Introduction

This report analyzes whether two different in-store advertisement types affect grape juice sales. Advertisement type 0 represents the natural-production advertisement, while advertisement type 1 represents the family-health advertisement. A Welch independent-samples t-test will be used to compare the average sales between the two groups.

Research Hypotheses

Null hypothesis (H₀): There is no difference in average grape juice sales between the natural-production advertisement and the family-health advertisement.

Alternative hypothesis (H₁): There is a difference in average grape juice sales between the two advertisement types.

Load the Data

data <- read.csv("grapeJuice.csv")
head(data)
##   X Sales price ad_type price_apple price_cookies
## 1 1   222  9.83       0        7.36          8.80
## 2 2   201  9.72       1        7.43          9.62
## 3 3   247 10.15       1        7.66          8.90
## 4 4   169 10.04       0        7.57         10.26
## 5 5   317  8.38       1        7.33          9.54
## 6 6   227  9.74       0        7.51          9.49
summary(data)
##        X             Sales           price           ad_type     price_apple   
##  Min.   : 1.00   Min.   :131.0   Min.   : 8.200   Min.   :0.0   Min.   :7.300  
##  1st Qu.: 8.25   1st Qu.:182.5   1st Qu.: 9.585   1st Qu.:0.0   1st Qu.:7.438  
##  Median :15.50   Median :204.5   Median : 9.855   Median :0.5   Median :7.580  
##  Mean   :15.50   Mean   :216.7   Mean   : 9.738   Mean   :0.5   Mean   :7.659  
##  3rd Qu.:22.75   3rd Qu.:244.2   3rd Qu.:10.268   3rd Qu.:1.0   3rd Qu.:7.805  
##  Max.   :30.00   Max.   :335.0   Max.   :10.490   Max.   :1.0   Max.   :8.290  
##  price_cookies   
##  Min.   : 8.790  
##  1st Qu.: 9.190  
##  Median : 9.515  
##  Mean   : 9.622  
##  3rd Qu.:10.140  
##  Max.   :10.580

Histogram of Sales

hist(data$Sales,
     main = "Histogram Plot for Sales Data",
     xlab = "Sales_Grape",
     prob = TRUE)

lines(density(data$Sales),
      lty = "dashed",
      lwd = 2.5,
      col = "blue")

The histogram shows that most grape juice sales are concentrated between approximately 150 and 250 units. The dashed blue density line shows the overall shape of the sales distribution.

Compare Average Sales by Advertisement Type

sales_ad_nature <- subset(data, ad_type == 0)
sales_ad_family <- subset(data, ad_type == 1)

mean(sales_ad_nature$Sales)
## [1] 186.6667
mean(sales_ad_family$Sales)
## [1] 246.6667

The average sales for the natural-production advertisement were 186.6667 units. The average sales for the family-health advertisement were 246.6667 units. Therefore, the family-health advertisement had the higher average sales.

Normality Tests

shapiro.test(sales_ad_nature$Sales)
## 
##  Shapiro-Wilk normality test
## 
## data:  sales_ad_nature$Sales
## W = 0.94255, p-value = 0.4155
shapiro.test(sales_ad_family$Sales)
## 
##  Shapiro-Wilk normality test
## 
## data:  sales_ad_family$Sales
## W = 0.89743, p-value = 0.08695

The Shapiro-Wilk test produced a p-value of 0.4155 for the natural-production advertisement and 0.08695 for the family-health advertisement. Both p-values are greater than 0.05, so the data do not show a significant violation of the normality assumption.

Welch Independent-Samples T-Test

t.test(Sales ~ ad_type, data = data)
## 
##  Welch Two Sample t-test
## 
## data:  Sales by ad_type
## t = -3.7515, df = 25.257, p-value = 0.0009233
## alternative hypothesis: true difference in means between group 0 and group 1 is not equal to 0
## 95 percent confidence interval:
##  -92.92234 -27.07766
## sample estimates:
## mean in group 0 mean in group 1 
##        186.6667        246.6667

The Welch independent-samples t-test produced a t-value of -3.7515, with 25.257 degrees of freedom and a p-value of 0.0009233. Because the p-value is less than 0.05, I reject the null hypothesis. There is a statistically significant difference in average grape juice sales between the two advertisement types.

The family-health advertisement had higher average sales at 246.6667 units compared with 186.6667 units for the natural-production advertisement. Based on these results, the family-health advertisement appears to be more effective.

Reflection: Why R Adds Career Value

R programming could add value to my career for several reasons. First, R can analyze large datasets more efficiently than manually completing calculations in Excel. Second, R makes analysis reproducible because the same script can be saved, checked, and reused when new data are added. Third, R provides powerful statistical tests and professional-quality graphs that can help businesses understand patterns and make better decisions. Learning R would improve my technical skills and make me more competitive for careers involving marketing analytics, business intelligence, research, or data analysis.