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The article contains a Statistical Inference quiz in R language with Answers. There are 16 questions in the “Statistical Inference Quiz in R Language”. The MCQs are from probability and regression models. Let us Start with the Statistical Inference Quiz in R.

Statistical Inference Quiz in R Language

1. Consider the mtcars data set. Fit a model with mpg as the outcome that includes numbers of cylinders as a factor variable and weight as a possible confounding variable. Compare the effect of 8 versus 4 cylinders on mpg for the adjusted and unadjusted by-weight models. Here, adjusted means including the weight variable as a term in the regression model and unadjusted means the model without weight included. What can be said about the effect comparing 8 and 4 cylinders after looking at models with and without weight included?

 Holding weight constant, cylinder appears to have less of an impact on mpg than if weight is disregarded

 Holding weight constant, cylinder appears to have more of an impact on mpg than if weight is disregarded

 Within a given weight, 8 cylinder vehicles have an expected 12 mpg drop in fuel efficiency

 Including or excluding weight does not appear to change anything regarding the estimated impact of number of cylinders on mpg

2. Consider the mtcars data set. Fit a model with mpg as the outcome that considers numbers of cylinders as a factor variable and weight as a confounder. Now fit a second model with mpg as the outcome model that considers the interaction between numbers of cylinders (as a factor variable) and weight. Give the P-value for the likelihood ratio test comparing the two models and suggest a model using 0.05 as a type I error rate significance benchmark.

 The P-value is small (less than 0.05). Thus it is surely true that there is no interaction term in the true model

 The P-value is larger than 0.05. So, according to our criterion, we would fail to reject, which suggests that the interaction terms is necessary

 The P-value is larger than 0.05. So, according to our criterion, we would fail to reject, which suggests that the interaction terms may not be necessary

 The P-value is small (less than 0.05). So, according to our criterion, we reject, which suggests that the interaction term is necessary

3. Consider the following data set

x <- c(0.586, 0.166, -0.042, -0.614, 11.72)
y <- c(0.549, -0.026, -0.127, -0.751, 1.344)

Give the slope dfbeta for the point with the highest hat value.

influence.measures(fit5)$infmat[which.max(abs(influence.measures(fit5)$infmat[, 2])), 2]


 $0.673$

 $-134$

 $-0.378$

 $-0.00134$

4. Suppose that diastolic blood pressures (DBPs) for men aged 35-44 are normally distributed with a mean of 80 (mm Hg) and a standard deviation of 10. About what is the probability that a random 35-44-year-old has a DBP less than 70?

 round(pnorm(70, mean = 80, sd=100)*10)

 round(pnorm(70, mean = 80, sd=10)*10)

 round(pnorm(70, mean = 80, sd=10)*100)

 round(pnorm(70, mean = 80, sd=10)/100)

5. Consider the mtcars data set. Fit a model with mpg as the outcome that includes the number of cylinders as a factor variable and weight included in the model as

lm(mpg ~ I(wt * 0.5) + factor(cyl), data = mtcars)

How is the wt coefficient interpreted?

 The estimated expected change in MPG per half ton increase in weight for a specific number of cylinders (4, 6, 8)

 The estimated expected change in MPG per one ton increase in weight for a specific number of cylinders (4, 6, 8)

 The estimated expected change in MPG per one ton increase in weight.

 The estimated expected change in MPG per half ton increase in weight for the average number of cylinders

6. Brain volume for adult women is normally distributed with a mean of about 1,100 cc for women with a standard deviation of 75 cc. What brain volume represents the 95th percentile?

 round(qnorm(95, mean = 1100, sd = 75))

 round(qnorm(0.95, mean = 1100, sd = 75))

 round(pnorm(0.95, mean = 1100, sd = 75))

 round(qnorm(0.95, mean = 1175, sd = 75))

7. The respiratory disturbance index (RDI), a measure of sleep disturbance, for a specific population has a mean of 15 (sleep events per hour) and a standard deviation of 10. They are not normally distributed. Give your best estimate of the probability that a sample average RDI of 100 people is between 14 and 16 events per hour.

