 d.

0.08 Feedback We may use simulations in order to compute the actual significance level of the test. Run the code: > lam <- 5 > n <- 15 > X.bar <- rep(0,10^5) > S <- rep(0,10^5) > for(i in 1:10^5) +{ + X <- rpois(n,lam) + X.bar[i] <- mean(X) + S[i] <- sd(X) +} > T <- (X.bar - 5)/(S/sqrt( n )) > mean(T > qt(0.95,n-1)) [1] 0.03954 The option 0.04 is the closest to the significance level that was obtained from the simulation.

The correct answer is: 0.04 Information

Flag question Information text Assume that you are interested in testing H0: E(X) = 1 versus H1: E(X) ≠ 1 using the t-test. Let the sample average, of a sample of size n = 64, be equal to x = 0.89 and the sample standard deviation be equal to s = 3.3.

Question 6 Correct Mark 1.00 out of 1.00 Flag question Question text The absolute value of the test statistic is equal to: (Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback The T statistic is the difference between the sample average and the null expectation, divided by the estimated standard deviation of the sample average. For the data provided in the question the statistic is (0.89 - 1)/(3.3/sqrt(64)) and is equal to -0.2666667. The absolute value of the statistic is 0.2666667.

The correct answer is: 0.2666667 Question 7 Correct Mark 1.00 out of 1.00 Flag question Question text The null hypothesis is: Select one: a.

Not rejected with a significance level of 5%. b.

Rejected with a significance level of 5%, but not with a significance level of 1%. c.

Rejected with a significance level of 1%, but not with a significance level of 0.1%. d.

Rejected with a significance level of 0.1% Feedback The thresholds for the two-sided test correspond to 97.5%-, 99.5%-, and 99.95%-percentiles of the t-distribution on 64-1=63 degrees of freedom. These percentiles can computed using the expression: qt(c(0.975,0.995,0.9995),63) that produce 1.998341, 2.656145, and 3.451769, respectively, as output. The statistic is less than the smallest of the thresholds. Therefore, the null hypothesis is not rejected at the 5% confidence level.

Alternatively, one may compute the p-value using the expression 2*(1-pt(0.2666667,63)), which turns out to be equal to 0.7905967. This value is larger than 0.05. Hence, again the null hypothesis is not rejected at the 5% confidence level.

The correct answer is: Not rejected with a significance level of 5%.

Question 8

Correct Mark 1.00 out of 1.00 Flag question Question text The p-value for testing the one-sided hypothesis H0: E(X) ≤ 1 versus H1: E(X) > 1 using the ttest is equal to: (Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback For the given one-sided alternative the null hypothesis is rejected when the statistic is larger than a threshold. In the computation of a p-value we use the observed value of the statistic as the threshold. This observed value is equal to -0.2666667. The probability that a T statistic is larger then the given threshold may be computed with the expression: 1-pt(-0.2666667,63). The outcome is 0.6047016, which is the p-value.

The correct answer is: 0.6047016 Information Flag question Information text Assume that you are interested in testing that the probability of an event is H0: p = 0.25 versus H1: p ≠ 0.25 using the test for proportions. Let the proportion of the event in a sample of size n = 240 be equal to 0.275.

Question 9 Correct Mark 1.00 out of 1.00 Flag question Question text The p-value for testing the given hypotheses is equal to: (Use the function "prop.test" with a continuity correction. Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback The number of "successes" in the Binomial distribution is k = 0.275 x 240 = 66. We apply the test for proportion to this data in order to test the null hypothesis that p = 0.25: > prop.test(66,240,p=0.25) 1-sample proportions test with continuity correction data: 66 out of 240, null probability 0.25 X-squared = 0.6722, df = 1, p-value = 0.4123 alternative hypothesis: true p is not equal to 0.25 95 percent confidence interval: 0.2204682 0.3368799 sample estimates: p 0.275 The p-value is 0.4123.

The correct answer is: 0.4123 Question 10

Correct Mark 1.00 out of 1.00 Flag question Question text The null hypothesis is: Select one: a.

Not rejected with a significance level of 5%. b.

Rejected with a significance level of 5%, but not with a significance level of 1%. c.

Rejected with a significance level of 1%, but not with a significance level of 0.1%. d.

Rejected with a significance level of 0.1% Feedback The p-value is equal to 0.4123, which is larger than 0.05. As a result, the null hypothesis is not rejected at the 5% significance level.

The correct answer is: Not rejected with a significance level of 5%.

Self-quiz unit five Started on Sunday, 8 May 2022, 10:17 AM State Finished Completed on Sunday, 8 May 2022, 10:52 AM Time taken 35 mins 18 secs Grade 10.00 out of 10.00 (100%) Information Flag question Information text In Question 11.1 of the Solved Exercises of Chapter 11 we introduced an experiment aimed at assessing the effect of rumors and prior reputation of the instructor on the evaluation of the instructor by her students. The data produced by the experiment are stored in the file "teacher.csv". The file can be found on the internet at this link. Use this data in order to answer the next 3 questions: Question 1 Correct Mark 1.00 out of 1.00 Flag question Question text The null hypothesis that states that there in no significant difference between the variance of the rating in the two conditions, "C" and "P" is: Select one: a.

Not rejected with a significance level of 5%. b.

Rejected with a significance level of 5%, but not with a significance level of 1%. c.

Rejected with a significance level of 1%, but not with a significance level of 0.1%.

d.

Rejected with a significance level of 0.1% Feedback Run the code: > teacher <- read.csv("teacher.csv") > plot(rating~condition,data=teacher) > var.test(rating~condition,data=teacher) F test to compare two variances data: rating by condition F = 0.9646, num df = 24, denom df = 23, p-value = 0.9288 alternative hypothesis: true ratio of variances is not equal to 1 95 percent confidence interval: 0.4196038 2.2013532 sample estimates: ratio of variances 0.96463 We upload the data in the first line and plot it in the second. The function "var.test" is applied in order to test the null hypothesis that the ratio is equal to 1 (hence, the difference equal to 0). The computed p-value is 0.9288. Consequently, the null hypothesis is not rejected at significance level of 5%.

The correct answer is: Not rejected with a significance level of 5%.

Question 2 Correct Mark 1.00 out of 1.00 Flag question Question text The conclusion of the test, with a significance level of 5%, is: Select one: a.

The sample variances in the two sub-samples are equal to each other. b.

The variance of rating measure is the same under both conditions. c.

The sample variances in the two sub-samples are not equal to each other. d.

The variance of rating measure are not the same under both conditions.

Feedback The null hypothesis states that the variance of the measurement "rating" is the same under both conditions. The null hypothesis was not rejected at the significance level of 5%. Therefore, the conclusion is that the variance of rating measure is the same under both conditions.

The correct answer is: The variance of rating measure is the same under both conditions.

Question 3 Correct Mark 1.00 out of 1.00 Flag question Question text

The value 1: Select one: a.

Is contained in a 95% confidence interval for the ratio between the variance of the rating in the two conditions. b.

Is not contained in a 95% confidence interval for the ratio between the variance of the rating in the two conditions.

Feedback Run the code: > teacher <- read.csv("teacher.csv") > plot(rating~condition,data=teacher) > var.test(rating~condition,data=teacher) F test to compare two variances data: rating by condition F = 0.9646, num df = 24, denom df = 23, p-value = 0.9288 alternative hypothesis: true ratio of variances is not equal to 1 95 percent confidence interval: 0.4196038 2.2013532 sample estimates: ratio of variances 0.96463 The confidence interval for the ratio of the variances is [0.4196038, 2.2013532]. This interval contains the value 1.

The correct answer is: Is contained in a 95% confidence interval for the ratio between the variance of the rating in the two conditions.

Information Flag question Information text We are interested in constructing a confidence interval for the ratio of the variance of the measurement in sub-population "a" divided by the variance of the measurement in subpopulation "b". It is assumed in the construction that the measurements are Normally distributed.

In the next 4 questions you are required to examine the robustness of confidence level to divergence from the assumption.

Assume there are na=29 observations in one group and nb = 21 observations in the other group.

Consider a confidence interval with a nominal 95% confidence level.

Question 4 Correct Mark 1.00 out of 1.00 Flag question Question text If the distribution of the measurement in sub-population "a" is Normal(4,16) and the distribution of the measurement in sub-population "b" is Normal(2,4) then the actual value of the estimated parameter is: Select one: a.

-2 b.

0.25 c.

2 d.

4 Feedback The variance in population "a" is 16 and the variance in population "b" is 4. The ratio between the two is 16/4=4.

The correct answer is: 4 Question 5 Correct Mark 1.00 out of 1.00 Flag question Question text If the distribution of the measurement in sub-population "a" is Normal(4,16) and the distribution of the measurement in sub-population "b" is Normal(2,4) then the actual confidence level of the confidence interval is: (Use simulations to compute the significance level. Choose the closest answer.)

Select one: a.

0.91 b.

0.93 c.

0.95 d.

0.97 Feedback Run the code: > mu.a <- 4 > sig.a <- 4 > mu.b <- 2 > sig.b <- 2 > n.a <- 29 > n.b <- 21 > S.a <- rep(0,10^5) > S.b <- rep(0,10^5) > for(i in 1:10^5) +{ + X.a <- rnorm(n.a,mu.a,sig.a) + X.b <- rnorm(n.b,mu.b,sig.b) + S.a[i] <- sd(X.a) + S.b[i] <- sd(X.b) +} > F <- S.a^2/S.b^2

> mean((F/qf(0.025,n.a-1,n.b-1) >= 4)&(F/qf(0.975,n.a-1,n.b-1) <= 4)) [1] 0.94985 The first 6 lines set the values of the parameters and the following two lines create the objects for storage. The "for" loop produces the sampling distribution of both estimators of standard deviations. The distribution of the ratio of estimators of the variances is kept in "F". The last line computes the coverage probability of the confidence interval. We obtain that the confidence level is practically equal to 0.95.

The correct answer is: 0.95 Question 6 Correct Mark 1.00 out of 1.00 Flag question Question text If the distribution of the measurement in sub-population "a" is Exponential(1/4) and the distribution of the measurement in sub-population "b" is Exponential(1/2) then the actual value of the estimated parameter is: Select one: a.

-2 b.

0.25 c.

2 d.

4 Feedback The variance of the measurement in population "a" is 42 = 16. The variance of the measurement in population "b" is 22 = 4. The ratio between the variances is 16/4 = 4.

The correct answer is: 4 Question 7 Correct Mark 1.00 out of 1.00 Flag question Question text If the distribution of the measurement in sub-population "a" is Exponential(1/4) and the distribution of the measurement in sub-population "b" is Exponential(1/2) then the actual confidence level of the confidence interval is: (Use simulations to compute the significance level.

Choose the closest answer.)

Select one: a.

0.70 b.

0.80 c.

0.90 d.

0.95

Feedback Run the code: > lam.a <- 1/4 > lam.b <- 1/2 > n.a <- 29 > n.b <- 21 > S.a <- rep(0,10^5) > S.b <- rep(0,10^5) > for(i in 1:10^5) +{ + X.a <- rexp(n.a,lam.a) + X.b <- rexp(n.b,lam.b) + S.a[i] <- sd(X.a) + S.b[i] <- sd(X.b) +} > F <- S.a^2/S.b^2 > mean((F/qf(0.025,n.a-1,n.b-1) >= 4)&(F/qf(0.975,n.a-1,n.b-1) <= 4)) [1] 0.72551 The first 4 lines set the values of the parameters and the following two lines create the objects for storage. The "for" loop produces the sampling distribution of both estimators of standard deviations. The distribution of the ratio of estimators of the variances is kept in "F". The last line computes the coverage probability of the confidence interval. We obtain that the confidence level is closest to the value 0.70.

The correct answer is: 0.70 Information Flag question Information text The sample average in one sub-sample is xa = 30.7 and the sample standard deviation is sa = 13.4.

The sample average in the second sub- sample is xb = 32.3 and the sample standard deviation is sb = 16.7. The size of the first sub-sample is na = 57 and the size of the second sub-sample is nb = 77. We are interested in testing that the expectations in the associated sub-populations are equal to each other. The following 3 questions refer to the given information.

Question 8 Correct Mark 1.00 out of 1.00 Flag question Question text The absolute value of the test statistic is equal to: (Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback Run the code: > n.a <- 57 > n.b <- 77 > x.bar.a <- 30.7 > x.bar.b <- 32.3 > s.a <- 13.4 > s.b <- 16.7

> t <- abs(x.bar.a-x.bar.b)/sqrt(s.a^2/n.a + s.b^2/n.b) >t [1] 0.6148335 The absolute value of the statistic is 0.6148335.

The correct answer is: 0.6148335 Question 9 Correct Mark 1.00 out of 1.00 Flag question Question text The p-value for testing the hypothesis is equal to: (Use the Normal approximation. Choose the closest answer.)

Select one: a.

0.5 b.

0.05 c.

0.005 d.

0.0005 Feedback Run the code: > 2*(1-pnorm(0.6148335)) [1] 0.5386647 The p-value for the two-sided test is equal to twice the upper tail. The (absolute) value of the statistic is 0.6148335. The code computes Normal probability above that value, multiplied by 2. The value closest to the computed p-value is 0.5.

The correct answer is: 0.5 Question 10 Correct Mark 1.00 out of 1.00 Flag question Question text The null hypothesis is: Select one: a.

Not rejected with a significance level of 5%. b.

Rejected with a significance level of 5%, but not with a significance level of 1%. c.

Rejected with a significance level of 1%, but not with a significance level of 0.1%. d.

Rejected with a significance level of 0.1% Feedback The p-value is larger than 0.05. Therefore, the null hypothesis is not rejected with a significance level of 5%.

The correct answer is: Not rejected with a significance level of 5%.

Self-quiz unit six

Started on Tuesday, 17 May 2022, 8:12 AM State Finished Completed on Tuesday, 17 May 2022, 8:42 AM Time taken 29 mins 13 secs Grade 10.00 out of 10.00 (100%) Information Flag question Information text In Figure 1 you may find a scatter plot with 10 points and 3 lines. Three of the points are marked by the letters a, b, and c. The lines are marked by the numbers 1, 2, and 3. The following 2 questions refer to this scatter plot.

Question 1 Correct Mark 1.00 out of 1.00 Flag question Question text

Match the coefficients to the lines:

Feedback Line 2 has a zero slope. Lines 1 and 3 have the same slope, but the intercept of line 1 is larger. It follows that: Line 1: Intercept = 3, Slope = 2.

Line 2: Intercept = 1.5, Slope = 0.

Line 3: Intercept = -1.5, Slope = 2.

The correct answer is: Line 1 → Intercept = 3, Slope = 2, Line 2 → Intercept = 1.5, Slope = 0, Line 3 → Intercept = -1.5, Slope = 2 Question 2 Correct Mark 1.00 out of 1.00 Flag question Question text Match the points:

Feedback Point a has the largest y value. Point b has the largest x value and point c has the smallest x value. Therefore: Point a => x = -0.1, y = 3.8 Point b => x = 1.6, y = -1.6 Point c => x = -0.6, y = 0.8 The correct answer is: Point a → x = -0.1, y = 3.8, Point b → x = 1.6, y = -1.6, Point c → x = -0.6, y = 0.8 Information Flag question

Information text Fit a regression model to the relation between "horsepower" as a response and "peak.rpm" as an explanatory variable. Both variables belong to the data set "cars.csv". The following 3 refer to the fitted model.

Question 3 Correct Mark 1.00 out of 1.00 Flag question Question text The estimated value of the slope is equal to: (Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback We use the function "lm" in order to fit the regression line between "peak.rpm" as an explanatory variable and "horsepower" as a response. Run the R code: > cars <- read.csv("cars.csv") > lm(horsepower~peak.rpm,data=cars) Call: lm(formula = horsepower ~ peak.rpm, data = cars) Coefficients: (Intercept) peak.rpm 48.63903 0.01085 The slope is equal to 0.01085.

