When creating the figure it was necessary to first calculate the summary data. This was done using the code below: d_sum % 1 group_by(location) 2 summarise(mean = mean(rating), 3 sd = sd(rating), 4 n = n(), 5 sderr = sd/sqrt(n)) %% 6 ungroup() 7 That code may have an error. If so, what line is the error in?

Question

When creating the figure it was necessary to first calculate the summary data. This was done using the code below:

d_sum <- d %>%                        1
  group_by(location)                  2
  summarise(mean = mean(rating),      3
            sd = sd(rating),          4
            n = n(),                  5
            sderr = sd/sqrt(n)) %>%   6
  ungroup()                           7

That code may have an error. If so, what line is the error in?

BlackTom AI · Accepted Answer

All lines are syntactically and functionally correct — there is no error.

1) Relevant knowledge points and concepts
- The dplyr pipe (%>%) passes the left-hand data frame into successive verbs.
- group_by(location) groups rows by the location column.
- summarise() computes summary statistics per group. Modern dplyr evaluates summary expressions left-to-right, so later summary columns can reference earlier ones in the same summarise call.
- sd(rating) gives the sample standard deviation, n() returns the group size, and the standard error is sd / sqrt(n).
- ungroup() removes grouping metadata from the resulting tibble.

2) Step-by-step reasoning
- Line 2: group_by(location) correctly creates groups by the location column.
- Lines 3–6: summarise(...) computes four new columns for each group: mean, sd, n, and sderr. Each expression is valid: mean(rating), sd(rating), and n() are standard dplyr summarising functions, and sderr uses the previously computed sd and n.
- Line 7: ungroup() is appropriate to drop grouping after summarisation.

3) Calculations and intermediate steps (per group)
- n = number of observations in that group.
- mean = (1/n) * sum_i rating_i.
- sd = sqrt( sum_i (rating_i − mean)^2 / (n − 1) ).
- sderr = sd / sqrt(n).

4) Why the correct answer is correct
Given the functions used and the left-to-right evaluation in summarise, computing sderr as sd/sqrt(n) is valid because sd and n have just been created in that same summarise call. The pipeline and syntax are consistent with typical dplyr usage, so no line contains an error.

5) Why the other options are incorrect
- Line 2 is a correct group_by call.
- Line 3 (mean = mean(rating)) correctly computes the group mean.
- Line 6 (sderr = sd/sqrt(n)) correctly computes standard error using earlier summarised columns.
- Line 7 ungroup() is a valid and common final step.

6) Conclusion
All lines are correct; the correct choice is “There is no error,” which matches the provided answer.

Similar Questions

na is a vector of numeric elements. You do not know how many elements na has. When you run the following, you are returned 1. sum(is.na(na)) If you execute the following, what would be the output? sum(is.na(na[which(is.na(na))] + na)) ==length(na)

Dharma has vector w in her active R session. There are no missing values in w. That is, there is no NA’s in w. She executed the following code: sum(w) == sum(w[which(w)]) R returned (outputted) TRUE on her R console. TRUE/FALSE: w is not a logical vector.

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