Quiz 4
Ethan Robbins
10/23/2020
For this quiz, you are going to use orange juice data. This data set is originally used in a machine learning (ML) class, with the goal to predict which of the two brands of orange juices the customers bought. Of course, you are not building a ML algorithm in this quiz. I just wanted to provide you with the context of the data.
The response variable (that ML algorithm is built to predict) is Purchase, which takes either CH (Citrus Hill) or MM (Minute Maid). The predictor variables (that ML algorithm uses to make predictions) are characteristics of the customer and the product itself. Together, the data set has 18 variables.WeekofPurchase is the week of purchase. LoyalCH is customer brand loyalty for CH (how loyal the customer is for CH on a scale of 0-1), and is the only variable that characterizes customers. All other variables are characteristics of the product or stores the sale occurred at. For more information on the data set, click the link below and scroll down to page 11. https://cran.r-project.org/web/packages/ISLR/ISLR.pdf
# Load the package
library(tidyverse)
# Import data
Orange <- read.csv('https://raw.githubusercontent.com/selva86/datasets/master/orange_juice_withmissing.csv', stringsAsFactors = TRUE) %>%
mutate(STORE = as.factor(STORE),
StoreID = as.factor(StoreID))
# Print the first 6 rows
head(Orange)## Purchase WeekofPurchase StoreID PriceCH PriceMM DiscCH DiscMM SpecialCH