 Left <- pnorm(q = 14, mean = 15, sd = 10/sqrt(100)) * 100
Right <- pnorm(q = 16, mean = 15, sd = 10/sqrt(100)) * 100
total <- Right * Left; total

 Left <- pnorm(q = 14, mean = 15, sd = 10/sqrt(100)) * 100
Right <- pnorm(q = 16, mean = 15, sd = 10/sqrt(100)) * 100
total <- Right / Left; total

 Left <- pnorm(q = 14, mean = 15, sd = 10/sqrt(100)) * 100
Right <- pnorm(q = 16, mean = 15, sd = 10/sqrt(100)) * 100
total <- Right + Left; total

 Left <- pnorm(q = 14, mean = 15, sd = 10/sqrt(100)) * 100
Right <- pnorm(q = 16, mean = 15, sd = 10/sqrt(100)) * 100
total <- Right – Left; total

8. Consider the following data set
x <- c(0.8, 0.47, 0.51, 0.73, 0.36, 0.58, 0.57, 0.85, 0.44, 0.42)
y <- c(1.39, 0.72, 1.55, 0.48, 1.19, -1.59, 1.23, -0.65, 1.49, 0.05)
Fit the regression through the origin and get the slope treating $y$ as the outcome and $x$ as the regressor.

(Hint, do not center the data since we want regression through the origin, not through the means of the data.)

 $0.8263$

 $-0.04462$

 $-1.713$

 $0.59915$

9. Consider the following data set
x <- c(0.586, 0.166, -0.042, -0.614, 11.72)
y <- c(0.549, -0.026, -0.127, -0.751, 1.344)
Give the hat diagonal for the most influential point

 0.9946

 0.2804

 0.2287

 0.2025

10. Consider the following PMF shown below in R
x <- 1:4
p <- x/sum(x)
temp <- rbind(x, p)
rownames(temp) <- c("X", "Prob")
temp
What is the mean?

 3

 2

 4

 1

11. Consider a standard uniform density. The mean for this density is 0.5 and the variance is 1 / 12. You sample 1,000 observations from this distribution and take the sample mean, what value would you expect it to be near?

 qnorm(p=0.5, mean = 0.5, sd = 1*12/sqrt(1000))

 qnorm(p=0.5, mean = 0.5, sd = 1/12/sqrt(100))

 qnorm(p=0.05, mean = 5, sd = 1/12/sqrt(1000))

 qnorm(p=0.5, mean = 0.5, sd = 1/12/sqrt(1000))

12. Consider the following data set. What is the intercept for fitting the model with $x$ as the predictor and $y$ as the outcome?
x <- c(0.8, 0.47, 0.51, 0.73, 0.36, 0.58, 0.57, 0.85, 0.44, 0.42)
y <- c(1.39, 0.72, 1.55, 0.48, 1.19, -1.59, 1.23, -0.65, 1.49, 0.05)

 $1.252$

 $1.567$

 $-1.713$

 $2.105$

13. The number of people showing up at a bus stop is assumed to be Poisson with a mean of 5 people per hour. You watch the bus stop for 3 hours. About what’s the probability of viewing 10 or fewer people?

 round(ppois(q=10, lambda = 3 * 5), digits=2)

 round(ppois(q=10, lambda = 3), digits=2)

 round(ppois(q=10, lambda = 3 + 5), digits=2)

 round(ppois(q=10, lambda = 5), digits=2)

14. Do data(mtcars) from the datasets package and fit the regression model with mpg as the outcome and weight as the predictor. Give the slope coefficient.

 $-9.559$

 $0.5591$

 $30.2851$

 $-5.344$

15. You flip a fair coin 5 times, about what’s the probability of getting 4 or 5 heads?

 round(pbinom(q = 3, size = 5, prob = 0.05, lower.tail = FALSE) * 100)

 round(pbinom(q = 3, size = 5, prob = 0.5, lower.tail = TRUE) * 100)

 round(pbinom(q = 3, size = 5, prob = 0.5, lower.tail = FALSE) * 10)

 round(pbinom(q = 3, size = 5, prob = 0.5, lower.tail = FALSE) * 100)

16. Consider the mtcars data set. Fit a model with mpg as the outcome that includes a number of cylinders as a factor variable and weight as a confounder. Give the adjusted estimate for the expected change in mpg comparing 8 cylinders to 4.

