Quiz3
Colton Petrosino
- Q1 Describe a situation when the median is more appropriate than the mean as a measure of centrality.
- Q2
SalePriceMMCalculate the mean price of Minute Maid orange joice. - Q3
SalePriceMMCalculate the median price of Minute Maid orange joice. - Q4
SalePriceMMPlot Minute Maid orange joice prices in a histogram. - Q5 Add the vertical lines of mean_pr and median_pr in the histogram.
- Q6 Which of the two measures would be more apprrpriate to represent the typical price? Why?
- Q7
Law of Large NumbersWe learned that the sample mean is not likley to be representative of the population mean when a sample is too small. Explain why? - Q8 Hide the messages and warnings, but display the code and its results on the webpage.
- Q9 Display the title and your name correctly at the top of the webpage.
- Q10 Use the correct slug.
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)## -- Attaching packages -------------------## v ggplot2 3.3.2 v purrr 0.3.4
## v tibble 3.0.3 v dplyr 1.0.2
## v tidyr 1.1.2 v stringr 1.4.0
## v readr 1.3.1 v forcats 0.5.0## -- Conflicts --- tidyverse_conflicts() --
## x dplyr::filter() masks stats::filter()
## x dplyr::lag() masks stats::lag()# 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 Describe a situation when the median is more appropriate than the mean as a measure of centrality.
I want to know how far the typical PSU student travels to come to campus. But there are to many students to collect data from. So I collected a sample from 10 students to get the mean. Would the mean of the 10 student sample be a good representation of all PSU students? What if there is a student from Alaska in the 10 student sample? Would that skew the mean? Would the Alaskan home bound person have same degree of influence had I collected a sample of 100 students?
The mean of the data can be skewed if the data set is very small and if on of the data point is way different from the rest of the data. This one data point will cause the mean to be way different than it really is due to the one out lair in the data, you would need to collect more data in order for this one out liar to not have an impact on the mean of the data.
This is when using the median of the data would be better, since the data set is smaller, you can use the median to get rid of the out lair to get a better answer to the measure of centrality.
Q2 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$SalePriceMM, na.rm = TRUE)
mean_pr## [1] 1.961934Q3 SalePriceMM Calculate the median price of Minute Maid orange joice.
Hint: Replace mean with median in the code in Q2. Save the result under median_pr.
median_pr <- median(Orange$SalePriceMM, na.rm = TRUE)
median_pr## [1] 2.09Q4 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 = SalePriceMM)) +
geom_histogram() +
labs(title = "Minute Maid Prices",
x = "price")## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
Q5 Add the vertical lines of mean_pr and median_pr in the histogram.
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 = SalePriceMM)) +
geom_histogram() +
labs(title = "Minute Maid Prices",
x = "price") + geom_vline(xintercept = mean_pr, color = "red") + geom_vline(xintercept = median_pr, color = "blue")## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
Q6 Which of the two measures would be more apprrpriate to represent the typical price? Why?
Median because the data is skewed, meaning the data has out liars which throws off the mean
Q7 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 because one out lairer in the data can through off the mean, this casues the data to be skewed. You need more data to get a good mean, escipally whens theres out liars in the data
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.