# Conditioning

STAT 20: Introduction to Probability and Statistics

## Agenda

• Concept Questions
• Data Pipelines
• Break
• Worksheet - Digital
• Lab 3.1: Flights

# Concept Questions

## Question 1

c("fruit", "fruit", "vegetable") == "fruit"

What will this line of code return?

Respond at pollev.com.

01:00

## Evaluating equivalence, cont.

In R, this evaluation happens element-wise when operating on vectors.

c("fruit", "fruit", "vegetable") == "fruit"
  TRUE  TRUE FALSE
c("fruit", "fruit", "vegetable") != "fruit"
 FALSE FALSE  TRUE
c("fruit", "vegetable", "boba") %in% c("fruit", "vegetable")
  TRUE  TRUE FALSE

## Question 2

Which observations will be included in the following data frame?

filter(class_survey,
coding_exp_scale < 3,
olympic_sport %in% c("Ice skating", "Speed skating"),
entrepreneur == TRUE)

Please respond at pollev.com.

01:00

## Question 3: Opinion

1. What are students’ perceptions of the chance that there is a new COVID variant that disrupts instruction in Spring 2023?

Do you think students in their first semester would be more likely or less likely to think we would remain in remote learning for the entire semester?

Answer at pollev.com.

## Question 4

Which data frame will have fewer rows?

class_survey <- mutate(class_survey,
first_sem = year == "This is my first semester!")

# this one
df_1 <- filter(class_survey, first_sem)

# or this one
df_2 <- filter(class_survey, year == "This is my first semester!")
01:00

# Building data pipelines

## Question 2 Redux

How do we extract the average of these students’ chance that class will be disrupted by a new COVID variant?

filter(class_survey,
coding_exp_scale < 3,
olympic_sport %in% c("Ice skating", "Speed skating"),
entrepreneur == TRUE)

## Question 2 Redux

How do we extract the average of these students’ chance that class will be disrupted by a new COVID variant?

select(filter(class_survey,
coding_exp_scale < 3,
olympic_sport %in% c("Ice skating", "Speed skating"),
entrepreneur == TRUE),
coding_exp_xcale,
olympic_sport,
entrepreneur,
new_COVID_variant)

## Question 2 Redux

How do we extract the average of these students’ chance that class will be disrupted by a new COVID variant?

summarize(select(filter(class_survey,
coding_exp_scale < 3,
olympic_sport %in% c("Ice skating", "Speed skating"),
entrepreneur == TRUE),
coding_exp_scale,
olympic_sport,
entrepreneur,
new_COVID_variant),
covid_avg = mean(new_COVID_variant))

## Question 2 Redux

How do we extract the average of these students’ chance that class will be disrupted by a new COVID variant?

summarize(select(filter(class_survey,
coding_exp_scale < 3,
olympic_sport %in% c("Ice skating", "Speed skating"),
entrepreneur == TRUE),
coding_exp_scale,
olympic_sport,
entrepreneur,
new_COVID_variant),
covid_avg = mean(new_COVID_variant))
# A tibble: 1 × 1
covid_avg
<dbl>
1      0.52

## Data Pipelines

Most claims about data start with a raw data set, undergo many subsetting, aggregating, and cleaning operations, then return a data product.

Let’s look at three equivalent ways to build a pipeline

## Nesting

summarize(select(filter(class_survey,
coding_exp_scale < 3,
olympic_sport %in% c("Ice skating", "Speed skating"),
entrepreneur == TRUE),
coding_exp_scale,
olympic_sport,
entrepreneur,
new_COVID_variant),
covid_avg = mean(new_COVID_variant))

Cons

• Must be read from inside out 👎
• Hard to keep track of arguments 👎

Pros

• All in one line of code 👍
• Only refer to one data frame 👍

## Step-by-step

df1 <- filter(class_survey,
coding_exp_scale < 3,
olympic_sport %in% c("Ice skating", "Speed skating"),
entrepreneur == TRUE)
df2 <- select(df1,
coding_exp_scale,
olympic_sport,
entrepreneur,
new_COVID_variant)
summarize(df2,
covid_avg = mean(new_COVID_variant))

