Assignment 2 Finalization
2024-11-08
Car Total Setting Working Directory
getwd()## [1] "/Users/markyuhasz/FallWinter 2024 Classes/MKTG3P98-Business analytics and intell/Assignment 2"setwd("/Users/markyuhasz/FallWinter 2024 Classes/MKTG3P98-Business analytics and intell/Assignment 2")Viewing Car Total
library(readxl)
library(ggplot2)
library(readr)
library(stringr)
library(dplyr)##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionCar_Total<-read.csv("Car_Total.csv")
names(Car_Total)## [1] "Resp" "Att_1" "Att_2" "Enj_1" "Enj_2"
## [6] "Perform_1" "Perform_2" "Perform_3" "WOM_1" "WOM_2"
## [11] "Futu_Pur_1" "Futu_Pur_2" "Valu_Percp_1" "Valu_Percp_2" "Pur_Proces_1"
## [16] "Pur_Proces_2" "Residence" "Pay_Meth" "Insur_Type" "Gender"
## [21] "Age" "Education" "X" "Region" "Model"
## [26] "MPG" "Cyl" "acc1" "C_cost." "H_Cost"
## [31] "Post.Satis"summary(Car_Total)## Resp Att_1 Att_2 Enj_1
## Length:1049 Min. :1.000 Min. :1.000 Min. :1.000
## Class :character 1st Qu.:4.000 1st Qu.:4.000 1st Qu.:4.000
## Mode :character Median :6.000 Median :6.000 Median :6.000
## Mean :4.882 Mean :5.287 Mean :5.378
## 3rd Qu.:6.000 3rd Qu.:6.000 3rd Qu.:7.000
## Max. :7.000 Max. :7.000 Max. :7.000
## NA's :4 NA's :4
## Enj_2 Perform_1 Perform_2 Perform_3
## Min. :1.000 Min. :1.000 Min. :1.000 Min. :1.000
## 1st Qu.:3.000 1st Qu.:4.000 1st Qu.:4.000 1st Qu.:3.000
## Median :5.000 Median :5.000 Median :5.000 Median :5.000
## Mean :4.575 Mean :4.947 Mean :4.831 Mean :4.217
## 3rd Qu.:6.000 3rd Qu.:6.000 3rd Qu.:6.000 3rd Qu.:6.000
## Max. :7.000 Max. :7.000 Max. :7.000 Max. :7.000
## NA's :4 NA's :2 NA's :4 NA's :1
## WOM_1 WOM_2 Futu_Pur_1 Futu_Pur_2 Valu_Percp_1
## Min. :1.000 Min. :1.00 Min. :1.000 Min. :1.000 Min. :1.000
## 1st Qu.:4.000 1st Qu.:4.00 1st Qu.:4.000 1st Qu.:5.000 1st Qu.:5.000
## Median :6.000 Median :6.00 Median :6.000 Median :6.000 Median :6.000
## Mean :5.286 Mean :5.35 Mean :5.321 Mean :5.371 Mean :5.411
## 3rd Qu.:7.000 3rd Qu.:6.00 3rd Qu.:6.000 3rd Qu.:6.000 3rd Qu.:6.000
## Max. :7.000 Max. :7.00 Max. :9.000 Max. :7.000 Max. :7.000
## NA's :1 NA's :3 NA's :5 NA's :2 NA's :4
## Valu_Percp_2 Pur_Proces_1 Pur_Proces_2 Residence
## Min. :1.000 Min. :1.000 Min. :1.000 Min. :1.000
## 1st Qu.:4.000 1st Qu.:5.000 1st Qu.:4.000 1st Qu.:1.000
## Median :5.000 Median :6.000 Median :5.000 Median :1.000
## Mean :5.114 Mean :5.256 Mean :4.923 Mean :1.474
## 3rd Qu.:6.000 3rd Qu.:6.000 3rd Qu.:6.000 3rd Qu.:2.000
## Max. :7.000 Max. :7.000 Max. :7.000 Max. :5.000
## NA's :1 NA's :3 NA's :4 NA's :5
## Pay_Meth Insur_Type