 -6.071

 -4.256

 -3.206

 33.991

 Loading …

Statistical Inference Quiz in R with Answers

• Consider the following PMF shown below in R
  x <- 1:4 p <- x/sum(x)
  temp <- rbind(x, p)
  rownames(temp) <- c(“X”, “Prob”)
  temp
  What is the mean?
• Suppose that diastolic blood pressures (DBPs) for men aged 35-44 are normally distributed with a mean of 80 (mm Hg) and a standard deviation of 10. About what is the probability that a random 35-44-year-old has a DBP less than 70?
• Brain volume for adult women is normally distributed with a mean of about 1,100 cc for women with a standard deviation of 75 cc. What brain volume represents the 95th percentile?
• You flip a fair coin 5 times, about what’s the probability of getting 4 or 5 heads?
• The respiratory disturbance index (RDI), a measure of sleep disturbance, for a specific population has a mean of 15 (sleep events per hour) and a standard deviation of 10. They are not normally distributed. Give your best estimate of the probability that a sample average RDI of 100 people is between 14 and 16 events per hour.
• Consider a standard uniform density. The mean for this density is 0.5 and the variance is 1 / 12. You sample 1,000 observations from this distribution and take the sample mean, what value would you expect it to be near?
• The number of people showing up at a bus stop is assumed to be Poisson with a mean of 5 people per hour. You watch the bus stop for 3 hours. About what’s the probability of viewing 10 or fewer people?
• Consider the mtcars data set. Fit a model with mpg as the outcome that includes a number of cylinders as a factor variable and weight as a confounder. Give the adjusted estimate for the expected change in mpg comparing 8 cylinders to 4.
• Consider the mtcars data set. Fit a model with mpg as the outcome that includes the number of cylinders as a factor variable and weight included in the model as
  lm(mpg ~ I(wt * 0.5) + factor(cyl), data = mtcars)
  How is the wt coefficient interpreted?
• Consider the following data set
  x <- c(0.586, 0.166, -0.042, -0.614, 11.72)
  y <- c(0.549, -0.026, -0.127, -0.751, 1.344)
  Give the hat diagonal for the most influential point
• Consider the following data set
  x <- c(0.586, 0.166, -0.042, -0.614, 11.72)
  y <- c(0.549, -0.026, -0.127, -0.751, 1.344)
  Give the slope dfbeta for the point with the highest hat value. influence.measures(fit5)$infmat[which.max(abs(influence.measures(fit5)$infmat[, 2])), 2]
• Consider the mtcars data set. Fit a model with mpg as the outcome that includes a number of cylinders as a factor variable and weight as a possible confounding variable. Compare the effect of 8 versus 4 cylinders on mpg for the adjusted and unadjusted by-weight models. Here, adjusted means including the weight variable as a term in the regression model and unadjusted means the model without weight included. What can be said about the effect comparing 8 and 4 cylinders after looking at models with and without weight included?
• Consider the mtcars data set. Fit a model with mpg as the outcome that considers a number of cylinders as a factor variable and weight as a confounder. Now fit a second model with mpg as the outcome model that considers the interaction between numbers of cylinders (as a factor variable) and weight. Give the P-value for the likelihood ratio test comparing the two models and suggest a model using 0.05 as a type I error rate significance benchmark.
• Consider the following data set
  x <- c(0.8, 0.47, 0.51, 0.73, 0.36, 0.58, 0.57, 0.85, 0.44, 0.42)
  y <- c(1.39, 0.72, 1.55, 0.48, 1.19, -1.59, 1.23, -0.65, 1.49, 0.05)
  Fit the regression through the origin and get the slope treating $y$ as the outcome and $x$ as the regressor. (Hint, do not center the data since we want regression through the origin, not through the means of the data.)
• Do data(mtcars) from the datasets package and fit the regression model with mpg as the outcome and weight as the predictor. Give the slope coefficient.
• Consider the following data set. What is the intercept for fitting the model with $x$ as the predictor and $y$ as the outcome?
  x <- c(0.8, 0.47, 0.51, 0.73, 0.36, 0.58, 0.57, 0.85, 0.44, 0.42)
y <- c(1.39, 0.72, 1.55, 0.48, 1.19, -1.59, 1.23, -0.65, 1.49, 0.05)

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