The correct answer is: 0.010851 Question 4 Correct Mark 1.00 out of 1.00 Flag question Question text Test the null hypothesis the intercept is equal to 0. This null hypothesis is: Select one: a.

Not rejected with a significance level of 5%. b.

Rejected with a significance level of 5%, but not with a significance level of 1%. c.

Rejected with a significance level of 1%, but not with a significance level of 0.1%. d.

Rejected with a significance level of 0.1% Feedback Tests for the coefficients of a regression model are part of the output of the application of the function "summary" to the output of the function "lm". Run the code: > cars <- read.csv("cars.csv") > summary(lm(horsepower~peak.rpm,data=cars)) Call:

lm(formula = horsepower ~ peak.rpm, data = cars) Residuals: Min 1Q Median 3Q Max -55.981 -31.768 -8.258 15.275 176.966 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 48.639025 29.823641 1.631 0.1045 peak.rpm 0.010851 0.005794 1.873 0.0625 .

--Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 39.47 on 201 degrees of freedom (2 observations deleted due to missingness) Multiple R-squared: 0.01715, Adjusted R-squared: 0.01226 F-statistic: 3.508 on 1 and 201 DF, p-value: 0.06252 The p-value for testing that the intercept is equal to zero is 0.1045, which is larger than 0.05. As a result, the null hypothesis is not rejected with a significance level of 5%.

The correct answer is: Not rejected with a significance level of 5%.

Question 5 Correct Mark 1.00 out of 1.00 Flag question Question text Use the function "confint" in order to construct a confidence interval to the slope of the regression line. The value 0: Select one: a.

Belongs to the 95% confidence interval. b.

Does not belongs to the 95% confidence interval.

Feedback We can produce confidence intervals for the coefficients of the regression model via the application of the function "confint" to the output of the function "lm". Run the code: > cars <read.csv("cars.csv") > confint(lm(horsepower~peak.rpm,data=cars)) 2.5 % 97.5 % (Intercept) -1.016832e+01 107.44636946 peak.rpm -5.728294e-04 0.02227551 The confidence interval for the slope is [-5.728294 x 10-4, 0.02227551]. The value 0 belongs to this interval.

The correct answer is: Belongs to the 95% confidence interval.

Information Flag question Information text The next 3 questions refer to the data given in following table:

Data x y 1 1.2 0.5 2 0.0 2.2 3 1.6 4.9 4 3.8 0.1 5 1.6 1.6 6 0.8 0.6 7 1.7 3.4 8 5.8 0.5 9 1.6 0.7 1 3.5 0 0.4 Question 6 Correct Mark 1.00 out of 1.00 Flag question Question text The covariance between x an y is equal to: (Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback We read the data into R and then apply the function "cov": > x <- c(-0.5,0.0,1.6,-0.1,1.6,-0.6,1.7,0.5,-0.7,-0.4) > y <- c(1.2,2.2,-4.9,3.8,-1.6,0.8,-3.4,5.8,1.6,3.5) > cov(x,y) [1] -2.772222 Alternatively, we may apply the formula for the computation of the covariance: > sum((x-mean(x))*(y-mean( y )))/(10-1) [1] -2.772222 The covariance is equal to -2.772222.

The correct answer is: -2.772222 Question 7 Correct Mark 1.00 out of 1.00 Flag question Question text

If y is the response and x is the explanatory variable then the slope of the regression line is equal to: (Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback The slope may be obtained by the application of the function "lm" to a formula that puts y as the response and x as the explanatory variable: > x <- c(-0.5,0.0,1.6,-0.1,1.6,-0.6,1.7,-0.5,-0.7,-0.4) > y <- c(1.2,2.2,-4.9,3.8,-1.6,0.8,-3.4,5.8,1.6,3.5) > lm(y~x) Call: lm(formula = y ~ x) Coefficients: (Intercept) x 1.476 -2.745 Alternatively, one may apply the formula for the computation of the slope: > cov(x,y)/var(x) [1] -2.745076 The slope is equal to -2.745.

The correct answer is: -2.745 Question 8 Correct Mark 1.00 out of 1.00 Flag question Question text If x is the response and y is the explanatory variable then the slope of the regression line is equal to: (Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback The slope may be obtained by the application of the function "lm" to a formula that puts x as the response and y as the explanatory variable: > x <- c(-0.5,0.0,1.6,-0.1,1.6,-0.6,1.7,-0.5,-0.7,-0.4) > y <- c(1.2,2.2,-4.9,3.8,-1.6,0.8,-3.4,5.8,1.6,3.5) > lm(x~y) Call: lm(formula = x ~ y) Coefficients: (Intercept) y 0.4349 -0.2499 Alternatively, one may apply the formula for the computation of the slope: > cov(x,y)/var( y ) [1] -0.2498998 The slope is equal to -0.2499.

The correct answer is: -0.2499 Question 9 Correct Mark 1.00 out of 1.00 Flag question Question text

Consider a response and an explanatory variable. Assume that the variance of the explanatory variable is positive. If the covariance between the two variables is positive then: Select one: a.

The regression line is decreasing b.

The regression line is increasing c.

The regression line may be increasing or decreasing. d.

The regression line has a zero slope.

Feedback The slope of the regression line is the ratio cov(x,y)/var(x) between the covariance and the variance of the explanatory variable. If the covariance is positive then the slope is positive. A positive slope corresponds to an increasing line. Therefore, the regrersion line is increasing.

The correct answer is: The regression line is increasing Question 10 Correct Mark 1.00 out of 1.00 Flag question Question text Consider a response and an explanatory variable. Assume that the variance of the explanatory variable is positive. If the slope of the regression line is equal to zero then covariance between x and y: Select one: a. is positive. b. is negative c. may be positive or negative. d. is equal to 0.

Feedback The slope of the regression line is the ratio cov(x,y)/var(x) between the covariance and the variance of the explanatory variable. If the slope is equal to zero then the covariance is also equal to zero.

The correct answer is: is equal to 0.

Graded-quiz unit six Started on Tuesday, 17 May 2022, 8:43 AM State Finished Completed on Tuesday, 17 May 2022, 9:33 AM Time taken 49 mins 51 secs Marks 20.00/20.00 Grade 100.00 out of 100.00

Information Flag question Information text

In Figure 2 you may find a scatter plot with 10 points and 3 lines. Three of the points are marked by the letters a, b, and c. The lines are marked by the numbers 1, 2, and 3. The following 2 questions refer to this scatter plot.

Information Flag question Information text Fit a regression model to the relation between "curb.weight" as a response and "engine.size" as an explanatory variable. Both variables belong to the data set "cars.csv". The following 3 refer to the fitted model.

Question 1 Correct Mark 1.00 out of 1.00 Flag question Question text The estimated value of the slope is equal to: (Provide an answer of the form xx.xxxx) Answer: Feedback The correct answer is: 10.6354 Question 2 Correct Mark 1.00 out of 1.00 Flag question Question text Test the null hypothesis the intercept is equal to 0. This null hypothesis is: Select one: a.

Not rejected with a significance level of 5%. b.

Rejected with a significance level of 5%, but not with a significance level of 1%. c.

Rejected with a significance level of 1%, but not with a significance level of 0.1%.

d.

Rejected with a significance level of 0.1% Feedback The correct answer is: Rejected with a significance level of 0.1% Question 3 Correct Mark 1.00 out of 1.00 Flag question Question text Use the function "confint" in order to construct a confidence interval to the slope of the regression line. The value 0: Select one: a.

Belongs to the 95% confidence interval. b.

Does not belongs to the 95% confidence interval.

Feedback The correct answer is: Does not belongs to the 95% confidence interval.

Information Flag question Information text The next 3 questions refer to the data given in following table: Data x y 1 0.5 3.4 2 1.5 1.6 3 2.5 1.5 4 3.5 0.1 5 1.5 1.8 6 2.5 0.9 - 7 0.5 4.1 8 1.5 1.3 - 9 0.5 1.2 1 1.5 0 2.9 Question 4 Correct

Mark 1.00 out of 1.00 Flag question Question text If y is the response and x is the explanatory variable then the slope of the regression line is equal to: (Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback The correct answer is: 0.8 Question 5 Correct Mark 1.00 out of 1.00 Flag question Question text If x is the response and y is the explanatory variable then the slope of the regression line is equal to: (Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback The correct answer is: 0.5799 Question 6 Correct Mark 1.00 out of 1.00 Flag question Question text Consider a response and an explanatory variable. Assume that the variance of the explanatory variable is positive. If the covariance between the two variables is negative then: Select one: a.

The regression line is decreasing b.

The regression line is increasing c.

The regression line may be increasing or decreasing. d.

The regression line has a zero slope.

Feedback The correct answer is: The regression line is decreasing Question 7 Correct Mark 1.00 out of 1.00 Flag question Question text Consider a response and an explanatory variable. Assume that the variance of the explanatory variable is positive. If the regression line is increasing then covariance between x and y: Select one: a. is positive.

b. is negative. c. may be positive or negative. d. is equal to 0.

Feedback The correct answers are: is positive., is equal to 0.

Information Flag question Information text A rectangular body is a "Golden Rectangle" if the ratio between the length and the width is equal to the golden ratio 1.618. Golden Rectangles are considered to be esthetic and are found, for example, in the structure of ancient Greek temples. In the quizzes of Unit 1 the possibility of testing that cars are square on the basis of the information in the data set "cars.csv" was explored.

In this exercise we want to test the null hypothesis that the body of a car is a Golden Rectangle.

Specifically, you are asked to apply the function "t.test" to the ratio between the length and the width of cars. (Hint: Use the code "r <- cars$length/cars$width" in order to produce "r" and apply the function to "r".)

Question 8 Correct Mark 1.00 out of 1.00 Flag question Question text The null hypothesis in the application of the function "t.test" to the variable "r" corresponds to the statement: Select one: a.

The ratio between the length of the car and the width of the car is equal to 1.618. b.

The ratio between the length of the car and the width of the car is not equal to 1.618. c.

The expectation of the ratio between the length of the car and the width of the car is equal to 1.618. d.

The expectation of the ratio between the length of the car and the width of the car is not equal to 1.618.

Feedback The correct answer is: The expectation of the ratio between the length of the car and the width of the car is equal to 1.618.

Information Flag question Information text It is assumed, when constructing the t-test, that the measurements are Normally distributed. In this exercise we examine the robustness of the test to divergence from the assumption. You are

required to compute the significance level of a one-sided t-test of H0: E(X) ≥ 5 versus H1: E(X) < 5. Assume there are n = 35 observations and use a t-test with a nominal 5% significance level.

Question 9 Correct Mark 1.00 out of 1.00 Flag question Question text Consider the test statistic T = [X-5]/[S/ √n], where X is the sample average, S2 is the sample variance and n is the sample size. The structure of the rejection region of the test is: Select one: a. {|T| > c} b. {T > c} c. {T < c} d. {T ≠ c} Feedback The correct answer is: {T < c} Question 10 Correct Mark 1.00 out of 1.00 Flag question Question text In order to obtain a significance level of 5% one should set the threshold c to be equal to: (Give an answer of the form x.xxx with 3 significance digits. Pay attention to the sign!)

Answer: Feedback The correct answer is: -1.690924 Question 11 Correct Mark 1.00 out of 1.00 Flag question Question text Assume that the distribution of the observations is Binomial(20,0.25). Then the actual significance level of the test is (Use simulations to compute the significance level. Choose the closest answer.): Select one: a.

0.01 b.

0.03 c.

0.05 d.

0.07 Feedback The correct answer is: 0.05 Information Flag question Information text Assume that you are interested in testing H0: E(X) ≤ 1 versus H1: E(X) >1 using the t-test. Let the sample average, of a sample of size n = 220, be equal to x = 1.36 and the sample standard deviation be equal to s = 2.20.

Question 12 Correct Mark 1.00 out of 1.00 Flag question Question text The value of the test statistic is equal to: (Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback The correct answer is: 2.42712 Question 13 Correct Mark 1.00 out of 1.00 Flag question Question text The p-value for testing the two-sided hypothesis H0: E(X) = 1 versus H1: E(X) ≠ 1 using the ttest is equal to: (Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback The correct answer is: 0.01602759 Information Flag question Information text Assume that you are interested in testing that the probability of an event is H0: p ≤ 0.25 versus H1: p > 0.25 using the test for proportions. Let the proportion of the event in a sample of size n = 2400 be equal to 0.275.

Question 14 Correct Mark 1.00 out of 1.00 Flag question Question text The p-value for testing the given hypotheses is equal to: (Use the function "prop.test" with a continuity correction. You may use the argument "alternative="greater"". Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback The correct answer is: 0.002517

Information Flag question Information text In Question 11.1 of the Solved Exercises of Chapter 11 we introduced an experiment aim at assessing the effect of rumors and prior reputation of the instructor on the evaluation of the instructor by her students. The data produced by the experiment are stored in the file "teacher.csv". The file can be found on the internet at this link. Use this data in order to answer the next 3 questions: Information Flag question Information text We are interested in constructing a confidence interval for the ratio of the variance of the measurement in sub-population "a" divided by the variance of the measurement in subpopulation "b". It is assumed in the construction that the measurements are Normally distributed.

In the next 4 questions you are required to examine the robustness of confidence level to divergence from the assumption.

Assume there are na=29 observations in one group and nb = 21 observations in the other group.

Consider a confidence interval with a nominal 95% confidence level.

Question 15 Correct Mark 1.00 out of 1.00 Flag question Question text If the distribution of the measurement in sub-population "a" is Uniform(0,4) and the distribution of the measurement in sub-population "b" is Uniform(0,5) then the actual value of the estimated parameter is: Select one: a.

0 b.

0.64 c.

0.8 d.

1 Feedback The correct answer is: 0.64 Question 16 Correct Mark 1.00 out of 1.00 Flag question Question text Subpopulation “a” has a Uniform(0,4) distribution, and subpopulation “b” has a Uniform (0,5) distribution.

Use simulation, using sample sizes of 29 and 21 respectively, to find a sample of 10^5 variances for

each of these subpopulations, and use these to find 10^5 values of F = variance of “a” / variance of “b”.

What proportion of these F-values falls in the middle .95 interval of the F distribution with the degrees of freedom appropriate for the sample sizes? Choose the closest answer.

Select one: a.

0.87 b.

0.91 c.

0.95 d.

0.99 Feedback The correct answers are: 0.91, 0.99 Question 17 Correct Mark 1.00 out of 1.00 Flag question Question text If the distribution of the measurement in sub-population "a" is Poisson(4) and the distribution of the measurement in sub-population "b" is Poisson(6) then the actual value of the estimated parameter is: Select one: a.

0 b.

0.444 c.

0.667 d.

1 Feedback The correct answer is: 0.667 Question 18 Correct Mark 1.00 out of 1.00 Flag question Question text The distribution of the measurement in sub-population "a" is Poisson(4) and the distribution of the measurement in sub-population "b" is Poisson(6). Using simulation, find 10^5 values of the variances and 10^5 values of F. What proportion of the F-values falls in the middle 95% interval of the appropriate F distribution? Choose the closest answer.

Select one: a.

0.87

b.

0.91 c.

0.95 d.

0.99 Feedback The correct answer is: 0.87 Information Flag question Information text The sample average in one sub-sample is xa = 4.3 and the sample standard deviation is sa = 5.4.

The sample average in the second sub- sample is xb = 2.1 and the sample standard deviation is sb = 6.7. The size of the first sub-sample is na = 83 and the size of the second sub-sample is nb = 92. We are interested in testing that the expectations in the associated sub-populations are equal to each other. The following 3 questions refer to the given information.

Question 19 Correct Mark 1.00 out of 1.00 Flag question Question text The absolute value of the test statistic is equal to: (Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback The correct answer is: 2.401455 Question 20 Correct Mark 1.00 out of 1.00 Flag question Question text The p-value for testing the hypothesis is equal to: (Use the Normal approximation. Choose the closest answer.)

