SalePriceMM Calculate the mean price of Minute Maid orange joice.SalePriceMM Calculate the median price of Minute Maid orange joice.SalePriceMM Plot Minute Maid orange joice prices in a histogram.Law of Large Numbers We learned that the sample mean is not likley to be representative of the population mean when a sample is too small. Explain why?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 :2The median is not affected by very large or very small values. So, when a data set has numbers that are drastically lower or drastically higher than the other numbers presented, median would be better to use that mode to finde centrality.
SalePriceMM Calculate the mean price of Minute Maid orange joice.Hint: Code it so that the outcome is a scalar, not a data frame. It’s the same code you learned in Quiz3-b. Save the result under mean_pr. mean_pr <-mean((Orange$SalePriceMM) , na.rm = TRUE) mean_pr ## [1] 1.961934
SalePriceMM Calculate the median price of Minute Maid orange joice.Hint: Replace mean with median in the code in Q2. Save the result under median_pr. median_pr <- median((Orange$SalePriceMM) , na.rm = TRUE) median_pr ## [1] 2.09
SalePriceMM Plot Minute Maid orange joice prices in a histogram.Hint: Refer to the code in Data Visualization with R: Ch3.2.1 Histogram. ggplot(Orange, aes(x = PriceMM)) + geom_histogram(fill = “black” color = “white”) + labs(title = “Minute maid OJ Prices”, x = “Sale Price”)
Hint: Copy the code from Q4 and add two lines of the geom_vline() function in the code for vertical lines of the mean and the median home prices. Google geom_vline() for its documentation. ggplot(Orange, aes(x = PriceMM)) + geom_histogram(fill = “black”, color = “white”) + labs(title=“MM OJ Prices”, x = “Sale Price”) + geom_vline(xintercept = mean_pr, color = ‘yellow’) + geom_vline(xintercept = median_pr, color = ‘green’)
In the case of Minute Maid Orange Juice Prices, it would be more effective to use the median price, $2.09, to represent typical price. This is because the outlying values on the lower end of the graph influenced the mean to misrepresent the data, as the majority of values in the set are well above it.
Law of Large Numbers We learned that the sample mean is not likley to be representative of the population mean when a sample is too small. Explain why?This is the inverse of the explanation of Question 6. Whereas in that question we don’t want the skew from outlier, a sample size too small would want them. If there are too few or too many, however, the produced representation may still be skewed. ## Q8 Hide the messages and warnings, 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.