Cons

• Have to repeat data frame names 👎
• Creates unnecessary objects 👎

Pros

• Stores intermediate objects 👍
• Can be read top to bottom 👍

## Using the pipe operator

class_survey %>%
filter(coding_exp_scale < 3,
olympic_sport %in% c("Ice skating", "Speed skating"),
entrepreneur == TRUE) %>%
select(coding_exp_scale,
olympic_sport,
entrepreneur,
new_COVID_variant) %>%
summarize(covid_avg = mean(new_COVID_variant))

Cons

• 🤷

Pros

• Can be read like an english paragraph 👍
• Only type the data once 👍
• No leftovers objects 👍

It’s good practice to understand the output of each line of code by breaking the pipe.

#| echo: true
#| eval: true
#| error: true
class_survey %>%
select(new_COVID_variant) %>%
filter(year == "It's my first year.")
class_survey %>%
select(new_COVID_variant)
# A tibble: 619 × 1
new_COVID_variant
<dbl>
1           0.25
2           0.1
3           0
4           0.2
5           0.9
6           0.2
7           0.4
8           0.00005
9           0.2
10           0.3
# … with 609 more rows

## Question 5

class_survey %>% # A #<<
filter(coding_exp_scale < 3,
olympic_sport %in% c("Ice skating",
"Speed skating"),
entrepreneur == TRUE) %>% # B #<<
select(coding_exp_scale,
olympic_sport,
entrepreneur,
new_COVID_variant) %>% # C #<<
summarize(covid_avg = mean(new_COVID_variant)) # D #<<

What are the dimensions (rows x columns) of the data frames output at each stage of this pipe?

01:00

## Question 2: Code

Do you think first year students would be more likely or less likely to think we would remain in remote learning for the entire semester?

Which commands are needed to help answer this question?

class_survey %>%
filter(year) %>%

## Describing new_COVID_variant

library(tidyverse)
library(stat20data)
class_survey <- class_survey %>%
select(year, new_COVID_variant) %>%
mutate(new_COVID_variant = round(new_COVID_variant, digits = 2))
ggplot(class_survey, aes(x = new_COVID_variant)) +
geom_histogram()

Aside: density plot

ggplot(class_survey, aes(x = new_COVID_variant)) +
geom_density() #<<

## Describing new_COVID_variant

ggplot(class_survey, aes(x = new_COVID_variant)) +
geom_histogram()
summarize(class_survey,
mean = mean(new_COVID_variant),
med = median(new_COVID_variant),
iqr = IQR(new_COVID_variant),
sd = sd(new_COVID_variant))
# A tibble: 1 × 4
mean   med   iqr    sd
<dbl> <dbl> <dbl> <dbl>
1 0.368   0.3  0.35 0.468

The distribution of probabilities of all students is right-skewed with a mean probability of 0.37 and a median probability of 0.3, an IQR of 0.35 and a SD of 0.47.

## Describing first year students

How can we focus our analysis on just first year students?

General goal: Identify whether the value in a variable meets a condition.

Here: Is the value in new_COVID_variant equal to "I'm in my first year."?

Our Tool, Comparison operators: A collection of operators that compare two values / vectors and return TRUE or FALSE.

## Evaluating equivalence

"fruit" == "vegetable"
 FALSE
"fruit" == "fruit"
 TRUE
"fruit" != "fruit"
 FALSE

== evaluates equality, != evaluates inequality.

class_survey <- mutate(class_survey,
first_year = year == "I'm in my first year.")
class_survey
# A tibble: 619 × 3
year                       new_COVID_variant first_year
<chr>                                  <dbl> <lgl>
1 I'm in my second year.                  0.25 FALSE
2 This is my first semester!              0.1  FALSE
3 This is my first semester!              0    FALSE
4 I'm in my second year.                  0.2  FALSE
5 I'm in my first year.                   0.9  TRUE
6 I'm in my second year.                  0.2  FALSE
7 I'm in my second year.                  0.4  FALSE
8 I'm in my second year.                  0    FALSE
9 I'm in my second year.                  0.2  FALSE
10 I'm in my first year.                   0.3  TRUE
# … with 609 more rows