Select one: a.

0.15 b.

0.04 c.

0.015 d.

0.004 Feedback The correct answer is: 0.015 Self-quiz unit seven Started on Tuesday, 24 May 2022, 12:18 PM

State Finished Completed on Tuesday, 24 May 2022, 12:59 PM Time taken 41 mins Marks 11.00/11.00 Grade 10.00 out of 10.00 (100%) Information Flag question Information text A regression model was fitted to data. The slope of the model was b = -0.2499 and the intercept was a = 0.4349. Let y be the response and x be the explanatory variable. The next 2 questions refer to this information: Question 1 Correct Mark 1.00 out of 1.00 Flag question Question text Match the observations to the estimated expectation: x = 1.2, y = -0.5 Estimated expectation = 0.135 x = 2.2, y = 0.0 Estimated expectation = -0.115 x = -4.9, y = 1.6 Estimated expectation = 1.659 Feedback The estimated expectation of the response for a given value of the explanatory variable is the level of the regression line for the given value. The regression line corresponds to the linear equation y = 0.4349 - 0.2499×x.

For x = 1.2 the estimated expectation is 0.4349 - 0.2499×1.2 = 0.13502 For x = 2.2 the estimated expectation is 0.4349 - 0.2499×2.2 = -0.11488 For x= -4.9 the estimated expectation is 0.4349 - 0.2499×(-4.9) = 1.65941 Consequently: The point x = 1.2, y = -0.5 is associated with the estimated expectation 0.135.

The point x = 2.2, y = 0.0 is associated with the estimated expectation -0.115.

The point x = -4.9, y = 1.6 is associated with the estimated expectation 1.659.

The correct answer is: x = 1.2, y = -0.5 → Estimated expectation = 0.135, x = 2.2, y = 0.0 → Estimated expectation = -0.115, x = -4.9, y = 1.6 → Estimated expectation = 1.659 Question 2 Correct Mark 1.00 out of 1.00 Flag question Question text Match the observations to the residuals: x = -3.4, y = 1.7 Residual = 0.415 x = 5.8, y = -0.5 Residual = 0.515 x = 1.6, y = -0.7 Residual = -0.735 Feedback The residual is the difference between the value of the response and the estimated expectation of the response. For the given regression line this difference corresponds to y - (0.4349 - 0.2499×x).

For x = -3.4, y = 1.7 the residual is equal to 1.7 - (0.4349 - 0.2499×(-3.4)) = 0.41544.

For x = 5.8, y = -0.5 the residual is equal to -0.5 - (0.4349 - 0.2499×5.8) = 0.51452.

For x = 1.6, y = -0.7 the residual is equal to -0.7 - (0.4349 - 0.2499×1.6) = -0.73506.

Consequently: The point x = -3.4, y = 1.7 is associated with the residual 0.415.

The point x = 5.8, y = -0.5 is associated with the residual 0.515.

The point x = 1.6, y = -0.7 is associated with the residual -0.735.

The correct answer is: x = -3.4, y = 1.7 → Residual = 0.415, x = 5.8, y = -0.5 → Residual = 0.515, x = 1.6, y = -0.7 → Residual = -0.735 Information Flag question Information text Fit a regression model to the relation between "height" as a response and "length" as an explanatory variable. Also, fit a regression model to the relation between "height" as a response and "width" as an explanatory variable. All 3 variables belong to the data set "cars.csv". The following 5 questions refer to the fitted models.

Question 3 Correct Mark 1.00 out of 1.00 Flag question Question text The relation between "height" and "length" is: Select one: a.

Not significant in a significant level of 5%. b.

Significant with a significance level of 5%, but not with a significance level of 1%. c.

Significant with a significance level of 1%, but not with a significance level of 0.1%. d.

Significant with a significance level of 0.1% Feedback Run the code: > cars <- read.csv("cars.csv") > summary(lm(height~length,data=cars)) Call: lm(formula = height ~ length, data = cars) Residuals: Min 1Q Median 3Q Max -7.6415 -1.2960 0.2438 1.2140 6.2235 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 36.79802 2.11297 17.415 < 2e-16 ***

length 0.09725 0.01211 8.031 7.73e-14 *** --Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 2.134 on 203 degrees of freedom Multiple R-squared: 0.2411, Adjusted R-squared: 0.2374 F-statistic: 64.5 on 1 and 203 DF, p-value: 7.732e-14 The p-value for testing the null hypothesis that the slope equals 0 is 7.73×10-14. This p-value is less than 0.001. Therefore, the null hypothesis is rejected with a significance level of 0.1%.

The correct answer is: Significant with a significance level of 0.1% Question 4 Correct Mark 1.00 out of 1.00 Flag question Question text The part of the variance of "height" explained by "length" is: (Choose the closest number) Select one: a.

0.012 b.

0.241 c.

0.491 d.

2.134 Feedback Run the code: > cars <- read.csv("cars.csv") > summary(lm(height~length,data=cars)) Call: lm(formula = height ~ length, data = cars) Residuals: Min 1Q Median 3Q Max -7.6415 -1.2960 0.2438 1.2140 6.2235 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 36.79802 2.11297 17.415 < 2e-16 *** length 0.09725 0.01211 8.031 7.73e-14 *** --Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 2.134 on 203 degrees of freedom Multiple R-squared: 0.2411, Adjusted R-squared: 0.2374

F-statistic: 64.5 on 1 and 203 DF, p-value: 7.732e-14 The value of R-squared is 0.2411. This is the part of the variance of the response that is explained by the regression model.

The correct answer is: 0.241 Question 5 Correct Mark 1.00 out of 1.00 Flag question Question text The relation between "height" and "width" is: Select one: a.

Not significant in a significant level of 5%. b.

Significant with a significance level of 5%, but not with a significance level of 1%. c.

Significant with a significance level of 1%, but not with a significance level of 0.1%. d.

Significant with a significance level of 0.1% Feedback Run the code: Call: lm(formula = height ~ width, data = cars) Residuals: Min 1Q Median 3Q Max -7.4172 -1.8362 0.4284 1.4455 6.4911 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 32.76369 5.06221 6.472 7.11e-10 *** width 0.31804 0.07677 4.143 5.03e-05 *** --Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 2.352 on 203 degrees of freedom Multiple R-squared: 0.07796, Adjusted R-squared: 0.07342 F-statistic: 17.16 on 1 and 203 DF, p-value: 5.031e-05 The p-value for testing the null hypothesis that the slope equals 0 is 5.03×10-5. This p-value is less than 0.001. Therefore, the null hypothesis is rejected with a significance level of 0.1%.

The correct answer is: Significant with a significance level of 0.1% Question 6 Correct Mark 1.00 out of 1.00 Flag question Question text The part of the variance of "height" explained by "width" is: (Choose the closest number) Select one:

a.

0.078 b.

0.277 c.

0.318 d.

2.352 Feedback Run the code: Call: lm(formula = height ~ width, data = cars) Residuals: Min 1Q Median 3Q Max -7.4172 -1.8362 0.4284 1.4455 6.4911 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 32.76369 5.06221 6.472 7.11e-10 *** width 0.31804 0.07677 4.143 5.03e-05 *** --Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 2.352 on 203 degrees of freedom Multiple R-squared: 0.07796, Adjusted R-squared: 0.07342 F-statistic: 17.16 on 1 and 203 DF, p-value: 5.031e-05 The value of R-squared is 0.07796. This is the part of the variance of the response that is explained by the regression model.

The correct answer is: 0.078 Question 7 Correct Mark 1.00 out of 1.00 Flag question Question text Of the two variables, the variable that explains better the variability of the response "height" is: Select one: a. "length" b. "width" Feedback The variable "length" explains 0.2411 of the variance of the response. The variable "width" explains only 0.07796 of the variance. Therefore, of the two explanatory variables, "length" produces a better explanation.

The correct answer is: "length" Information Flag question

Information text Consider the relation between the response "drive.wheels" and the explanatory variable "fuel.type". Both variables belong to the data set "cars.csv". The next two questions refer to these variables.

Question 8 Correct Mark 1.00 out of 1.00 Flag question Question text Plot the mosaic plot that describes the relation between the response "drive.wheels" and the explanatory variable "fuel.type". The height of the rectangle with the lightest shade of grey above the value "gas" corresponds to: Select one: a.

The relative frequency of cars that run on gas. b.

The relative frequency of cars that use "rwd" among cars that run on gas. c.

The relative frequency of cars that use "rwd". d.

The relative frequency of cars that run on gas among cars that use "rwd".

Feedback The width of a rectangle corresponds to the relative frequency of a level of the explanatory variable. The hight of a rectangle of a given shade of grey above a level of the explanatory variable corresponds to the relative frequency of a level of the response among the sub-sample that is associated with the given level of the explanatory variable.

In the current case we consider rectangles above the level "gas" of the explanatory variable. The said shade of grey is associated with the level "rwd" of the response. Therefore, the height of the rectangle with the lightest shade of grey above the value "gas" corresponds to the relative frequency of cars that use "rwd" among cars that run on gas.

The correct answer is: The relative frequency of cars that use "rwd" among cars that run on gas.

Question 9 Correct Mark 1.00 out of 1.00 Flag question Question text Test the null hypothesis that the probability that cars that run on diesel use "fwd" is equal to the probability that cars that run on gas use "fwd". Test this hypothesis versus the two-sided alternative. The p-value computed for the test is equal to: (Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback Run the code: > table(cars$fuel.type,cars$drive.wheels=="fwd") FALSE TRUE

diesel 11 9 gas 74 111 > prop.test(table(cars$fuel.type,cars$drive.wheels=="fwd")) 2-sample test for equality of proportions with continuity correction data: table(cars$fuel.type, cars$drive.wheels == "fwd") X-squared = 1.1122, df = 1, p-value = 0.2916 alternative hypothesis: two.sided 95 percent confidence interval: -0.1068785 0.4068785 sample estimates: prop 1 prop 2 0.55 0.40 The first line produces the table of (joint) frequencies of the response and the explanatory variable. The response is a logical sequence that indicates, for each car, does it use "fwd" or not. The explanatory variable is the fuel type, a factor. In the second line the function "prop.test" is applies to the table of frequencies.

The output of the function "prop.test" includes a test of the null hypothesis the probability that cars that run on diesel use "fwd" is equal to the probability that cars that run on gas use "fwd".

The resulting p-value is equal to 0.2916.

The correct answer is: 0.2916 Information Flag question Information text Consider the relation between the response "drive.wheels" and the explanatory variable "engine.size". Both variables belong to the data set "cars.csv". The next two questions refer to these variables.

Question 10 Correct Mark 1.00 out of 1.00 Flag question Question text Plot the mosaic plot that describes the relation between the response "drive.wheels" and the explanatory variable "engine.size". The width of the second rectangle from the left corresponds to: Select one: a.

150 - 100 = 50. b.

The frequency of cars that use "rwd". c.

The frequency of cars with engine size between 100 and 150. d.

The frequency of cars that use "rwd" among cars with engine size between 100 and 150.

Feedback

Run the expression "plot(drive.wheels~engine.size,data=cars)" in order to produce the mosaic plot. The width of a rectangle corresponds to the relative frequency of the range of levels of the explanatory variable. The hight of a rectangle of a given shade of grey above a a range of levels of the explanatory variable corresponds to the relative frequency of a level of the response among the sub-sample that is associated with the given range of levels of the explanatory variable.

The second rectangle from the left is associate with the range of engine sizes between 100 and 150. The width of this rectangle corresponds to the frequency of cars with engine size between 100 and 150.

The correct answer is: The frequency of cars with engine size between 100 and 150.

Question 11 Correct Mark 1.00 out of 1.00 Flag question Question text Test the null hypothesis that there is no relation between the probability of using "fwd" and the size of the engine. The null hypothesis is: Select one: a.

Not rejected with a significance level of 5%. b.

Rejected with a significance level of 5%, but not with a significance level of 1%. c.

Rejected with a significance level of 1%, but not with a significance level of 0.1%. d.

Rejected with a significance level of 0.1% Feedback Run the code: > summary(glm(drive.wheels=="fwd"~engine.size,binomial,data=cars)) Call: glm(formula = drive.wheels == "fwd" ~ engine.size, family = binomial, data = cars) Deviance Residuals: Min 1Q Median 3Q Max -2.3756 -0.8398 0.5447 0.7786 2.1225 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 5.851567 0.865354 6.762 1.36e-11 *** engine.size -0.044160 0.006948 -6.356 2.08e-10 *** --Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for gaussian family taken to be 0.1792372)

Null deviance: 278.19 on 204 degrees of freedom Residual deviance: 206.57 on 203 degrees of freedom AIC: 233.35 Number of Fisher Scoring iterations: 5 This code implements logistic regression to describe the relation between the response of car using or not "fwd" and the engine size as an explanatory variable.

The null hypothesis that states that there no relation between the response and the explanatory variable corresponds to the hypothesis that the coefficient of the variable "engine.size" is equal to zero. The p-value for testing this hypothesis is equal to 2.08×10-10. This p-value is less than 0.001. Therefore, the null hypothesis is rejected with a significance level of 0.1%.

The correct answer is: Rejected with a significance level of 0.1% Self-quiz unit eight Started on Friday, 27 May 2022, 7:30 PM State Finished Completed on Friday, 27 May 2022, 7:52 PM Time taken 21 mins 36 secs Marks 7.00/7.00 Grade 10.00 out of 10.00 (100%) Question 1 Correct Mark 1.00 out of 1.00 Flag question Question text Let X be a measurement with the Normal distribution. The measurement is taken for a sample of size 20 under two conditions. For half the sample the measurement is taken under one condition and for the other half the measurement is taken under a different condition. Denote the sample variance computed for the first half by Sa2 and the sample variance computed for the second half

by Sb2. Assume that the variance of the measurement is the same under both conditions. A region that contains 90% of the distribution of the statistic F = Sa2/Sb2 is: (Choose the closest option.)

Select one: a.

[-1.65,1.65] b.

[-1.83,1.83] c.

[0.31, 3.18] d.

[3.33, 16.92] Feedback The distribution of the ratio of two estimators of the same variance, when the observations come from the Normal distribution, is the F distribution. This distribution is not symmetric. The region between that 5%- and the 95%-percentiles of the F distribution contain 90% of the distribution.

These percentiles can be computed using the code "qf(c(0.05,0.95),9,9)", that produces [0.3145749, 3.1788931] as an output.

The correct answer is: [0.31, 3.18] Question 2 Correct Mark 1.00 out of 1.00 Flag question Question text The maximum of 15 Uniform(0,b) observations is used in order to estimate the parameter "b".

Bellow we present a simulation that assesses the sampling distribution of the estimator in the case where the value of the parameter is 7. The mean squared error (MSE) of the estimator, based on the output of the simulation, is equal to: (Give an answer of the form x.xxx, i.e. with 3 significance digits)

Answer: Feedback The MSE is equal to the sum of the variance and the square of the bias. The variance is equal to the square of the standard deviation 0.4111911. The bias is equal to the difference between the expectation and the target value of the parameter: 6.561692 - 7. Therefore, the MSE is equal to 0.41119112 + (6.561692 - 7)2 = 0.361192.

The correct answer is: 0.361192 Question 3 Correct Mark 1.00 out of 1.00 Flag question Question text A production batch contains 2,500 items. 30 items were randomly selected for inspection from the batch. 5 of the selected items did not pass the inspection. Let p be the probability that a selected item does not pass inspection. A 95% confidence interval for p is: (Choose the interval closest to the answer) Select one: a.

[0.063, 0.355] b.

[0.072, 0.415]

c.

[0.0082, 0.0173] d.

