Skills Check Solutions
PART 1: Data Wrangling
library(tidyverse)1) Make a new data set that has the average depth and price of the diamonds in the data set.
avgPandD = diamonds %>% summarise(mean(price), mean(depth))
avgPandD## # A tibble: 1 x 2
## `mean(price)` `mean(depth)`
## <dbl> <dbl>
## 1 3933. 61.72) Add a new column to the data set that records each diamond’s price per carat.
diamonds_ppc = diamonds %>% mutate(price_per_c = price/carat)
diamonds_ppc## # A tibble: 53,940 x 11
## carat cut color clarity depth table price x y z price_per_c
## <dbl> <ord> <ord> <ord> <dbl> <dbl> <int> <dbl> <dbl> <dbl> <dbl>
## 1 0.23 Ideal E SI2 61.5 55 326 3.95 3.98 2.43 1417.
## 2 0.21 Premium E SI1 59.8 61 326 3.89 3.84 2.31 1552.
## 3 0.23 Good E VS1 56.9 65 327 4.05 4.07 2.31 1422.
## 4 0.290 Premium I VS2 62.4 58 334 4.2 4.23 2.63 1152.
## 5 0.31 Good J SI2 63.3 58 335 4.34 4.35 2.75 1081.
## 6 0.24 Very Good J VVS2 62.8 57 336 3.94 3.96 2.48 1400
## 7 0.24 Very Good I VVS1 62.3 57 336 3.95 3.98 2.47 1400
## 8 0.26 Very Good H SI1 61.9 55 337 4.07 4.11 2.53 1296.
## 9 0.22 Fair E VS2 65.1 61 337 3.87 3.78 2.49 1532.
## 10 0.23 Very Good H VS1 59.4 61 338 4 4.05 2.39 1470.
## # … with 53,930 more rows3) Create a new data set that groups diamonds by their cut and displays the average price of each group.
cp = diamonds %>% group_by(cut) %>% summarise(mean(price))## `summarise()` ungrouping output (override with `.groups` argument)cp## # A tibble: 5 x 2
## cut `mean(price)`
## <ord> <dbl>
## 1 Fair 4359.
## 2 Good 3929.
## 3 Very Good 3982.
## 4 Premium 4584.
## 5 Ideal 3458.4) Create a new data set that groups diamonds by color and displays the average depth and average table for each group.
cd = diamonds %>% group_by(color) %>% summarise(mean(depth), mean(table))## `summarise()` ungrouping output (override with `.groups` argument)cd## # A tibble: 7 x 3
## color `mean(depth)` `mean(table)`
## <ord> <dbl> <dbl>
## 1 D 61.7 57.4
## 2 E 61.7 57.5
## 3 F 61.7 57.4
## 4 G 61.8 57.3
## 5 H 61.8 57.5
## 6 I 61.8 57.6
## 7 J 61.9 57.8Extra-Credit) Add two columns to the diamonds data set. The first column should display the average depth of diamonds in the diamond’s color group. The second column should display the average table of diamonds in the diamonds color group
df1 = diamonds %>% group_by(color) %>% summarise(mean(depth))## `summarise()` ungrouping output (override with `.groups` argument)df2 = diamonds %>% group_by(color) %>% summarise(mean(table))## `summarise()` ungrouping output (override with `.groups` argument)main = left_join(diamonds, df1)## Joining, by = "color"main## # A tibble: 53,940 x 11
## carat cut color clarity depth table price x y z `mean(depth)`
## <dbl> <ord> <ord> <ord> <dbl> <dbl> <int> <dbl> <dbl> <dbl> <dbl>
## 1 0.23 Ideal E SI2 61.5 55 326 3.95 3.98 2.43 61.7
## 2 0.21 Premium E SI1 59.8 61 326 3.89 3.84 2.31 61.7
## 3 0.23 Good E VS1 56.9 65 327 4.05 4.07 2.31 61.7
## 4 0.290 Premium I VS2 62.4 58 334 4.2 4.23 2.63 61.8
## 5 0.31 Good J SI2 63.3 58 335 4.34 4.35 2.75 61.9
## 6 0.24 Very G… J VVS2 62.8 57 336 3.94 3.96 2.48 61.9
## 7 0.24 Very G… I VVS1 62.3 57 336 3.95 3.98 2.47 61.8
## 8 0.26 Very G… H SI1 61.9 55 337 4.07 4.11 2.53 61.8
## 9 0.22 Fair E VS2 65.1 61 337 3.87 3.78 2.49 61.7
## 10 0.23 Very G… H VS1 59.4 61 338 4 4.05 2.39 61.8
## # … with 53,930 more rows5) Which color diamonds seem to be largest on average (in terms of carats)?
size_v_col = diamonds %>% group_by(color) %>% summarise(mean(carat))## `summarise()` ungrouping output (override with `.groups` argument)size_v_col## # A tibble: 7 x 2
## color `mean(carat)`
## <ord> <dbl>
## 1 D 0.658
## 2 E 0.658
## 3 F 0.737
## 4 G 0.771
## 5 H 0.912
## 6 I 1.03
## 7 J 1.16Color J seems to be the largest on average.
6) What color of diamonds occurs the most frequently among diamonds with ideal cuts?
