empirical rule
Your favorite orange juice brand is Minute Maid. You went to a local market and found that it is sold at $2.50. You are surprised at the price tag, which seems too pricey. You wonder how rare it would be to encounter a Minute Maid orange juice at this price. Fortunately, you have a dataset of the Minute Maid juice prices. How would you use the dataset to answer your question? Describe.PriceMM
Is the data normally distributed? Plot Minute Maid orange juice price in a histogram.PriceMM
Calculate the mean of price.PriceMM
Calculate the standard deviation of price.empirical rule
Base on your analysis in Q2, would it be appropriate to use the empirical rule for Q1? Why? Why not?empirical rule
Let’s just assume that the prices are normally distributed for the sake of discussion. Would you pay $2.50 for the Minute Maid orange juice? Or walk away?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 :2
empirical rule
Your favorite orange juice brand is Minute Maid. You went to a local market and found that it is sold at $2.50. You are surprised at the price tag, which seems too pricey. You wonder how rare it would be to encounter a Minute Maid orange juice at this price. Fortunately, you have a dataset of the Minute Maid juice prices. How would you use the dataset to answer your question? Describe.Hint: Discuss all of the following topics in your answer: the normal distribution, the mean, and the standard deviation.
I would use the data set to calculate what the mean or average price of Minute Made (PriceMM). This would show me what the average price is to compare it to the $2.50 price I saw at the store. Next I would plot the data for the price of MM to see how the distribution is. That would allow me to see how different the prices really are between locations.
PriceMM
Is the data normally distributed? Plot Minute Maid orange juice price in a histogram.Orange %>%
ggplot(aes(PriceMM)) +
geom_histogram()
PriceMM
Calculate the mean of price.Hint: You may add the na.rm = TRUE
argument in the mean() function, if the function returns NA. It means that the variable has at least one row with NA.
mean(Orange$PriceMM, na.rm = TRUE)
## [1] 2.085038
PriceMM
Calculate the standard deviation of price.Hint: You may add the na.rm = TRUE
argument in the mean() function, if the function returns NA. It means that the variable has at least one row with NA.
sd(Orange$PriceMM, na.rm = TRUE)
## [1] 0.1344285
empirical rule
Base on your analysis in Q2, would it be appropriate to use the empirical rule for Q1? Why? Why not?Hint: Discuss characterstics of the normal distribution.
It would be appropriate to use the empirical rule because the mean price for the data is $2.08. The empirical rule can be used to find out the chances of finding a specific price.
empirical rule
Let’s just assume that the prices are normally distributed for the sake of discussion. Would you pay $2.50 for the Minute Maid orange juice? Or walk away?Hint: Discuss in terms of the mean, the standard deviation, and the probability.
I would probably walk away and find another juice. The mean price of the Juice is 2.08 and the price at the store was was 2.50. the standard deviation from the mean makes it so that you are better off finding another juice.
Hint: Discuss in terms of the unit of the variance and the standard deviation.
The empirical rule used the standard deviation as a measure of spread instead of the variance because standard deviation is simpler to interpret and determine than variance. Variance weighs outliers more heavily than the data closer to the mean, which shows why the empirical rule uses standard deviations because it relates to how far data is from the mean instead of to variance which relies more on outliers.
Hint: Use message
, echo
and results
in the chunk options. Refer to the RMarkdown Reference Guide.