[0.0007, 0.0050] Feedback The parameter of interest is the probability of selecting a "bad" item. Of the 30 items that were selected, 5 were bad. A confidence interval for the probability of selecting a bad item can be obtained as part of the output of the function "prop.test": > prop.test(5,30) 1-sample proportions test with continuity correction data: 5 out of 30, null probability 0.5 X-squared = 12.0333, df = 1, p-value = 0.0005226 alternative hypothesis: true p is not equal to 0.5 95 percent confidence interval: 0.06303555 0.35451084 sample estimates: p 0.1666667 This function produces [0.06303555, 0.35451084] as the 95%-confidence interval. It applies a continuity correction in the computation of the interval.

Observe that the sample size is small. Therefore, the application of a continuity correction is recommended. A confidence interval without a continuity correction may be computed with the code: > p.hat <- 5/30 > p.hat + qnorm(c(0.025,0.975))*sqrt(p.hat*(1-p.hat)/30) [1] 0.03330801 0.30002532 The resulting confidence interval is still best approximated by the interval [0.063, 0.355].

The correct answer is: [0.063, 0.355] Question 4 Correct Mark 1.00 out of 1.00 Flag question Question text Assume that you are interested in testing H0: E(X) = 100 versus H1: E(X) ≥ 100 using the t-test.

Let the sample average, of a sample of size n = 17, be equal to x= 111.0 and the sample standard deviation be equal to s = 19.3. The null hypothesis is: Select one: a.

Not rejected with a significance level of 5%. b.

Rejected with a significance level of 5%, but not with a significance level of 1%. c.

Rejected with a significance level of 1%, but not with a significance level of 0.1%. d.

Rejected with a significance level of 0.1% Feedback

The statistic for testing the one-sided hypothesis is equal to (x - 100)/(s/√n) = (111.0 - 100)/ (19.3/√17) = 2.349957.

The thresholds for the one-sided test correspond to 95%-, 99%-, and 99.9%-percentiles of the tdistribution on 17-1=16 degrees of freedom. These percentiles can computed using the expression: qt(c(0.95,0.99,0.999),16) that produce 1.745884, 2.583487, and 3.686155, respectively, as output.

The statistic is larger than the smallest threshold but smaller that the threshold that is associated with a 1%-significance level. Therefore, the null hypothesis is rejected at the 5% confidence level but is not rejected at the 1%-significance level.

Alternatively, one may compute the p-value using the expression 1-pt((111.0-100)/ (19.3/sqrt(17)),16), which turns out to be equal to 0.01596924. This value is less than 0.05 and larger than 0.01. Hence, again, the null hypothesis is rejected at the 5% confidence level but is not rejected at the 1%-significance level.

The correct answer is: Rejected with a significance level of 5%, but not with a significance level of 1%.

Question 5 Correct Mark 1.00 out of 1.00 Flag question Question text The data set "cars.csv" contains information on various models of cars. The file can be found on the internet at this link. Two of the variables in this data set are "highway.mpg", the number of miles per gallon in highway driving conditions, and "fuel.type", the type of fuel the car is using.

Apply the t-test with "highway.mpg" as the response and "fuel.type" as the explanatory variable.

The conclusion of the default application of the test, with a significance level of 5%, is: Select one: a.

The expectation of the response for cars that use diesel is larger than the expectation for cars that use gas. b.

The expectation of the response for cars that use diesel is not equal to the expectation for cars that use gas. c.

The expectation of the response for cars that use diesel is smaller than the expectation for cars that use gas. d.

The expectation of the response for cars that use diesel is equal to the expectation for cars that use gas.

Feedback Run the code: > cars <- read.csv("cars.csv") > plot(highway.mpg~fuel.type,data=cars) > t.test(highway.mpg~fuel.type,data=cars)

Welch Two Sample t-test data: highway.mpg by fuel.type t = 2.2288, df = 21.438, p-value = 0.03665 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 0.3016753 8.5604869 sample estimates: mean in group diesel mean in group gas 34.75000 30.31892 We upload the data in the first line and plot it in the second. The function "t.test" is applied in order to test the null hypothesis that the difference between the expectations of the response in the two groups is equal to 0 against the two-sided alternative.

The computed p-value is 0.03665. Consequently, the null hypothesis is rejected with a significance level of 5%. The conclusion of the test is that the expectation of the response for cars that use diesel is not equal to the expectation for cars that use gas.

The correct answer is: The expectation of the response for cars that use diesel is not equal to the expectation for cars that use gas.

Question 6 Correct Mark 1.00 out of 1.00 Flag question Question text Consider data regarding a response y and an explanatory variable x, both numeric: Data x y 2.

1 0.7 8 6.

2 1.3 8 7. 3 0 1.1 4.

4 0.2 7 5.

5 1.1 0 5. 6 0 0.1 2.

7 1.7 9 3.

8 1.0 7 9 3. 0.8

0 1 2.

0.8 0 6 The null hypothesis that the expected value of the response is constant for all values of the explanatory variable is: Select one: a.

Not rejected with a significance level of 5%. b.

Rejected with a significance level of 5%, but not with a significance level of 1%. c.

Rejected with a significance level of 1%, but not with a significance level of 0.1%. d.

Rejected with a significance level of 0.1% Feedback Run the code: > x <- c(2.8,6.8,7.0,4.7,5.0,5.0,2.9,3.7,3.0,2.6) > y <- c(0.7,1.3,-1.1,0.2,1.1,-0.1,1.7,1.0,0.8,0.8) > summary(lm(y~x)) Call: lm(formula = y ~ x) Residuals: Min 1Q Median 3Q Max -1.1066 -0.3449 -0.2105 0.5127 1.2456 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 1.6798 0.6987 2.404 0.0429 * x -0.2390 0.1514 -1.579 0.1530 --Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.7386 on 8 degrees of freedom Multiple R-squared: 0.2376, Adjusted R-squared: 0.1423 F-statistic: 2.493 on 1 and 8 DF, p-value: 0.153 The first two lines of code produce the explanatory variable and the response, respectively. A linear regression analysis of the relation between the two variables is produced by the third line.

The situation when the expected value of the response is constant for all values of the explanatory variable corresponds to the case where the slope of the regression model is equal to zero. The p-value for testing that the slope is equal to zero is equal to 0.1530. This value is larger than 0.05. Consequently, the null hypothesis is not rejected with a significance level of 5%.

The correct answer is: Not rejected with a significance level of 5%.

Question 7 Correct

Mark 1.00 out of 1.00 Flag question Question text The data set "magnets.csv" contains the results of an experiment that examines the effectiveness of magnets as a pain treatment. This data was presented in Section 9.5 of the book. The file can be found on the internet at this link.

Consider the relation between the response "change" and the explanatory variable "active", the type of treatment that was used. Both variables belong to the data set "magnets.csv". Test the null hypothesis that the probability that the value of "change" is equal to zero is the same for both levels of the explanatory variable. The null hypothesis is: Select one: a.

Not rejected with a significance level of 5%. b.

Rejected with a significance level of 5%, but not with a significance level of 1%. c.

Rejected with a significance level of 1%, but not with a significance level of 0.1%. d.

Rejected with a significance level of 0.1% Feedback Run the code: > magnets <- read.csv("magnets.csv") > plot(factor(change==0)~active,data=magnets) > prop.test(table(magnets$active,magnets$change==0)) 2-sample test for equality of proportions with continuity correction data: table(magnets$active, magnets$change == 0) X-squared = 6.8966, df = 1, p-value = 0.008636 alternative hypothesis: two.sided 95 percent confidence interval: 0.09708026 0.67467672 sample estimates: prop 1 prop 2 0.8620690 0.4761905 The data is uploaded in the first line. The relation between the response and the explanatory variable is plotted in the second line and tested in the third.

The response is the indicator of no change and the explanatory variable is a factor with two levels. We apply the function "prop.test" in order to test the hypothesis that the probability of no change is the same for both levels. Recall that the input to the function is a 2×2 table of frequencies, with the response entering as the second variable in this table. The p-value for testing the null hypothesis that the two probabilities are identical (against the two-sided alternative) is 0.008636. This number is between 0.01 and 0.001. Therefore, the null hypothesis is rejected with a significance level of 1%, but not with a significance level of 0.1%.

The correct answer is: Rejected with a significance level of 1%, but not with a significance level of 0.1%.

Assignment unit 8 – quiz Started on Friday, 27 May 2022, 7:16 PM State Finished Completed on Friday, 27 May 2022, 7:51 PM Time taken 34 mins 41 secs Marks 8.00/8.00 Grade 100.00 out of 100.00 Information Flag question Information text In this assignment we analyze data that was collected as part of an experiment that investigated the mental relation between weapon and aggression. The experiment involved 32 undergraduate students ranging in age between 18 and 24 years. They were told that the purpose of this study was to test reading ability of various words. Each student was exposed to 192 trials. In each trial a computer presented a priming stimulus word for 1.25 seconds, a blank screen for 0.5 seconds, and then a target word. The experimenter instructed the subjects to read the first word to themselves and then to read the second word out loud as quickly as they could. The computer recorded reaction times.

The 192 trials were divided into 4 groups of size 48 each. In one group the priming stimulus word was a weapon word (such as "dagger" or "bullet") and the target word was an aggressive word (such as "destroy" or "wound"). In another group the target word was instead a nonaggressive word (such as "evacuate" or "ashamed"). The third and fourth groups involved a name of an animal (such as "rabbit" or "dog") as the priming stimulus word and the target, again, was either an aggressive or a nonaggressive word.

The file "weapons.csv" contains, for each subject and for each group of trials, the average of the 48 reaction times. The file can be found here. The file contains 5 variables: sex = The sex of the subject. aw = Aggressive target word following a weapon prime. an = Aggressive target word following a nonweapon (animal) prime. cw = Control target word (nonaggressive) following a weapon prime. cn = Control target word (nonaggressive) following a nonweapon prime.

(This data set is taken from the Rice Virtual Lab in Statistics. More details on this experiment can be found in the case study Weapons and Aggression that quotes as a source the paper: Anderson, C.A., Benjamin, A.J., & Bartholow, B.D. (1998). Does the gun pull the trigger?

Automatic priming effects of weapon pictures and weapon names. Psychological Science, 9, 308-314.)

Question 1 Correct Mark 1.00 out of 1.00 Flag question Question text Mark all the variables of type "factor": Select one or more: a.     

sex b. aw c. an d. cw e. cn Feedback The correct answer is: sex Question 2 Correct Mark 1.00 out of 1.00 Flag question Question text The total number of missing values is: Answer: Feedback The correct answer is: 0 Question 3 Correct Mark 1.00 out of 1.00 Flag question Question text Mark all the null hypotheses that are not rejected with a significance level of 5%: Select one or more: a.

The expected value of "an" is the same both for males and for females. b.

The expected value of "aw" is the same both for males and for females. c.

The expected value of "cn" is the same both for males and for females. d.

The expected value of "cw" is the same both for males and for females. e.

All of the given null hypotheses are rejected.

Feedback The correct answers are: The expected value of "an" is the same both for males and for females., The expected value of "aw" is the same both for males and for females., The expected value of "cn" is the same both for males and for females., The expected value of "cw" is the same both for males and for females.

Question 4 Correct Mark 1.00 out of 1.00 Flag question

Question text Mark all the null hypotheses that are not rejected with a significance level of 5%: Select one or more: a.

The variance of "an" is the same both for males and for females. b.

The variance of "aw" is the same both for males and for females. c.

The variance of "cn" is the same both for males and for females. d.

The variance of "cw" is the same both for males and for females. e.

All of the given null hypotheses are rejected.

Feedback The correct answers are: The variance of "aw" is the same both for males and for females., The variance of "cn" is the same both for males and for females., The variance of "cw" is the same both for males and for females.

Question 5 Correct Mark 1.00 out of 1.00 Flag question Question text One may measure the effect of a "weapon" primer on the speed in naming a word associated with aggression by considering the difference "aw - an". Call this difference "da". Likewise, one may measure the effect of a "weapon" primer on the speed in naming a neutral word by considering the difference "cw - cn". Call this other difference "dc". An overall contrast that may be used in order to test the relation between a "weapon" primer and a target associated with aggression is the difference "da - dc". Apply the t-test in test the null hypothesis that the expectation of the contrast is zero against a two-sided alternative. The resulting p-value is: (Give an answer of the form x.xxx, i.e. with 3 significance digits) Answer: Feedback The correct answer is: 0.03759 Question 6 Correct Mark 1.00 out of 1.00 Flag question Question text Construct a 95%-confidence for "aw - an" and a 95%-confidence interval for "cw - cn". The value 0 is contained in: Select one: a.

Only in the confidence interval for "aw - an". b.

Only in the confidence interval for "cw - cn". c.

Both confidence intervals. d.

Neither of the confidence intervals.

Feedback The correct answer is: Only in the confidence interval for "cw - cn".

Question 7 Correct Mark 1.00 out of 1.00 Flag question Question text The variable "aw" is the reaction time for a target word associated with aggression when the primer is a word associated with weapons. The variability of this variable between the subjects may be attributed in part to variability in reading ability among people. One may propose the variable "an" as measure of reading ability of words associated with aggression when the "weapon" effect is not present. Let "an" be an explanatory variable and let "aw" be the response.

The percentage of the variance that is explained by the explanatory variable is: Select one: a.

Less than 25%. b.

Between 25% and 50%. c.

Between 50% and 75%. d.

More than 75%.

Feedback The correct answer is: More than 75%.

Question 8 Correct Mark 1.00 out of 1.00 Flag question Question text Consider a regression model where "an" is the explanatory variable and "aw" is the response.

Consider the 95%-confidence intervals for the intercept and the slope. Then: Select one: a.

The value 0 belongs to the confidence interval for the intercept and the value 1 belongs to the confidence interval for the slope. b.

The value 0 belongs to the confidence interval for the intercept but the value 1 does not belong to the confidence interval for the slope. c.

The value 0 does not belongs to the confidence interval for the intercept but the value 1 belong to the confidence interval for the slope. d.

The value 0 does not belong to the confidence interval for the intercept and the value 1 does not belong to the confidence interval for the slope.

Feedback The correct answer is: The value 0 belongs to the confidence interval for the intercept and the value 1 belongs to the confidence interval for the slope.

Review quiz unit 9 Started on Friday, 27 May 2022, 7:33 PM State Finished Completed on Friday, 27 May 2022, 8:42 PM Time taken 1 hour 8 mins Marks 77.00/77.00 Grade 100.00 out of 100.00 Information Flag question Information text The data set "cars.csv" was introduced in Chapter 9 of the book. This data set contains information on various models of cars. Download this file to your computer and store it in the working directory of R. The file can be downloaded from MATH 1281 Data Files in the course page for MATH 1281.

The variable "width" in the data set contains the width of the body of the car in inches. The variable "length" contains the length of the body of the car in inches. Produce a new variable by the name "area" that contains the area of the body of the car. This variable should be produced by multiplying the the width of the body by the length of the body. Answer the following questions with respect to the variable "area": Question 1 Correct Mark 1.00 out of 1.00 Flag question Question text The variable "area" is a: Select one: a.

Numeric b.

Factor Feedback Upload the file "cars.csv" to a data set by the name "cars" using the code: cars <read.csv("cars.csv"). Form the variable "area" using the code: area <- cars$length*cars$width.

The variable "area" is a product of two numeric variables and it obtains numeric values as well.

Question 2 Correct Mark 1.00 out of 1.00 Flag question Question text The average area of the body of a car is:

Answer: 11493.36 Feedback Apply the code: mean(area). The sample average turn out to be equal to 11493.36. Alternatively, you may apply the code: summary(area) and obtain the value of the mean from the produced report.

Question 3 Correct Mark 1.00 out of 1.00 Flag question Question text The median area of the body of a car is: Answer: 11314.2 Feedback Apply the code: median(area). The sample median turn out to be equal to 11314.2. Alternatively, you may apply the code: summary(area) and obtain the value of the median from the produced report.