cut_color = diamonds %>% filter(cut == "Ideal") %>% count(color)
cut_color## # A tibble: 7 x 2
## color n
## <ord> <int>
## 1 D 2834
## 2 E 3903
## 3 F 3826
## 4 G 4884
## 5 H 3115
## 6 I 2093
## 7 J 896## You can also do
cut_color2 = diamonds %>%
filter(cut == "Ideal")%>%
group_by(color)%>%
summarise(n=n())## `summarise()` ungrouping output (override with `.groups` argument)cut_color2## # A tibble: 7 x 2
## color n
## <ord> <int>
## 1 D 2834
## 2 E 3903
## 3 F 3826
## 4 G 4884
## 5 H 3115
## 6 I 2093
## 7 J 896There are the most ideally cut diamonds with a color of G.
7) Which clarity of diamonds has the largest average table per carats?
tpc = diamonds %>% group_by(clarity) %>% summarise(mean(table/carat))## `summarise()` ungrouping output (override with `.groups` argument)tpc## # A tibble: 8 x 2
## clarity `mean(table/carat)`
## <ord> <dbl>
## 1 I1 56.3
## 2 SI2 69.1
## 3 SI1 89.6
## 4 VS2 103.
## 5 VS1 107.
## 6 VVS2 127.
## 7 VVS1 141.
## 8 IF 140.VVS1 has the highest average table per carats.
8) What is the average price per carat of diamonds that cost more than $10000?
avg_ppc = diamonds %>% filter(price > 10000) %>% summarise(mean(price/carat))
avg_ppc## # A tibble: 1 x 1
## `mean(price/carat)`
## <dbl>
## 1 8044.9) Of the diamonds that cost more than $10000, what is the most common clarity?
comclar = diamonds %>% filter(price > 10000) %>% count(clarity)
comclar## # A tibble: 8 x 2
## clarity n
## <ord> <int>
## 1 I1 30
## 2 SI2 1239
## 3 SI1 1184
## 4 VS2 1155
## 5 VS1 747
## 6 VVS2 452
## 7 VVS1 247
## 8 IF 168## YOU CAN ALSO DO
comclar2 = diamonds %>%
filter(price > 10000) %>%
group_by(clarity)%>%
summarise(n=n())## `summarise()` ungrouping output (override with `.groups` argument)comclar2## # A tibble: 8 x 2
## clarity n
## <ord> <int>
## 1 I1 30
## 2 SI2 1239
## 3 SI1 1184
## 4 VS2 1155
## 5 VS1 747
## 6 VVS2 452
## 7 VVS1 247
## 8 IF 168The most common clarity for diamonds over $10000 is SI2.
PART 2: Data Exploration and Visualization
data("ToothGrowth")
?ToothGrowth
str(ToothGrowth)## 'data.frame': 60 obs. of 3 variables:
## $ len : num 4.2 11.5 7.3 5.8 6.4 10 11.2 11.2 5.2 7 ...
## $ supp: Factor w/ 2 levels "OJ","VC": 2 2 2 2 2 2 2 2 2 2 ...
## $ dose: num 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 ...1) What do the rows of this dataset represent?
Individual Guinea Pigs.
2) What do the columns of this dataset represent? Indicate whether each variable in the study is numerical or categorical. If numerical, identify as continuous or discrete. If categorical, indicate if the variable is ordinal.
The first column represents tooth length. This variable is numerical and continuous. The second represents supplement type. This is categorical but not ordinal. The final column represents dose. This is numeric and discrete.
3) What are the response and explanatory variables in this study?
The response variable is tooth length in this study. The explanatory variable are supplement type and dose.
4) Create a hypothesis about supplement treatment and dosage levels, without first looking at the data.
Vitamin C supplementation does not impact the length of odontoblasts in Guinea Pigs.
5) Use ggplot to create a side-by-side boxplot, which illustrates the distribution of each supplement treatment and allows for both visual comparison across and within treatments.
ggplot(ToothGrowth, aes(supp, len, fill = supp)) + geom_boxplot()
6) Now add facets to your data to compare across dosage as well.
ggplot(ToothGrowth, aes(supp, len, fill = supp)) + geom_boxplot() + facet_wrap(~dose)
7) Describe any possible trends in these data. Explain in the context of this study.
When viewing the box plot from question 6 you can immediately see that as the supplement dosage increased so did the tooth length. Additionally, the Guinea Pigs that received OJ had significantly a higher response value in Guinea Pigs that received a dosage of 0.5 or 1 milligrams a day. In Guinea Pigs that received 2 mg/day, the medians are about even.
8) Did you see anything surprising? Does your hypothesis appear to be supported? Note that this is not a formal hypothesis test but rather an exploration.
Initially, it looks like vitamin C supplementation could impact odontoblast lengths in Guinea Pigs. This would be a rejection of the null hypothesis. Additionally, it appears orange juice could be a more effective way of delivering the vitamin. When the dosage is lowered, it seems the delivery method is more impactful, potentially due to asorbic being a less efficient method of delivery.