## 1 CH 237 1 1.75 1.99 0.00 0.0 0
## 2 CH 239 1 1.75 1.99 0.00 0.3 0
## 3 CH 245 1 1.86 2.09 0.17 0.0 0
## 4 MM 227 1 1.69 1.69 0.00 0.0 0
## 5 CH 228 7 1.69 1.69 0.00 0.0 0
## 6 CH 230 7 1.69 1.99 0.00 0.0 0
## SpecialMM LoyalCH SalePriceMM SalePriceCH PriceDiff Store7 PctDiscMM
## 1 0 0.500000 1.99 1.75 0.24 No 0.000000
## 2 1 0.600000 1.69 1.75 -0.06 No 0.150754
## 3 0 0.680000 2.09 1.69 0.40 No 0.000000
## 4 0 0.400000 1.69 1.69 0.00 No 0.000000
## 5 0 0.956535 1.69 1.69 0.00 Yes 0.000000
## 6 1 0.965228 1.99 1.69 0.30 Yes 0.000000
## PctDiscCH ListPriceDiff STORE
## 1 0.000000 0.24 1
## 2 0.000000 0.24 1
## 3 0.091398 0.23 1
## 4 0.000000 0.00 1
## 5 0.000000 0.00 0
## 6 0.000000 0.30 0# Get a sense of the dataset
glimpse(Orange)## Rows: 1,070
## Columns: 18
## $ Purchase <fct> CH, CH, CH, MM, CH, CH, CH, CH, CH, CH, CH, CH, CH, ...
## $ WeekofPurchase <int> 237, 239, 245, 227, 228, 230, 232, 234, 235, 238, 24...
## $ StoreID <fct> 1, 1, 1, 1, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 1, 2...
## $ PriceCH <dbl> 1.75, 1.75, 1.86, 1.69, 1.69, 1.69, 1.69, 1.75, 1.75...
## $ PriceMM <dbl> 1.99, 1.99, 2.09, 1.69, 1.69, 1.99, 1.99, 1.99, 1.99...
## $ DiscCH <dbl> 0.00, 0.00, 0.17, 0.00, 0.00, 0.00, 0.00, 0.00, 0.00...
## $ DiscMM <dbl> 0.00, 0.30, 0.00, 0.00, 0.00, 0.00, 0.40, 0.40, 0.40...
## $ SpecialCH <int> 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0...
## $ SpecialMM <int> 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1...
## $ LoyalCH <dbl> 0.500000, 0.600000, 0.680000, 0.400000, 0.956535, 0....
## $ SalePriceMM <dbl> 1.99, 1.69, 2.09, 1.69, 1.69, 1.99, 1.59, 1.59, 1.59...
## $ SalePriceCH <dbl> 1.75, 1.75, 1.69, 1.69, 1.69, 1.69, 1.69, 1.75, 1.75...
## $ PriceDiff <dbl> 0.24, -0.06, 0.40, 0.00, 0.00, 0.30, -0.10, -0.16, -...
## $ Store7 <fct> No, No, No, No, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Y...
## $ PctDiscMM <dbl> 0.000000, 0.150754, 0.000000, 0.000000, 0.000000, 0....
## $ PctDiscCH <dbl> 0.000000, 0.000000, 0.091398, 0.000000, 0.000000, 0....
## $ ListPriceDiff <dbl> 0.24, 0.24, 0.23, 0.00, 0.00, 0.30, 0.30, 0.24, 0.24...
## $ STORE <fct> 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2...summary(Orange)## Purchase WeekofPurchase StoreID PriceCH PriceMM
## CH:653 Min. :227.0 1 :157 Min. :1.690 Min. :1.690
## MM:417 1st Qu.:240.0 2 :222 1st Qu.:1.790 1st Qu.:1.990
## Median :257.0 3 :196 Median :1.860 Median :2.090
## Mean :254.4 4 :139 Mean :1.867 Mean :2.085
## 3rd Qu.:268.0 7 :355 3rd Qu.:1.990 3rd Qu.:2.180
## Max. :278.0 NA's: 1 Max. :2.090 Max. :2.290
## NA's :1 NA's :4
## DiscCH DiscMM SpecialCH SpecialMM
## Min. :0.00000 Min. :0.0000 Min. :0.000 Min. :0.0000
## 1st Qu.:0.00000 1st Qu.:0.0000 1st Qu.:0.000 1st Qu.:0.0000
## Median :0.00000 Median :0.0000 Median :0.000 Median :0.0000
## Mean :0.05196 Mean :0.1234 Mean :0.147 Mean :0.1624
## 3rd Qu.:0.00000 3rd Qu.:0.2300 3rd Qu.:0.000 3rd Qu.:0.0000
## Max. :0.50000 Max. :0.8000 Max. :1.000 Max. :1.0000
## NA's :2 NA's :4 NA's :2 NA's :5
## LoyalCH SalePriceMM SalePriceCH PriceDiff Store7
## Min. :0.000011 Min. :1.190 Min. :1.390 Min. :-0.6700 No :714
## 1st Qu.:0.320000 1st Qu.:1.690 1st Qu.:1.750 1st Qu.: 0.0000 Yes:356
## Median :0.600000 Median :2.090 Median :1.860 Median : 0.2300
## Mean :0.565203 Mean :1.962 Mean :1.816 Mean : 0.1463
## 3rd Qu.:0.850578 3rd Qu.:2.130 3rd Qu.:1.890 3rd Qu.: 0.3200
## Max. :0.999947 Max. :2.290 Max. :2.090 Max. : 0.6400
## NA's :5 NA's :5 NA's :1 NA's :1
## PctDiscMM PctDiscCH ListPriceDiff STORE
## Min. :0.00000 Min. :0.00000 Min. :0.000 0 :356
## 1st Qu.:0.00000 1st Qu.:0.00000 1st Qu.:0.140 1 :157
## Median :0.00000 Median :0.00000 Median :0.240 2 :222
## Mean :0.05939 Mean :0.02732 Mean :0.218 3 :194
## 3rd Qu.:0.11268 3rd Qu.:0.00000 3rd Qu.:0.300 4 :139
## Max. :0.40201 Max. :0.25269 Max. :0.440 NA's: 2
## NA's :5 NA's :2Q1 empirical rule Your favorite orange juice brand is Citrus Hill. You went to a local market and found that it is sold at $2.20. You are surprised at the price tag, which seems too pricey. You wonder how rare it would be to encounter a Citrus Hill orange juice at this price. Fortunately, you have a dataset of the Citrus Hill juice prices. How would you use the dataset to answer your question? Describe.
you would use the empirical rule, you see if the data is distributed normally or not. If so you can determine how many standard deviations the price of the $2.20 orange juice if from the mean.
Q2 PriceCH Is the data normally distributed? Plot Citrus Hill orange juice price in a histogram.
ggplot(Orange, aes(x = PriceCH)) +
geom_histogram()## Warning: Removed 1 rows containing non-finite values (stat_bin).
No, the Data is not normally distributed
Q3 PriceCH Calculate the mean of price.
mean(Orange$PriceCH, na.rm = TRUE)## [1] 1.867428Q4 PriceCH Calculate the standard deviation of price.
sd(Orange$PriceCH, na.rm = TRUE)## [1] 0.1020172Q5 empirical rule Base on your analysis in Q2, would it be appropriate to use the empirical rule for Q1? Why? Why not?
No, it would not be appropriate to use empirical rule. You can only use this when the data is normally distributed, in this case it is not.
Q6 empirical rule Let’s just assume that the prices are normally distributed for the sake of discussion. Would you pay $2.20 for the Citrus Hill orange juice? Or walk away?
I would walk away, the mean price is less than the price listed for the orange juice.
Q7 Why do you think the empirical rule uses the standard deviation as a measure of spread instead of the variance?
This is because variance is basically the same as the standard deviation but variance is squared. So everything would be square rooted.
Q8 Hide the messages, but display the code and its results on the webpage.
Hint: Use message, echo and results in the chunk options. Refer to the RMarkdown Reference Guide.