Question 4 Correct Mark 1.00 out of 1.00 Flag question Question text The interquartile range of the area is: Answer: 1566.18 Feedback Apply the code: quantile(area,c(0.25,0.75)). The first quartile (Q1) is equal to 10709.72 and the third quartile (Q3) is equal to 12275.90. The difference between Q3 and Q1 is the interquartile range and is equal to IQR = Q3 - Q1 = 1566.18.

Alternatively, you may apply the code: summary(area) and obtain the values of Q1 and Q3 from the produced report.

Question 5 Correct Mark 1.00 out of 1.00 Flag question Question text The number of outlier observations in the variable "area" is: Select one: a.

0 b.

1 c.

2 d.

3 e.

More than 3 Feedback You may produce the box-plot of the variable using the code: boxplot(area). One point in the upper part of the plot represents the outliers. In order to make sure that this point represents a single data point you may produce the table with the code: table(area). The frequency of the largest value is 1.

An alternative method for counting the number of data points above the threshold Q3 + 1.5 (Q3Q1) may use the code: sum(12275.9 + 1.5*1566.18 < area). The count is 1.

Question 6 Correct Mark 1.00 out of 1.00 Flag question Question text The number of missing observations in the variable "area" is: Select one: a.

0 b.

1 c.

2 d.

3 Feedback There are no missing values in this variable.

Information Flag question Information text A rectangular body is square if the length is equal to the width. Consider, for each car, the difference between the length and the width of the body. Call this variable X. If cars tend to be square then we would expect the expectation of this variable to be equal to 0.

In Chapter 12 we will present a statistical test for testing that the expectation of a variable is equal to 0. The test statistic for this problem is given by: T = X/[S/ √n], where X is the sample average, S is the sample standard deviation, and n is the sample size.

In the following questions you are required to determine the region that contains 95% of the values of the test statistic if indeed cars tend to be square in their shape and to check whether or not the evaluation of the test statistic for the "cars" data set falls within this region.

Question 7 Correct Mark 1.00 out of 1.00 Flag question Question text Let X1, X2, ..., X205 be a sample of Normal random variables with expectation μ=0 and variance σ2=100. (Namely, standard deviation of σ=10.) The region that contains 95% of the distribution of the statistic T = X/[S/ √205] is best described by:

Select one: a.

[-19.7,19.7] b.

[88.3,127.7] c.

[-1.97,1.97] d.

[106,110] Feedback The interval that contains 95% of the sampling distribution of the test statistic may be identified via simulations. Run the code: mu <- 0 sig <- 10 test.stat <- rep(0,10^5) for(i in 1:10^5) { X <- rnorm(205,mu,sig) test.stat[i] <- (mean(X))/sqrt(var(X)/205) } quantile(test.stat,c(0.025,0.975)) 2.5% 97.5% -1.958288 1.978535 The result is best described by [-1.97,1.97] Question 8 Correct Mark 1.00 out of 1.00 Flag question Question text The standard deviation of the statistic T is Answer: 1 Feedback Run the code: mu <- 0 sig <- 10 test.stat <- rep(0,10^5) for(i in 1:10^5) { X <- rnorm(205,mu,sig) test.stat[i] <- (mean(X))/sqrt(var(X)/205) } sd(test.stat) [1] 1.000478 The standard deviation is approximately equal to 1.

Question 9 Correct Mark 1.00 out of 1.00 Flag question Question text

Produce a variable by the name "x" that contains the difference between the length and the width of a car and compute the value of the test statistic for this variable. (Hint: Use the code "x <cars$length - cars$width" in order to produce "x" and apply the formula for the computation of the statistic to "x".) The value of the statistic for the observed data set "cars" is Answer: 146.1173 Feedback Produce the variable with the code: x <- cars$length - cars$width. The test statistic may be evaluated for the data using the code: mean(x)/sqrt(var(x)/205) The value of the test statistic is 146.1173.

Question 10 Correct Mark 1.00 out of 1.00 Flag question Question text The value of the test statistic for the data set "cars" Select one: a. is inside the region that contains 95% of the sampling distribution of the statistics when cars are square b. is outside of the region that contains 95% of the sampling distribution of the statistics when cars are square Feedback The region is [-1.97,1.97]. The value 146.1173 is outside the region.

Information Flag question Information text In the book we compared two different estimators of the variance to each other. One estimator was the sample variance S2, an unbiased estimator of the variance. This estimator was computed using the function "var", which involves dividing the sum of squared deviations from the average by the number of observations minus 1 (n-1). The alternative estimator was obtained by dividing the sum of squares by the number of observations ( n ). This alternative estimator was equal to [(n-1)/n]S2.

In this exercise we compare two estimators of the standard deviation that are derived from the estimators of the variance. The first is S, the square root of S2, and the second is the square root of the alternative estimator of the variance. The first estimator of the standard deviation may be computed with the function "sd". The alternative estimator may be computed by the multiplying the first estimator by √ [(n-1)/n].

We consider the bias, variance and mean square error of both estimators of the standard deviation. We assume a Normal sample of n=20 observations. The expectation of an observation is μ=5 and the variance is σ2 = 3. The next 5 questions refer to the following R code that simulates the sampling distribution of the two estimators:

Question 11 Correct Mark 1.00 out of 1.00 Flag question Question text In the context of the simulation, the value of the parameter that is being estimated is: (Choose the number closest to the answer) Select one: a.

1.71 b.

3.00 c.

1.73 d.

1.69 Feedback The variance of the measurement is equal to 3. The standard deviation, the value that is being estimated, is equal to √ 3 = 1.732051. After rounding up we obtain 1.73 as the solution.

Question 12 Correct Mark 1.00 out of 1.00 Flag question Question text The bias of the first estimator of the standard deviation is: (Choose the number closest to the answer) Select one: a.

1.71 b.

-0.023 c.

1.73 d.

0.00 Feedback The bias of the first estimator is the difference between its expectation, 1.708774, and the value of the estimated parameter, √ 3. The difference is equal to -0.02327681 ≅ -0.023.

Question 13 Correct Mark 1.00 out of 1.00 Flag question Question text The bias of the alternative estimator of the standard deviation is: (Choose the number closest to the answer) Select one: a.

1.71 b.

-0.023 c.

1.73 d.

-0.067 Feedback The bias of the second estimator is the difference between its expectation, 1.665507, and the value of the estimated parameter, √ 3. The difference is equal to -0.06654381 ≅ -0.067.

Question 14 Correct Mark 1.00 out of 1.00 Flag question Question text The mean squared error (MSE) of the first estimator of the standard deviation is: (Choose the number closest to the answer) Select one: a.

0.0792 b.

0.0780 c.

-0.023 d.

0.00 Feedback

The mean square error (MSE) of the first estimator is equal to the sum of the variance of the estimator and the square of the bias. Run the code: > 0.07870757 + (1.708774-sqrt(3))^2 [1] 0.07924938 After rounding up we get 0.0792.

Question 15 Correct Mark 1.00 out of 1.00 Flag question Question text The mean squared error (MSE) of the second estimator of the standard deviation is: (Choose the number closest to the answer) Select one: a.

0.0792 b.

0.0780 c.

-0.023 d.

0.00 Feedback The mean square error (MSE) of the first estimator is equal to the sum of the variance of the estimator and the square of the bias. Run the code: > 0.07477219 + (1.665507-sqrt(3))^2 [1] 0.07920027 After rounding up we get 0.0792.

Information Flag question Information text In the book it was proposed to use 1/X (one over the sample average) as an estimator of the rate λ of the Exponential distribution. In the following 3 questions you are requested to compute the bias, the variance and the mean square error (MSE) of the given estimator for an Exponential sample of size n=25 observations and when λ=3. You may carry out these computations by conducting a simulation in R: Question 16 Correct Mark 1.00 out of 1.00 Flag question Question text The expectation of the estimator is: (Give an answer of the form x.xxx, i.e. with 3 significance digits) Answer: 3.126 Feedback Run the simulation: > lam <- 3 > lam.hat <- rep(0,10^5) > for(i in 1:10^5) +{ + X <- rexp(25,lam) + lam.hat[i] <- 1/mean(X)

+} > mean(lam.hat) [1] 3.126287 . The expectation of the estimator is approximately equal to 3.126.

Question 17 Correct Mark 1.00 out of 1.00 Flag question Question text The mean square error (MSE) of the estimator is: (Give an answer of the form x.xxx, i.e. with 3 significance digits) Answer: 0.439 Feedback Run the simulation: > lam <- 3 > lam.hat <- rep(0,10^5) > for(i in 1:10^5) +{ + X <- rexp(25,lam) + lam.hat[i] <- 1/mean(X) +} > var(lam.hat)+(mean(lam.hat)-3)^2 [1] 0.4386786 . The MSE of the estimator is approximately equal to 0.439.

Question 18 Correct Mark 1.00 out of 1.00 Flag question Question text The estimator is an unbiased estimator: Select one: a.

TRUE b.

FALSE Feedback The estimator is biased since its expectation, 3.126, is not equal to the value of the parameter, 3.000, that it tries to estimate.

Information Flag question Information text The next 2 questions refer to the following information: The variable "num.of.doors" in the data set "cars" counts the number of cars in the sample that have four doors and the number that have two doors. In the given data set the number of cars with four doors is 114 and the number of cars with two doors is 89. Two observations have a missing value for this variable. Denote by p the probability that a car has two doors.

Question 19 Correct Mark 1.00 out of 1.00

Flag question Question text The estimated value of p, ignoring missing values, is: (Give an answer of the form x.xxx with 3 significance digits) Answer: 0.4384236 Feedback The estimator of a probability of an event is the relative frequency of the event in the sample.

The event is "having 2 doors". This occurs in the sample 89, out of 89+114=203 times. The relative frequency is 89/203=0.4384236 Question 20 Correct Mark 1.00 out of 1.00 Flag question Question text If the value of the parameter p is 0.45 then the standard deviation of the estimator of the parameter is: (Consider, as before, the sample size without the missing observations and give an answer of the form x.xxx with 3 significance digits) Answer: 0.0349 Feedback The variance of the estimator is p(1-p)/n, where n=203 is the sample size. The standard deviation is the square root of the variance. If p=0.45 then the variance is 0.45*0.55/203 = 0.001219212 and the standard deviation is 0.03491721.

Information Flag question Information text A case study that deals with the effect of prior reputation on the evaluation of an instructor is presented in Question 11.1 in the book (Section 11.5). Students were exposed to either one of two conditions, C and P, and were asked to rate a lecture given by the instructor. It was found that the average rating among the 25 students that were exposed to condition C was 2.613 and the average rating among the 24 students that were exposed to condition P was 2.236. The sample standard deviations were 0.533 and 0.543, respectively.

Question 21 Correct Mark 1.00 out of 1.00 Flag question Question text You want to construct a confidence interval for the expected rating of students that were exposed to condition P. The confidence level of the interval should be 85%. The quantile of the tdistribution that should be used for the construction is: (Assume a Normal distribution for the rating) Select one: a. qt(0.85,24) b. qt(0.925,24) c.

qt(0.85,23) d. qt(0.925,23) Feedback The central part should include 85% of the distribution. The tails should include 15%. Half of this probability, or 7.5%, goes to the upper tail and half to the lower tail. The probability above the quantile is 0.075. Therefore, the probability below the quantile should be 0.925.

The number of degrees of freedom is the number of observations minus 1. The number of observations is 24. Hence, the number of degrees of freedom is 23.

Combining the two arguments we get that quantile should be qt(0.925,23).

Question 22 Correct Mark 1.00 out of 1.00 Flag question Question text The confidence interval with confidence level of 85% for the expected rating of students that were exposed to condition P is: (Assume a Normal distribution for the rating. Choose the interval closest to the answer) Select one: a.

[2.07, 2.40] b. [ 2.12, 2.35] c.

[2.50, 2.73] d.

[2.59, 2.64] Feedback The sample average is 2.236 and the sample standard deviation is 0.543. The estimated standard deviation of the sample average is 0.543/√24, where 24 is the sample size. We apply the formula for computing a confidence interval for the expectation of a Normal measurement: 2.236 + qt(c(0.075,0.925),23)*0.543/sqrt(24) . The output of the computation is: [2.070929, 2.401071], which after rounding up gives [2.07,2.40].

Question 23 Correct Mark 1.00 out of 1.00 Flag question Question text You want to construct a confidence interval for the variance of the rating of students that were exposed to condition P. The confidence level of the interval should be 80%. A quantile of the chisquare distribution that should be used for the construction is: (Assume a Normal distribution for the rating) Select one: a. qchisq(0.85,23) b.

qchisq(0.10,23) c. qchisq(0.90,24) d. qchisq(0.95,23) Feedback The central part should contain 80% of the distribution. The tails should contain 20%. Half of this probability, or 10%, goes to the upper tail and other 10% goes to the lower tail. It follows that the quantiles of interest are those associated with 10% and with 90%.

The number of degrees of freedom is the number of observations minus 1. The number of observations is 24. Hence, the number of degrees of freedom is 23.

Combining the two arguments we get that quantile should be considered are qchisq(0.10,23) and qchisq(0.90,23). Only the former appears among the options.

Question 24 Correct Mark 1.00 out of 1.00 Flag question Question text The confidence interval with confidence level of 80% for the variance of the rating of students that were exposed to condition P is: (Assume a Normal distribution for the rating. Choose the interval closest to the answer) Select one: a.

[0.212, 0.457] b.

[0.390, 0.841] c.

[0.239, 0.395] d.

[0.439, 0.727] Feedback The sample standard deviation is 0.543 and the sample variance is the square of this number. We apply the formula for computing a confidence interval for the variance of a Normal measurement: 23/qchisq(c(0.90,0.10),23)*0.543^2. The output of the computation is: [0.2118770, 0.4567314], which after rounding up gives [0.212, 0.457].

Information Flag question Information text Confidence intervals are computed for the expectation and the variance of a measurement on the basis of 20 observations. The confidence interval are constructed under the assumption that the measurement has a Normal distribution. In reality, the distribution of the observations is the Uniform(0.5,5.5) distribution. Recall that the expectation of the measurement is 3 and the variance is 2.083333. Use simulations to answer the following questions: Question 25 Correct Mark 1.00 out of 1.00

Flag question Question text Consider a confidence interval for the expectation with a nominal confidence level of 95%. The actual confidence level is: (Select the closest number.)

Select one: a.

0.93 b.

0.95 c.

0.97 d.

0.99 Feedback Run the simulation: > a <- 0.5 > b <- 5.5 > n <- 20 > X.bar <- rep(0,10^5) > S <- rep(0,10^5) > for(i in 1:10^5) +{ + X <- runif(n,a,b) + X.bar[i] <- mean(X) + S[i] <- sd(X) +} > mu.LCL <- X.bar - qt(0.975,n-1)*S/sqrt( n ) > mu.UCL <- X.bar + qt(0.975,n-1)*S/sqrt( n ) > mean((3 >= mu.LCL) & (3 <= mu.UCL)) [1] 0.94833 The closest answer is 0.95.

Question 26 Correct Mark 1.00 out of 1.00 Flag question Question text Consider a confidence interval for the variance with a nominal confidence level of 95%. The actual confidence level is: (Select the closest number.)

Select one: a.

0.93 b.

0.95 c.

0.97 d.

0.99 Feedback

Run the simulation: > a <- 0.5 > b <- 5.5 > n <- 20 > X.bar <- rep(0,10^5) > S <- rep(0,10^5) > for(i in 1:10^5) +{ + X <- runif(n,a,b) + X.bar[i] <- mean(X) + S[i] <- sd(X) +} > var.LCL <- (n-1)*S^2/qchisq(0.975,n-1) > var.UCL <- (n-1)*S^2/qchisq(0.025,n-1) > mean((2.083333 >= var.LCL) & (2.083333 <= var.UCL)) [1] 0.99603 The closest answer is 0.99.

Information Flag question Information text A poll to determine the popularity of a candidate for election was conducted among 500 people.

