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summary(cars)
##      speed           dist       
##  Min.   : 4.0   Min.   :  2.00  
##  1st Qu.:12.0   1st Qu.: 26.00  
##  Median :15.0   Median : 36.00  
##  Mean   :15.4   Mean   : 42.98  
##  3rd Qu.:19.0   3rd Qu.: 56.00  
##  Max.   :25.0   Max.   :120.00

install.packages(“readr”) library(readr) data <- read_csv(“ab_testing1.csv”) ls(data) head(data)

data\(Ads <- factor(data\)Ads) is.factor(data$Ads)

summary(lm(Purchase~Ads, data = data))

cols(ads = col_double(), Purchases = col_double())

display <- read_csv(“ab_testing1.csv”) ls(display) head(display)

display\(Ads <- factor(display\)Ads) is.factor(display$Ads)

summary(lm(Purchase~Ads, data = display))

Hypothesis: I believe that Ad 0 will generate the least number of sales because no ad i shown to the group for ad 0. I also think Ad 2 will generate the most since it’s the second ad the company is showing. I think ad 1 will be in the middle, it will generate sales but not as much as Ad 2.

Outcomes: Ad=1 generated the most sales from the targeted group, Ad=1 had a Std. of 69.71 Ad=2 generated less sales than ad=1 with a Std. of 24.75. There were not as many purchases from the group after watching ad=2 Ad=0 generated the least number of sales.

My hypothesis was partially correct.

Recommendation: Judging by the data I would suggest that the company uses Ad=1 since it produces the greatest number of purchases. When marketing products the company should stick with Ad=1 to maximize purchase volume.

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