R Language Quiz

The post is about multiple-choice questions about the package dplyr in R Language. There are 20 MCQs about the package and its use. Let us start with the Quiz on dplyr in R Language.

MCQs dplyr in R Language

• What is the function of the dplyr verb Filter?
• What is the function of the dplyr verb Select?
• What is the function of the dplyr verb Group By?
• How does Summarise work?
• What does the dplyr verb mutate do?
• The dplyr verb Arrange is responsible for what action?
• The dplyr verb ‘Filter‘ does what to a data frame?
• The dplyr verb ‘Select‘ does?
• What does the dplyr verb ‘Group By‘ do?
• What does the dplyr verb ‘Arrange‘ do?
• What does the dplyr verb ‘Mutate‘ do?
• What symbol is used in dplyr that holds verbs together in a single phrase?
• Example tools for reproducible report writing are:
• Reproducibility tools for reports like knitr help with:
• What is the purpose of the distinct() function in dplyr?
• In dplyr, what is the purpose of the %>% operator (known as pipe operator)
• ———– function is similar to the existing subset() function in R but is quite a bit faster.
• What is the purpose of ungroup() function in dplyr?
• In dplyr, what does the slice() function do?
• How can a new column/variable (total_price) be created in dplyr with the sum of two existing columns/variables price1 and price2?

An Introduction to dplyr Package

The dplyr package is used for data manipulation and transformation. It gives a set of functions that make it easy to perform common data manipulation tasks, which include (1) filtering, (2) grouping, (3) summarizing, (4) arranging, and (5) joining data frames.

The package is part of the tidyverse, a collection of R packages designed to work together seamlessly for data analysis and visualization.

Some key functions available in dplyr R Package include:

• filter(): Used to subset rows based on specified conditions.
• select(): Used to choose specific columns from a data frame.
• arrange(): Used to reorder rows based on one or more columns.
• mutate(): Used to create new columns or modify existing ones.
• group_by(): Used to group data by one or more variables.
• summarize(): Used to compute summary statistics for groups of data.
• join(): Used to merge data frames based on common keys.

The dplyr package provides a powerful and efficient toolkit for data manipulation in R.

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The post is about R Language Quiz with Answers. The quiz covers, MCQs about Data Structure, Data Analysis in R, and Some Basics of R Programming Languages. Let us start with the R Programming MCQs Quiz.

Please go to R Language Quiz 15: Important MCQs to view the test

R Language Quiz with Answers

• What is NaN called?
• what is the output of the following y <- c(TRUE, 2)
• what is the class of object $y$ y <- c(2, “t”)
• What is the class of the object $y$ y <- c(FALSE, 2)
• Which of them is not a basic datatype in R?
• How one can create an integer say 5?
• The dimension attribute is itself an integer vector having length ——-.
• Matrices can be created by row-binding using the function
• What function is used to test objects if they are NaN?
• What is the function to give names to columns for a matrix?
• How many atomic vector types does R have
• What is the output of the following d <- diag(5, nrow = 2, ncol = 2); d
• Which of the following can be used to display the names of (most of) the objects that are currently stored within R?
• A ——– is a variable that holds one value at a time
• What is the meaning of “<-” in R
• Which of the following is not an R object?
• Dataframes can be converted into a matrix by calling the following function data ———-
• R files have an extension ———-.
• R language has a superficial similarity with ———.
• Which of the following is an alternative to ‘?’ symbol ———.

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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.

Please go to Best Statistical Inference Quiz in R 14 to view the test

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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