It was found that 25 of those people are not eligible for voting, and among those that are eligible to vote, 207 support the candidate. Let p be the probability that an eligible voter supports the candidate.

Question 27 Correct Mark 1.00 out of 1.00 Flag question Question text How many people eligible for voting should have been included in the survey in order to assure that the radius of a 95% confidence interval for p is no more than 0.01: (Choose the smallest sufficient number among the given choices.)

Select one: a.

9,100 b.

9,300 c.

9,500 d.

9,700 Feedback The radius of the confidence interval is bounded by qnorm(0.975)/(2√n). Therefore, , in order to assure a radius of 0.01, the sample size should be at least (qnorm(0.975)/0.02)^2 This expression produces 9603.647. We get that 9,700 is the smallest of the options still larger than the minimal required sample size.

Question 28 Correct

Mark 1.00 out of 1.00 Flag question Question text A 95% confidence interval for p is: (Choose the interval closest to the answer) Select one: a.

[0.391, 0.480] b.

[0.520, 0.609] c.

[0.527, 0.602] d.

[0.400, 0.473] Feedback We use the formula for the computation of a confidence interval for a proportion and run the code: p.hat+qnorm(c(0.025,0.975))*sqrt(p.hat*(1-p.hat)/475). The resulting confidence interval is [0.3911971, 0.4803818]. After rounding up we get the interval [0.391, 0.480].

Question 29 Correct Mark 1.00 out of 1.00 Flag question Question text The estimated value of p is: (Give an answer of the form x.xxx with 3 significance digits) Answer: 0.436 Feedback The number of eligible voters is 500-25=475. Of them, 207 support the candidate. Therefore, the sample proportion is 207/475=0.4357895 Information Flag question Information text A rectangular body is square if the length is equal to the width. In the quizzes of Unit 1 the possibility of testing that cars are square on the basis of the information in the data set "cars.csv" was explored. In this exercise we re-do the test using the tools that were introduced in this unit.

Specifically, you are asked to apply the function "t.test" to the difference between the length and the width of cars. (Hint: Use the code "x <- cars$length - cars$width" in order to produce "x" and apply the function to "x".)

Question 30 Correct Mark 1.00 out of 1.00 Flag question Question text The (default) null hypothesis in the application of the function "t.test" to the variable "x" corresponds to the statement: Select one: a.

The length of the car is equal to the width of the car.

b.

The length of the car is not equal to the width of the car. c.

The expected difference between the length and the width of the car is equal to 0. d.

The expected difference between the length and the width of the car is not equal to 0.

Feedback Hypothesis are formulated in terms of parameters of the distribution and not in terms of the observations themselves. The default hypothesis of the function "t.test" is "mu = 0", which corresponds to the expectation of the measurement being equal to zero. In the current setting the measurement is the difference between the length and the width. The default null is that the expectation of this difference is equal to zero.

Question 31 Correct Mark 1.00 out of 1.00 Flag question Question text The null hypothesis is: Select one: a.

Not rejected with a significance level of 5%. b.

Rejected with a significance level of 5%, but not with a significance level of 1%. c.

Rejected with a significance level of 1%, but not with a significance level of 0.1%. d.

Rejected with a significance level of 0.1% Feedback When we run the code: x <- cars$length - cars$width we get that the p-value is smaller than 2.2 x 10-16. In particular, the p-value is less than 0.1%. Therefore, the null hypothesis is rejected at the the 0.1% significance level.

Information Flag question Information text It is assumed, when constructing the t-test, that the measurements are Normally distributed. In this exercise we examine the robustness of the test to divergence from the assumption. You are required to compute the significance level of a one-sided t-test of H0: E(X) ≤ 5 versus H1: E(X) > 5. Assume there are n = 15 observations and use a t-test with a nominal 5% significance level.

Question 32 Correct Mark 1.00 out of 1.00 Flag question Question text Consider the test statistic T = [X-5]/[S/ √n], where X is the sample average, S2 is the sample variance and n is the sample size. The structure of the rejection region of the test is: Select one:

a. {|T| > c} b. {T > c} c. {T < c} d. {T ≠ c} Feedback The alternative hypothesis corresponds to expectations larger than 5. The T statistic measured the difference between the sample average and the value 5. Larger values of T are an indication for the validity of the alternative hypothesis and an evidence against the validity of the null hypothesis. Therefore, we reject the null hypothesis in favor of the alternative hypothesis for values of the T statistic that are larger than a threshold.

Question 33 Correct Mark 1.00 out of 1.00 Flag question Question text In order to obtain a significance level of 5% one should set the threshold c to be equal to: (Give an answer of the form x.xxx with 3 significance digits) Answer: 1.76131 Feedback If the measurement has a Normal distribution then the distribution of the T statistic is the tdistribution on 15-1=14 degrees of freedom. The rejection region involves the upper right tail of this distribution. Consequently, the threshold is the 95-percentile of the t-distribution, which can be computed using the expression: qt(0.95,14). It turns out that the threshold is equal to 1.76131.

Question 34 Correct Mark 1.00 out of 1.00 Flag question Question text Assume that the distribution of the observations is Poisson(5). Then the actual significance level of the test is (Use simulations to compute the significance level. Choose the closest answer.): Select one: a.

0.02 b.

0.04 c.

0.06 d.

0.08 Feedback We may use simulations in order to compute the actual significance level of the test. Run the code: > lam <- 5

> n <- 15 > X.bar <- rep(0,10^5) > S <- rep(0,10^5) > for(i in 1:10^5) +{ + X <- rpois(n,lam) + X.bar[i] <- mean(X) + S[i] <- sd(X) +} > T <- (X.bar - 5)/(S/sqrt( n )) > mean(T > qt(0.95,n-1)) [1] 0.03954 The option 0.04 is the closest to the significance level that was obtained from the simulation.

Information Flag question Information text Assume that you are interested in testing H0: E(X) = 1 versus H1: E(X) ≠ 1 using the t-test. Let the sample average, of a sample of size n = 64, be equal to x = 0.89 and the sample standard deviation be equal to s = 3.3.

Question 35 Correct Mark 1.00 out of 1.00 Flag question Question text The absolute value of the test statistic is equal to: (Give an answer of the form x.xxx with 3 significance digits) Answer: 0.267 Feedback The T statistic is the difference between the sample average and the null expectation, divided by the estimated standard deviation of the sample average. For the data provided in the question the statistic is (0.89 - 1)/(3.3/sqrt(64)) and is equal to -0.2666667. The absolute value of the statistic is 0.2666667.

Question 36 Correct Mark 1.00 out of 1.00 Flag question Question text The null hypothesis is: Select one: a.

Not rejected with a significance level of 5%. b.

Rejected with a significance level of 5%, but not with a significance level of 1%. c.

Rejected with a significance level of 1%, but not with a significance level of 0.1%. d.

Rejected with a significance level of 0.1% Feedback The thresholds for the two-sided test correspond to 97.5%-, 99.5%-, and 99.95%-percentiles of the t-distribution on 64-1=63 degrees of freedom. These percentiles can computed using the expression: qt(c(0.975,0.995,0.9995),63) that produce 1.998341, 2.656145, and 3.451769, respectively, as output. The statistic is less than the smallest of the thresholds. Therefore, the null hypothesis is not rejected at the 5% confidence level.

Alternatively, one may compute the p-value using the expression 2*(1-pt(0.2666667,63)), which turns out to be equal to 0.7905967. This value is larger than 0.05. Hence, again the null hypothesis is not rejected at the 5% confidence level.

Question 37 Correct Mark 1.00 out of 1.00 Flag question Question text The p-value for testing the one-sided hypothesis H0: E(X) ≤ 1 versus H1: E(X) > 1 using the ttest is equal to: (Give an answer of the form x.xxx with 3 significance digits) Answer: 0.605 Feedback For the given one-sided alternative the null hypothesis is rejected when the statistic is larger than a threshold. In the computation of a p-value we use the observed value of the statistic as the threshold. This observed value is equal to -0.2666667. The probability that a T statistic is larger then the given threshold may be computed with the expression: 1-pt(-0.2666667,63). The outcome is 0.6047016, which is the p-value.

Information Flag question Information text Assume that you are interested in testing that the probability of an event is H0: p = 0.25 versus H1: p ≠ 0.25 using the test for proportions. Let the proportion of the event in a sample of size n = 240 be equal to 0.275.

Question 38 Correct Mark 1.00 out of 1.00 Flag question Question text The p-value for testing the given hypotheses is equal to: (Use the function "prop.test" with a continuity correction. Give an answer of the form x.xxx with 3 significance digits) Answer: 0.412 Feedback The number of "successes" in the Binomial distribution is k = 0.275 x 240 = 66. We apply the test for proportion to this data in order to test the null hypothesis that p = 0.25: > prop.test(66,240,p=0.25) 1-sample proportions test with continuity correction

data: 66 out of 240, null probability 0.25 X-squared = 0.6722, df = 1, p-value = 0.4123 alternative hypothesis: true p is not equal to 0.25 95 percent confidence interval: 0.2204682 0.3368799 sample estimates: p 0.275 The p-value is 0.4123.

Question 39 Correct Mark 1.00 out of 1.00 Flag question Question text The null hypothesis is: Select one: a.

Not rejected with a significance level of 5%. b.

Rejected with a significance level of 5%, but not with a significance level of 1%. c.

Rejected with a significance level of 1%, but not with a significance level of 0.1%. d.

Rejected with a significance level of 0.1% Feedback The p-value is equal to 0.4123, which is larger than 0.05. As a result, the null hypothesis is not rejected at the 5% significance level.

Information Flag question Information text In Question 11.1 of the Solved Exercises of Chapter 11 we introduced an experiment aimed at assessing the effect of rumors and prior reputation of the instructor on the evaluation of the instructor by her students. The data produced by the experiment are stored in the file "teacher.csv". The file can be found on the internet at this link. Use this data in order to answer the next 3 questions: Question 40 Correct Mark 1.00 out of 1.00 Flag question Question text The null hypothesis that states that there in no significant difference between the variance of the rating in the two conditions, "C" and "P" is: Select one: a.

Not rejected with a significance level of 5%. b.

Rejected with a significance level of 5%, but not with a significance level of 1%. c.

Rejected with a significance level of 1%, but not with a significance level of 0.1%. d.

Rejected with a significance level of 0.1% Feedback Run the code: > teacher <- read.csv("teacher.csv") > plot(rating~condition,data=teacher) > var.test(rating~condition,data=teacher) F test to compare two variances data: rating by condition F = 0.9646, num df = 24, denom df = 23, p-value = 0.9288 alternative hypothesis: true ratio of variances is not equal to 1 95 percent confidence interval: 0.4196038 2.2013532 sample estimates: ratio of variances 0.96463 We upload the data in the first line and plot it in the second. The function "var.test" is applied in order to test the null hypothesis that the ratio is equal to 1 (hence, the difference equal to 0). The computed p-value is 0.9288. Consequently, the null hypothesis is not rejected at significance level of 5%.

Question 41 Correct Mark 1.00 out of 1.00 Flag question Question text The conclusion of the test, with a significance level of 5%, is: Select one: a.

The sample variances in the two sub-samples are equal to each other. b.

The variance of rating measure is the same under both conditions. c.

The sample variances in the two sub-samples are not equal to each other. d.

The variance of rating measure are not the same under both conditions.

Feedback The null hypothesis states that the variance of the measurement "rating" is the same under both conditions. The null hypothesis was not rejected at the significance level of 5%. Therefore, the conclusion is that the variance of rating measure is the same under both conditions.

Question 42 Correct Mark 1.00 out of 1.00 Flag question

Question text The value 1: Select one: a.

Is contained in a 95% confidence interval for the ratio between the variance of the rating in the two conditions. b.

Is not contained in a 95% confidence interval for the ratio between the variance of the rating in the two conditions.

Feedback Run the code: > teacher <- read.csv("teacher.csv") > plot(rating~condition,data=teacher) > var.test(rating~condition,data=teacher) F test to compare two variances data: rating by condition F = 0.9646, num df = 24, denom df = 23, p-value = 0.9288 alternative hypothesis: true ratio of variances is not equal to 1 95 percent confidence interval: 0.4196038 2.2013532 sample estimates: ratio of variances 0.96463 The confidence interval for the ratio of the variances is [0.4196038, 2.2013532]. This interval contains the value 1.

Information Flag question Information text We are interested in constructing a confidence interval for the ratio of the variance of the measurement in sub-population "a" divided by the variance of the measurement in subpopulation "b". It is assumed in the construction that the measurements are Normally distributed.

In the next 4 questions you are required to examine the robustness of confidence level to divergence from the assumption.

Assume there are na=29 observations in one group and nb = 21 observations in the other group.

Consider a confidence interval with a nominal 95% confidence level.

Question 43 Correct Mark 1.00 out of 1.00 Flag question Question text If the distribution of the measurement in sub-population "a" is Normal(4,16) and the distribution of the measurement in sub-population "b" is Normal(2,4) then the actual value of the estimated parameter is: Select one: a.

-2

b.

0.25 c.

2 d.

4 Feedback The variance in population "a" is 16 and the variance in population "b" is 4. The ratio between the two is 16/4=4.

Question 44 Correct Mark 1.00 out of 1.00 Flag question Question text If the distribution of the measurement in sub-population "a" is Normal(4,16) and the distribution of the measurement in sub-population "b" is Normal(2,4) then the actual confidence level of the confidence interval is: (Use simulations to compute the significance level. Choose the closest answer.)

Select one: a.

0.91 b.

0.93 c.

0.95 d.

0.97 Feedback Run the code: > mu.a <- 4 > sig.a <- 4 > mu.b <- 2 > sig.b <- 2 > n.a <- 29 > n.b <- 21 > S.a <- rep(0,10^5) > S.b <- rep(0,10^5) > for(i in 1:10^5) +{ + X.a <- rnorm(n.a,mu.a,sig.a) + X.b <- rnorm(n.b,mu.b,sig.b) + S.a[i] <- sd(X.a) + S.b[i] <- sd(X.b) +} > F <- S.a^2/S.b^2 > mean((F/qf(0.025,n.a-1,n.b-1) >= 4)&(F/qf(0.975,n.a-1,n.b-1) <= 4)) [1] 0.94985 The first 6 lines set the values of the parameters and the following two lines create

the objects for storage. The "for" loop produces the sampling distribution of both estimators of standard deviations. The distribution of the ratio of estimators of the variances is kept in "F". The last line computes the coverage probability of the confidence interval. We obtain that the confidence level is practically equal to 0.95.

Question 45 Correct Mark 1.00 out of 1.00 Flag question Question text If the distribution of the measurement in sub-population "a" is Exponential(1/4) and the distribution of the measurement in sub-population "b" is Exponential(1/2) then the actual value of the estimated parameter is: Select one: a.

-2 b.

0.25 c.

2 d.

4 Feedback The variance of the measurement in population "a" is 42 = 16. The variance of the measurement in population "b" is 22 = 4. The ratio between the variances is 16/4 = 4.

Question 46 Correct Mark 1.00 out of 1.00 Flag question Question text If the distribution of the measurement in sub-population "a" is Exponential(1/4) and the distribution of the measurement in sub-population "b" is Exponential(1/2) then the actual confidence level of the confidence interval is: (Use simulations to compute the significance level.

Choose the closest answer.)

Select one: a.

0.70 b.

0.80 c.

0.90 d.

0.95 Feedback Run the code: > lam.a <- 1/4 > lam.b <- 1/2 > n.a <- 29

> n.b <- 21 > S.a <- rep(0,10^5) > S.b <- rep(0,10^5) > for(i in 1:10^5) +{ + X.a <- rexp(n.a,lam.a) + X.b <- rexp(n.b,lam.b) + S.a[i] <- sd(X.a) + S.b[i] <- sd(X.b) +} > F <- S.a^2/S.b^2 > mean((F/qf(0.025,n.a-1,n.b-1) >= 4)&(F/qf(0.975,n.a-1,n.b-1) <= 4)) [1] 0.72551 The first 4 lines set the values of the parameters and the following two lines create the objects for storage. The "for" loop produces the sampling distribution of both estimators of standard deviations. The distribution of the ratio of estimators of the variances is kept in "F". The last line computes the coverage probability of the confidence interval. We obtain that the confidence level is closest to the value 0.70.