# Filtering data using logical vectors

## Filtering rows

first_yr_df <- filter(class_survey, first_year)
first_yr_df
# A tibble: 245 × 3
year                  new_COVID_variant first_year
<chr>                             <dbl> <lgl>
1 I'm in my first year.               0.9 TRUE
2 I'm in my first year.               0.3 TRUE
3 I'm in my first year.               0.6 TRUE
4 I'm in my first year.               0.3 TRUE
5 I'm in my first year.               0.3 TRUE
6 I'm in my first year.               0.1 TRUE
7 I'm in my first year.               0.7 TRUE
8 I'm in my first year.               0.2 TRUE
9 I'm in my first year.               0.5 TRUE
10 I'm in my first year.               0.5 TRUE
# … with 235 more rows

## Describing new_COVID_variant with statistics

Statistics from all students

summarize(class_survey,
mean = mean(new_COVID_variant),
med = median(new_COVID_variant),
iqr = IQR(new_COVID_variant),
sd = sd(new_COVID_variant))
# A tibble: 1 × 4
mean   med   iqr    sd
<dbl> <dbl> <dbl> <dbl>
1 0.368   0.3  0.35 0.468

Statistics from first year students

summarize(first_yr_df,
mean = mean(new_COVID_variant),
med = median(new_COVID_variant),
iqr = IQR(new_COVID_variant),
sd = sd(new_COVID_variant))
# A tibble: 1 × 4
mean   med   iqr    sd
<dbl> <dbl> <dbl> <dbl>
1 0.398   0.3   0.3 0.561

## Describing new_COVID_variant with graphics

Histogram for all students

ggplot(class_survey, aes(x = new_COVID_variant)) +
geom_histogram()

Histograms from first year and non-first year students

ggplot(class_survey, aes(x = new_COVID_variant)) +
geom_histogram() +
facet_wrap(vars(first_year))

## Example 1

What is the mean probability of new_COVID_variant for students who were very confident that we could engineer our way out of the effects of climate change (6 or above on climate_change)?

class_survey
# A tibble: 619 × 3
year                       new_COVID_variant first_year
<chr>                                  <dbl> <lgl>
1 I'm in my second year.                  0.25 FALSE
2 This is my first semester!              0.1  FALSE
3 This is my first semester!              0    FALSE
4 I'm in my second year.                  0.2  FALSE
5 I'm in my first year.                   0.9  TRUE
6 I'm in my second year.                  0.2  FALSE
7 I'm in my second year.                  0.4  FALSE
8 I'm in my second year.                  0    FALSE
9 I'm in my second year.                  0.2  FALSE
10 I'm in my first year.                   0.3  TRUE
# … with 609 more rows
optimist_df <- filter(class_survey, climate_change >= 6)
summarize(optimist_df, mean(new_COVID_variant))
summarize(class_survey, mean(new_COVID_variant))
# A tibble: 1 × 1
mean(new_COVID_variant)
<dbl>
1                     0.368

## Example 2

What is the mean probability of new_COVID_variant for first-year students who were very confident that we could engineer our way out of the effects of climate change (6 or above on climate_change)?

data("class_survey")
optimist_df <- filter(class_survey,
climate_change >= 6,
year == "I'm in my first year.")
summarize(optimist_df, mean(new_COVID_variant))
# A tibble: 1 × 1
mean(new_COVID_variant)
<dbl>
1                     0.370

## > You can string together conditions by adding them as arguments to filter() separated by commas.

What else can logical vectors be used for?

summarize(class_survey, mean(year == "I'm in my first year."))

What is will this line of code return?

Respond at pollev.com.

## Boolean Algebra

Logical vectors have a dual representation as TRUE FALSE and 1, 0, so you can do math on logicals accordingly.

TRUE + TRUE
 2
TRUE * TRUE
 1

Taking the mean of a logical vector is equivalent to find the proportion of rows that are TRUE (i.e. the proportion of rows that meet the condition).

# Worksheet - Digital

25:00