Information Flag question Information text The sample average in one sub-sample is xa = 30.7 and the sample standard deviation is sa = 13.4.

The sample average in the second sub- sample is xb = 32.3 and the sample standard deviation is sb = 16.7. The size of the first sub-sample is na = 57 and the size of the second sub-sample is nb = 77. We are interested in testing that the expectations in the associated sub-populations are equal to each other. The following 3 questions refer to the given information.

Question 47 Correct Mark 1.00 out of 1.00 Flag question Question text The absolute value of the test statistic is equal to: (Give an answer of the form x.xxx with 3 significance digits) Answer: 0.615 Feedback Run the code: > n.a <- 57 > n.b <- 77 > x.bar.a <- 30.7 > x.bar.b <- 32.3 > s.a <- 13.4 > s.b <- 16.7 > t <- abs(x.bar.a-x.bar.b)/sqrt(s.a^2/n.a + s.b^2/n.b) >t [1] 0.6148335 The absolute value of the statistic is 0.6148335.

Question 48 Correct

Mark 1.00 out of 1.00 Flag question Question text The p-value for testing the hypothesis is equal to: (Use the Normal approximation. Choose the closest answer.)

Select one: a.

0.5 b.

0.05 c.

0.005 d.

0.0005 Feedback Run the code: > 2*(1-pnorm(0.6148335)) [1] 0.5386647 The p-value for the two-sided test is equal to twice the upper tail. The (absolute) value of the statistic is 0.6148335. The code computes Normal probability above that value, multiplied by 2. The value closest to the computed p-value is 0.5.

Question 49 Correct Mark 1.00 out of 1.00 Flag question Question text The null hypothesis is: Select one: a.

Not rejected with a significance level of 5%. b.

Rejected with a significance level of 5%, but not with a significance level of 1%. c.

Rejected with a significance level of 1%, but not with a significance level of 0.1%. d.

Rejected with a significance level of 0.1% Feedback The p-value is larger than 0.05. Therefore, the null hypothesis is not rejected with a significance level of 5%.

Information Flag question Information text In Figure 1 you may find a scatter plot with 10 points and 3 lines. Three of the points are marked by the letters a, b, and c. The lines are marked by the numbers 1, 2, and 3. The following 2 questions refer to this scatter plot.

Question 50 Correct Mark 1.00 out of 1.00 Flag question Question text Match the coefficients to the lines: Line 1 Intercept = 3, Slope = 2 Line 2 Intercept = 1.5, Slope = 0 Line 3 Intercept = -1.5, Slope = 2 Feedback Line 2 has a zero slope. Lines 1 and 3 have the same slope, but the intercept of line 1 is larger. It follows that: Line 1: Intercept = 3, Slope = 2.

Line 2: Intercept = 1.5, Slope = 0.

Line 3: Intercept = -1.5, Slope = 2.

Question 51 Correct Mark 1.00 out of 1.00 Flag question Question text Match the points: Point a X = -0.1, y = 3.8 Point b X = 1.6, y = -1.6 Point c X = -0.6, y = 0.8 Feedback

Point a has the largest y value. Point b has the largest x value and point c has the smallest x value. Therefore: Point a => x = -0.1, y = 3.8 Point b => x = 1.6, y = -1.6 Point c => x = -0.6, y = 0.8 Information Flag question Information text Fit a regression model to the relation between "horsepower" as a response and "peak.rpm" as an explanatory variable. Both variables belong to the data set "cars.csv". The following 3 refer to the fitted model.

Question 52 Correct Mark 1.00 out of 1.00 Flag question Question text The estimated value of the slope is equal to: (Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback We use the function "lm" in order to fit the regression line between "peak.rpm" as an explanatory variable and "horsepower" as a response. Run the R code: > cars <- read.csv("cars.csv") > lm(horsepower~peak.rpm,data=cars) Call: lm(formula = horsepower ~ peak.rpm, data = cars) Coefficients: (Intercept) peak.rpm 48.63903 0.01085 The slope is equal to 0.01085.

Question 53 Correct Mark 1.00 out of 1.00 Flag question Question text Test the null hypothesis the intercept is equal to 0. This null hypothesis is: Select one: a.

Not rejected with a significance level of 5%. b.

Rejected with a significance level of 5%, but not with a significance level of 1%. c.

Rejected with a significance level of 1%, but not with a significance level of 0.1%. d.

Rejected with a significance level of 0.1% Feedback

Tests for the coefficients of a regression model are part of the output of the application of the function "summary" to the output of the function "lm". Run the code: > cars <- read.csv("cars.csv") > summary(lm(horsepower~peak.rpm,data=cars)) Call: lm(formula = horsepower ~ peak.rpm, data = cars) Residuals: Min 1Q Median 3Q Max -55.981 -31.768 -8.258 15.275 176.966 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 48.639025 29.823641 1.631 0.1045 peak.rpm 0.010851 0.005794 1.873 0.0625 .

--Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 39.47 on 201 degrees of freedom (2 observations deleted due to missingness) Multiple R-squared: 0.01715, Adjusted R-squared: 0.01226 F-statistic: 3.508 on 1 and 201 DF, p-value: 0.06252 The p-value for testing that the intercept is equal to zero is 0.1045, which is larger than 0.05. As a result, the null hypothesis is not rejected with a significance level of 5%.

Question 54 Correct Mark 1.00 out of 1.00 Flag question Question text Use the function "confint" in order to construct a confidence interval to the slope of the regression line. The value 0: Select one: a.

Belongs to the 95% confidence interval. b.

Does not belongs to the 95% confidence interval.

Feedback We can produce confidence intervals for the coefficients of the regression model via the application of the function "confint" to the output of the function "lm". Run the code: > cars <read.csv("cars.csv") > confint(lm(horsepower~peak.rpm,data=cars)) 2.5 % 97.5 % (Intercept) -1.016832e+01 107.44636946 peak.rpm -5.728294e-04 0.02227551 The confidence interval for the slope is [-5.728294 x 10-4, 0.02227551]. The value 0 belongs to this interval.

Information Flag question Information text The next 3 questions refer to the data given in following table: Data x y 1 1.2 0.5 2 0.0 2.2 3 1.6 4.9 4 3.8 0.1 5 1.6 1.6 6 0.8 0.6 7 1.7 3.4 8 5.8 0.5 9 1.6 0.7 1 3.5 0 0.4 Question 55 Correct Mark 1.00 out of 1.00 Flag question Question text The covariance between x an y is equal to: (Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback We read the data into R and then apply the function "cov": > x <- c(-0.5,0.0,1.6,-0.1,1.6,-0.6,1.7,0.5,-0.7,-0.4) > y <- c(1.2,2.2,-4.9,3.8,-1.6,0.8,-3.4,5.8,1.6,3.5) > cov(x,y) [1] -2.772222 Alternatively, we may apply the formula for the computation of the covariance: > sum((x-mean(x))*(y-mean( y )))/(10-1) [1] -2.772222 The covariance is equal to -2.772222.

Question 56 Correct Mark 1.00 out of 1.00

Flag question Question text If y is the response and x is the explanatory variable then the slope of the regression line is equal to: (Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback The slope may be obtained by the application of the function "lm" to a formula that puts y as the response and x as the explanatory variable: > x <- c(-0.5,0.0,1.6,-0.1,1.6,-0.6,1.7,-0.5,-0.7,-0.4) > y <- c(1.2,2.2,-4.9,3.8,-1.6,0.8,-3.4,5.8,1.6,3.5) > lm(y~x) Call: lm(formula = y ~ x) Coefficients: (Intercept) x 1.476 -2.745 Alternatively, one may apply the formula for the computation of the slope: > cov(x,y)/var(x) [1] -2.745076 The slope is equal to -2.745.

Question 57 Correct Mark 1.00 out of 1.00 Flag question Question text If x is the response and y is the explanatory variable then the slope of the regression line is equal to: (Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback The slope may be obtained by the application of the function "lm" to a formula that puts x as the response and y as the explanatory variable: > x <- c(-0.5,0.0,1.6,-0.1,1.6,-0.6,1.7,-0.5,-0.7,-0.4) > y <- c(1.2,2.2,-4.9,3.8,-1.6,0.8,-3.4,5.8,1.6,3.5) > lm(x~y) Call: lm(formula = x ~ y) Coefficients: (Intercept) y 0.4349 -0.2499 Alternatively, one may apply the formula for the computation of the slope: > cov(x,y)/var( y ) [1] -0.2498998 The slope is equal to -0.2499.

Question 58 Correct Mark 1.00 out of 1.00 Flag question Question text

Consider a response and an explanatory variable. Assume that the variance of the explanatory variable is positive. If the covariance between the two variables is positive then: Select one: a.

The regression line is decreasing b.

The regression line is increasing c.

The regression line may be increasing or decreasing. d.

The regression line has a zero slope.

Feedback The slope of the regression line is the ratio cov(x,y)/var(x) between the covariance and the variance of the explanatory variable. If the covariance is positive then the slope is positive. A positive slope corresponds to an increasing line. Therefore, the regrersion line is increasing.

Question 59 Correct Mark 1.00 out of 1.00 Flag question Question text Consider a response and an explanatory variable. Assume that the variance of the explanatory variable is positive. If the slope of the regression line is equal to zero then covariance between x and y: Select one: a. is positive. b. is negative c. may be positive or negative. d. is equal to 0.

Feedback The slope of the regression line is the ratio cov(x,y)/var(x) between the covariance and the variance of the explanatory variable. If the slope is equal to zero then the covariance is also equal to zero.

Information Flag question Information text A regression model was fitted to data. The slope of the model was b = -0.2499 and the intercept was a = 0.4349. Let y be the response and x be the explanatory variable. The next 2 questions refer to this information: Question 60 Correct Mark 1.00 out of 1.00 Flag question

Question text Match the observations to the estimated expectation: x = 1.2, y = -0.5 Estimated expectation = 0.135 x = 2.2, y = 0.0 Estimated expectation = -0.115 x = -4.9, y = 1.6 Estimated expectation = 1.659 Feedback The estimated expectation of the response for a given value of the explanatory variable is the level of the regression line for the given value. The regression line corresponds to the linear equation y = 0.4349 - 0.2499×x.

For x = 1.2 the estimated expectation is 0.4349 - 0.2499×1.2 = 0.13502 For x = 2.2 the estimated expectation is 0.4349 - 0.2499×2.2 = -0.11488 For x= -4.9 the estimated expectation is 0.4349 - 0.2499×(-4.9) = 1.65941 Consequently: The point x = 1.2, y = -0.5 is associated with the estimated expectation 0.135.

The point x = 2.2, y = 0.0 is associated with the estimated expectation -0.115.

The point x = -4.9, y = 1.6 is associated with the estimated expectation 1.659.

Question 61 Correct Mark 1.00 out of 1.00 Flag question Question text Match the observations to the residuals: x = -3.4, y = 1.7 Residual = 0.415 x = 5.8, y = -0.5 Residual = 0.515 x = 1.6, y = -0.7 Residual = -0.735 Feedback The residual is the difference between the value of the response and the estimated expectation of the response. For the given regression line this difference corresponds to y - (0.4349 - 0.2499×x).

For x = -3.4, y = 1.7 the residual is equal to 1.7 - (0.4349 - 0.2499×(-3.4)) = 0.41544.

For x = 5.8, y = -0.5 the residual is equal to -0.5 - (0.4349 - 0.2499×5.8) = 0.51452.

For x = 1.6, y = -0.7 the residual is equal to -0.7 - (0.4349 - 0.2499×1.6) = -0.73506.

Consequently: The point x = -3.4, y = 1.7 is associated with the residual 0.415.

The point x = 5.8, y = -0.5 is associated with the residual 0.515.

The point x = 1.6, y = -0.7 is associated with the residual -0.735.

Information Flag question Information text Fit a regression model to the relation between "height" as a response and "length" as an explanatory variable. Also, fit a regression model to the relation between "height" as a response and "width" as an explanatory variable. All 3 variables belong to the data set "cars.csv". The following 5 questions refer to the fitted models.

Question 62 Correct Mark 1.00 out of 1.00 Flag question

Question text The relation between "height" and "length" is: Select one: a.

Not significant in a significant level of 5%. b.

Significant with a significance level of 5%, but not with a significance level of 1%. c.

Significant with a significance level of 1%, but not with a significance level of 0.1%. d.

Significant with a significance level of 0.1% Feedback Run the code: > cars <- read.csv("cars.csv") > summary(lm(height~length,data=cars)) Call: lm(formula = height ~ length, data = cars) Residuals: Min 1Q Median 3Q Max -7.6415 -1.2960 0.2438 1.2140 6.2235 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 36.79802 2.11297 17.415 < 2e-16 *** length 0.09725 0.01211 8.031 7.73e-14 *** --Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 2.134 on 203 degrees of freedom Multiple R-squared: 0.2411, Adjusted R-squared: 0.2374 F-statistic: 64.5 on 1 and 203 DF, p-value: 7.732e-14 The p-value for testing the null hypothesis that the slope equals 0 is 7.73×10-14. This p-value is less than 0.001. Therefore, the null hypothesis is rejected with a significance level of 0.1%.

Question 63 Correct Mark 1.00 out of 1.00 Flag question Question text The part of the variance of "height" explained by "length" is: (Choose the closest number) Select one: a.

0.012 b.

0.241

c.

0.491 d.

2.134 Feedback Run the code: > cars <- read.csv("cars.csv") > summary(lm(height~length,data=cars)) Call: lm(formula = height ~ length, data = cars) Residuals: Min 1Q Median 3Q Max -7.6415 -1.2960 0.2438 1.2140 6.2235 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 36.79802 2.11297 17.415 < 2e-16 *** length 0.09725 0.01211 8.031 7.73e-14 *** --Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 2.134 on 203 degrees of freedom Multiple R-squared: 0.2411, Adjusted R-squared: 0.2374 F-statistic: 64.5 on 1 and 203 DF, p-value: 7.732e-14 The value of R-squared is 0.2411. This is the part of the variance of the response that is explained by the regression model.

Question 64 Correct Mark 1.00 out of 1.00 Flag question Question text The relation between "height" and "width" is: Select one: a.

Not significant in a significant level of 5%. b.

Significant with a significance level of 5%, but not with a significance level of 1%. c.

Significant with a significance level of 1%, but not with a significance level of 0.1%. d.

Significant with a significance level of 0.1% Feedback Run the code: Call: lm(formula = height ~ width, data = cars)

Residuals: Min 1Q Median 3Q Max -7.4172 -1.8362 0.4284 1.4455 6.4911 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 32.76369 5.06221 6.472 7.11e-10 *** width 0.31804 0.07677 4.143 5.03e-05 *** --Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 2.352 on 203 degrees of freedom Multiple R-squared: 0.07796, Adjusted R-squared: 0.07342 F-statistic: 17.16 on 1 and 203 DF, p-value: 5.031e-05 The p-value for testing the null hypothesis that the slope equals 0 is 5.03×10-5. This p-value is less than 0.001. Therefore, the null hypothesis is rejected with a significance level of 0.1%.

Question 65 Correct Mark 1.00 out of 1.00 Flag question Question text The part of the variance of "height" explained by "width" is: (Choose the closest number) Select one: a.

0.078 b.

0.277 c.

0.318 d.

2.352 Feedback Run the code: Call: lm(formula = height ~ width, data = cars) Residuals: Min 1Q Median 3Q Max -7.4172 -1.8362 0.4284 1.4455 6.4911 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 32.76369 5.06221 6.472 7.11e-10 *** width 0.31804 0.07677 4.143 5.03e-05 *** --Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 2.352 on 203 degrees of freedom Multiple R-squared: 0.07796, Adjusted R-squared: 0.07342 F-statistic: 17.16 on 1 and 203 DF, p-value: 5.031e-05 The value of R-squared is 0.07796. This is the part of the variance of the response that is explained by the regression model.

Question 66 Correct Mark 1.00 out of 1.00 Flag question Question text Of the two variables, the variable that explains better the variability of the response "height" is: Select one: a. "length" b. "width" Feedback The variable "length" explains 0.2411 of the variance of the response. The variable "width" explains only 0.07796 of the variance. Therefore, of the two explanatory variables, "length" produces a better explanation.

Information Flag question Information text Consider the relation between the response "drive.wheels" and the explanatory variable "fuel.type". Both variables belong to the data set "cars.csv". The next two questions refer to these variables.

Question 67 Correct Mark 1.00 out of 1.00 Flag question Question text Plot the mosaic plot that describes the relation between the response "drive.wheels" and the explanatory variable "fuel.type". The height of the rectangle with the lightest shade of grey above the value "gas" corresponds to: Select one: a.

The relative frequency of cars that run on gas. b.

The relative frequency of cars that use "rwd" among cars that run on gas. c.

The relative frequency of cars that use "rwd". d.

The relative frequency of cars that run on gas among cars that use "rwd".

Feedback The width of a rectangle corresponds to the relative frequency of a level of the explanatory variable. The hight of a rectangle of a given shade of grey above a level of the explanatory variable corresponds to the relative frequency of a level of the response among the sub-sample

that is associated with the given level of the explanatory variable.

In the current case we consider rectangles above the level "gas" of the explanatory variable. The said shade of grey is associated with the level "rwd" of the response. Therefore, the height of the rectangle with the lightest shade of grey above the value "gas" corresponds to the relative frequency of cars that use "rwd" among cars that run on gas.

Question 68 Correct Mark 1.00 out of 1.00 Flag question Question text Test the null hypothesis that the probability that cars that run on diesel use "fwd" is equal to the probability that cars that run on gas use "fwd". Test this hypothesis versus the two-sided alternative. The p-value computed for the test is equal to: (Give an answer of the form x.xxx with 3 significance digits) Answer: Feedback Run the code: > table(cars$fuel.type,cars$drive.wheels=="fwd") FALSE TRUE diesel 11 9 gas 74 111 > prop.test(table(cars$fuel.type,cars$drive.wheels=="fwd")) 2-sample test for equality of proportions with continuity correction data: table(cars$fuel.type, cars$drive.wheels == "fwd") X-squared = 1.1122, df = 1, p-value = 0.2916 alternative hypothesis: two.sided 95 percent confidence interval: -0.1068785 0.4068785 sample estimates: prop 1 prop 2 0.55 0.40 The first line produces the table of (joint) frequencies of the response and the explanatory variable. The response is a logical sequence that indicates, for each car, does it use "fwd" or not. The explanatory variable is the fuel type, a factor. In the second line the function "prop.test" is applies to the table of frequencies.

The output of the function "prop.test" includes a test of the null hypothesis the probability that cars that run on diesel use "fwd" is equal to the probability that cars that run on gas use "fwd".

The resulting p-value is equal to 0.2916.

Information Flag question Information text

Consider the relation between the response "drive.wheels" and the explanatory variable "engine.size". Both variables belong to the data set "cars.csv". The next two questions refer to these variables.

Question 69 Correct Mark 1.00 out of 1.00 Flag question Question text Plot the mosaic plot that describes the relation between the response "drive.wheels" and the explanatory variable "engine.size". The width of the second rectangle from the left corresponds to: Select one: a.

150 - 100 = 50. b.

The frequency of cars that use "rwd". c.

The frequency of cars with engine size between 100 and 150. d.

The frequency of cars that use "rwd" among cars with engine size between 100 and 150.

Feedback Run the expression "plot(drive.wheels~engine.size,data=cars)" in order to produce the mosaic plot. The width of a rectangle corresponds to the relative frequency of the range of levels of the explanatory variable. The hight of a rectangle of a given shade of grey above a a range of levels of the explanatory variable corresponds to the relative frequency of a level of the response among the sub-sample that is associated with the given range of levels of the explanatory variable.

The second rectangle from the left is associate with the range of engine sizes between 100 and 150. The width of this rectangle corresponds to the frequency of cars with engine size between 100 and 150.

Question 70 Correct Mark 1.00 out of 1.00 Flag question Question text Test the null hypothesis that there is no relation between the probability of using "fwd" and the size of the engine. The null hypothesis is: Select one: a.

Not rejected with a significance level of 5%. b.

Rejected with a significance level of 5%, but not with a significance level of 1%. c.

Rejected with a significance level of 1%, but not with a significance level of 0.1%. d.

Rejected with a significance level of 0.1% Feedback Run the code: > summary(glm(drive.wheels=="fwd"~engine.size,binomial,data=cars)) Call: glm(formula = drive.wheels == "fwd" ~ engine.size, family = binomial, data = cars) Deviance Residuals: Min 1Q Median 3Q Max -2.3756 -0.8398 0.5447 0.7786 2.1225 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 5.851567 0.865354 6.762 1.36e-11 *** engine.size -0.044160 0.006948 -6.356 2.08e-10 *** --Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for gaussian family taken to be 0.1792372) Null deviance: 278.19 on 204 degrees of freedom Residual deviance: 206.57 on 203 degrees of freedom AIC: 233.35 Number of Fisher Scoring iterations: 5 This code implements logistic regression to describe the relation between the response of car using or not "fwd" and the engine size as an explanatory variable.

The null hypothesis that states that there no relation between the response and the explanatory variable corresponds to the hypothesis that the coefficient of the variable "engine.size" is equal to zero. The p-value for testing this hypothesis is equal to 2.08×10-10. This p-value is less than 0.001. Therefore, the null hypothesis is rejected with a significance level of 0.1%.

Question 71 Correct Mark 1.00 out of 1.00 Flag question Question text Let X be a measurement with the Normal distribution. The measurement is taken for a sample of size 20 under two conditions. For half the sample the measurement is taken under one condition and for the other half the measurement is taken under a different condition. Denote the sample variance computed for the first half by Sa2 and the sample variance computed for the second half

by Sb2. Assume that the variance of the measurement is the same under both conditions. A region that contains 90% of the distribution of the statistic F = Sa2/Sb2 is: (Choose the closest option.)

Select one: a.

[-1.65,1.65]

b.

[-1.83,1.83] c.

[0.31, 3.18] d.

[3.33, 16.92] Feedback The distribution of the ratio of two estimators of the same variance, when the observations come from the Normal distribution, is the F distribution. This distribution is not symmetric. The region between that 5%- and the 95%-percentiles of the F distribution contain 90% of the distribution.

These percentiles can be computed using the code "qf(c(0.05,0.95),9,9)", that produces [0.3145749, 3.1788931] as an output.

Question 72 Correct Mark 1.00 out of 1.00 Flag question Question text The maximum of 15 Uniform(0,b) observations is used in order to estimate the parameter "b".

Bellow we present a simulation that assesses the sampling distribution of the estimator in the case where the value of the parameter is 7. The mean squared error (MSE) of the estimator, based on the output of the simulation, is equal to: (Give an answer of the form x.xxx, i.e. with 3 significance digits)

Answer: Feedback The MSE is equal to the sum of the variance and the square of the bias. The variance is equal to the square of the standard deviation 0.4111911. The bias is equal to the difference between the expectation and the target value of the parameter: 6.561692 - 7. Therefore, the MSE is equal to 0.41119112 + (6.561692 - 7)2 = 0.361192.

Question 73 Correct Mark 1.00 out of 1.00 Flag question Question text

A production batch contains 2,500 items. 30 items were randomly selected for inspection from the batch. 5 of the selected items did not pass the inspection. Let p be the probability that a selected item does not pass inspection. A 95% confidence interval for p is: (Choose the interval closest to the answer) Select one: a.

[0.063, 0.355] b.

[0.072, 0.415] c.

[0.0082, 0.0173] d.

[0.0007, 0.0050] Feedback The parameter of interest is the probability of selecting a "bad" item. Of the 30 items that were selected, 5 were bad. A confidence interval for the probability of selecting a bad item can be obtained as part of the output of the function "prop.test": > prop.test(5,30) 1-sample proportions test with continuity correction data: 5 out of 30, null probability 0.5 X-squared = 12.0333, df = 1, p-value = 0.0005226 alternative hypothesis: true p is not equal to 0.5 95 percent confidence interval: 0.06303555 0.35451084 sample estimates: p 0.1666667 This function produces [0.06303555, 0.35451084] as the 95%-confidence interval. It applies a continuity correction in the computation of the interval.

Observe that the sample size is small. Therefore, the application of a continuity correction is recommended. A confidence interval without a continuity correction may be computed with the code: > p.hat <- 5/30 > p.hat + qnorm(c(0.025,0.975))*sqrt(p.hat*(1-p.hat)/30) [1] 0.03330801 0.30002532 The resulting confidence interval is still best approximated by the interval [0.063, 0.355].

Question 74 Correct Mark 1.00 out of 1.00 Flag question Question text Assume that you are interested in testing H0: E(X) = 100 versus H1: E(X) ≥ 100 using the t-test.

Let the sample average, of a sample of size n = 17, be equal to x= 111.0 and the sample standard deviation be equal to s = 19.3. The null hypothesis is: Select one: a.

Not rejected with a significance level of 5%. b.

Rejected with a significance level of 5%, but not with a significance level of 1%. c.

Rejected with a significance level of 1%, but not with a significance level of 0.1%. d.

Rejected with a significance level of 0.1% Feedback The statistic for testing the one-sided hypothesis is equal to (x - 100)/(s/√n) = (111.0 - 100)/ (19.3/√17) = 2.349957.

The thresholds for the one-sided test correspond to 95%-, 99%-, and 99.9%-percentiles of the tdistribution on 17-1=16 degrees of freedom. These percentiles can computed using the expression: qt(c(0.95,0.99,0.999),16) that produce 1.745884, 2.583487, and 3.686155, respectively, as output.

The statistic is larger than the smallest threshold but smaller that the threshold that is associated with a 1%-significance level. Therefore, the null hypothesis is rejected at the 5% confidence level but is not rejected at the 1%-significance level.

Alternatively, one may compute the p-value using the expression 1-pt((111.0-100)/ (19.3/sqrt(17)),16), which turns out to be equal to 0.01596924. This value is less than 0.05 and larger than 0.01. Hence, again, the null hypothesis is rejected at the 5% confidence level but is not rejected at the 1%-significance level.

Question 75 Correct Mark 1.00 out of 1.00 Flag question Question text The data set "cars.csv" contains information on various models of cars. The file can be found on the internet at this link. Two of the variables in this data set are "highway.mpg", the number of miles per gallon in highway driving conditions, and "fuel.type", the type of fuel the car is using.

Apply the t-test with "highway.mpg" as the response and "fuel.type" as the explanatory variable.

The conclusion of the default application of the test, with a significance level of 5%, is: Select one: a.

The expectation of the response for cars that use diesel is larger than the expectation for cars that use gas. b.

The expectation of the response for cars that use diesel is not equal to the expectation for cars that use gas. c.

The expectation of the response for cars that use diesel is smaller than the expectation for cars that use gas. d.

The expectation of the response for cars that use diesel is equal to the expectation for cars that use gas.

Feedback Run the code: > cars <- read.csv("cars.csv") > plot(highway.mpg~fuel.type,data=cars) > t.test(highway.mpg~fuel.type,data=cars) Welch Two Sample t-test data: highway.mpg by fuel.type t = 2.2288, df = 21.438, p-value = 0.03665 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 0.3016753 8.5604869 sample estimates: mean in group diesel mean in group gas 34.75000 30.31892 We upload the data in the first line and plot it in the second. The function "t.test" is applied in order to test the null hypothesis that the difference between the expectations of the response in the two groups is equal to 0 against the two-sided alternative.

The computed p-value is 0.03665. Consequently, the null hypothesis is rejected with a significance level of 5%. The conclusion of the test is that the expectation of the response for cars that use diesel is not equal to the expectation for cars that use gas.

Question 76 Correct Mark 1.00 out of 1.00 Flag question Question text Consider data regarding a response y and an explanatory variable x, both numeric: Data x y 2.

1 0.7 8 6.

2 1.3 8 7. 3 0 1.1 4.

4 0.2 7 5.

5 1.1 0 5. 6 0 0.1 7 2. 1.7

9 3.

8 1.0 7 3.

9 0.8 0 1 2.

0.8 0 6 The null hypothesis that the expected value of the response is constant for all values of the explanatory variable is: Select one: a.

Not rejected with a significance level of 5%. b.

Rejected with a significance level of 5%, but not with a significance level of 1%. c.

Rejected with a significance level of 1%, but not with a significance level of 0.1%. d.

Rejected with a significance level of 0.1% Feedback Run the code: > x <- c(2.8,6.8,7.0,4.7,5.0,5.0,2.9,3.7,3.0,2.6) > y <- c(0.7,1.3,-1.1,0.2,1.1,-0.1,1.7,1.0,0.8,0.8) > summary(lm(y~x)) Call: lm(formula = y ~ x) Residuals: Min 1Q Median 3Q Max -1.1066 -0.3449 -0.2105 0.5127 1.2456 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 1.6798 0.6987 2.404 0.0429 * x -0.2390 0.1514 -1.579 0.1530 --Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.7386 on 8 degrees of freedom Multiple R-squared: 0.2376, Adjusted R-squared: 0.1423 F-statistic: 2.493 on 1 and 8 DF, p-value: 0.153 The first two lines of code produce the explanatory variable and the response, respectively. A linear regression analysis of the relation between the two variables is produced by the third line.

The situation when the expected value of the response is constant for all values of the explanatory variable corresponds to the case where the slope of the regression model is equal to

zero. The p-value for testing that the slope is equal to zero is equal to 0.1530. This value is larger than 0.05. Consequently, the null hypothesis is not rejected with a significance level of 5%.

Question 77 Correct Mark 1.00 out of 1.00 Flag question Question text The data set "magnets.csv" contains the results of an experiment that examines the effectiveness of magnets as a pain treatment. This data was presented in Section 9.5 of the book. The file can be found on the internet at this link.

Consider the relation between the response "change" and the explanatory variable "active", the type of treatment that was used. Both variables belong to the data set "magnets.csv". Test the null hypothesis that the probability that the value of "change" is equal to zero is the same for both levels of the explanatory variable. The null hypothesis is: Select one: a.

Not rejected with a significance level of 5%. b.

Rejected with a significance level of 5%, but not with a significance level of 1%. c.

Rejected with a significance level of 1%, but not with a significance level of 0.1%. d.

Rejected with a significance level of 0.1% Feedback Run the code: > magnets <- read.csv("magnets.csv") > plot(factor(change==0)~active,data=magnets) > prop.test(table(magnets$active,magnets$change==0)) 2-sample test for equality of proportions with continuity correction data: table(magnets$active, magnets$change == 0) X-squared = 6.8966, df = 1, p-value = 0.008636 alternative hypothesis: two.sided 95 percent confidence interval: 0.09708026 0.67467672 sample estimates: prop 1 prop 2 0.8620690 0.4761905 The data is uploaded in the first line. The relation between the response and the explanatory variable is plotted in the second line and tested in the third.

The response is the indicator of no change and the explanatory variable is a factor with two levels. We apply the function "prop.test" in order to test the hypothesis that the probability of no change is the same for both levels. Recall that the input to the function is a 2×2 table of frequencies, with the response entering as the second variable in this table. The p-value for testing the null hypothesis that the two probabilities are identical (against the two-sided

alternative) is 0.008636. This number is between 0.01 and 0.001. Therefore, the null hypothesis is rejected with a significance level of 1%, but not with a significance level of 0.1%.

Final exam