Gender Age
## Min. :1.000 Length:1049 Length:1049 Min. :18.00
## 1st Qu.:1.000 Class :character Class :character 1st Qu.:23.00
## Median :2.000 Mode :character Mode :character Median :34.00
## Mean :2.153 Mean :35.22
## 3rd Qu.:3.000 3rd Qu.:48.00
## Max. :3.000 Max. :60.00
##
## Education X Region Model
## Min. :1.000 Mode:logical Length:1049 Length:1049
## 1st Qu.:2.000 NA's:1049 Class :character Class :character
## Median :2.000 Mode :character Mode :character
## Mean :1.989
## 3rd Qu.:2.000
## Max. :3.000
##
## MPG Cyl acc1 C_cost. H_Cost
## Min. :14.00 Min. :4.0 Min. :3.600 Min. : 7.00 Min. : 6.000
## 1st Qu.:17.00 1st Qu.:4.0 1st Qu.:5.100 1st Qu.:10.00 1st Qu.: 8.000
## Median :19.00 Median :6.0 Median :6.500 Median :12.00 Median :10.000
## Mean :19.58 Mean :5.8 Mean :6.202 Mean :11.35 Mean : 9.634
## 3rd Qu.:22.00 3rd Qu.:6.0 3rd Qu.:7.500 3rd Qu.:13.00 3rd Qu.:11.000
## Max. :26.00 Max. :8.0 Max. :8.500 Max. :16.00 Max. :14.000
##
## Post.Satis
## Min. :2.00
## 1st Qu.:5.00
## Median :6.00
## Mean :5.28
## 3rd Qu.:6.00
## Max. :7.00
## Cleaning all NA Values for proper manipulation
numeric_cols <- sapply(Car_Total, is.numeric)
Car_Total[, numeric_cols] <- lapply(Car_Total[, numeric_cols], function(x) {
mean_val <- mean(x, na.rm = TRUE)
x[is.na(x)] <- mean_val
return(x)
})
summary(Car_Total)## Resp Att_1 Att_2 Enj_1
## Length:1049 Min. :1.000 Min. :1.000 Min. :1.000
## Class :character 1st Qu.:4.000 1st Qu.:4.000 1st Qu.:5.000
## Mode :character Median :5.000 Median :6.000 Median :6.000
## Mean :4.882 Mean :5.287 Mean :5.378
## 3rd Qu.:6.000 3rd Qu.:6.000 3rd Qu.:7.000
## Max. :7.000 Max. :7.000 Max. :7.000
## Enj_2 Perform_1 Perform_2 Perform_3
## Min. :1.000 Min. :1.000 Min. :1.000 Min. :1.000
## 1st Qu.:3.000 1st Qu.:4.000 1st Qu.:4.000 1st Qu.:3.000
## Median :5.000 Median :5.000 Median :5.000 Median :5.000
## Mean :4.575 Mean :4.947 Mean :4.831 Mean :4.217
## 3rd Qu.:6.000 3rd Qu.:6.000 3rd Qu.:6.000 3rd Qu.:6.000
## Max. :7.000 Max. :7.000 Max. :7.000 Max. :7.000
## WOM_1 WOM_2 Futu_Pur_1 Futu_Pur_2 Valu_Percp_1
## Min. :1.000 Min. :1.00 Min. :1.000 Min. :1.000 Min. :1.000
## 1st Qu.:4.000 1st Qu.:4.00 1st Qu.:5.000 1st Qu.:5.000 1st Qu.:5.000
## Median :6.000 Median :6.00 Median :6.000 Median :6.000 Median :6.000
## Mean :5.286 Mean :5.35 Mean :5.321 Mean :5.371 Mean :5.411
## 3rd Qu.:7.000 3rd Qu.:6.00 3rd Qu.:6.000 3rd Qu.:6.000 3rd Qu.:6.000
## Max. :7.000 Max. :7.00 Max. :9.000 Max. :7.000 Max. :7.000
## Valu_Percp_2 Pur_Proces_1 Pur_Proces_2 Residence
## Min. :1.000 Min. :1.000 Min. :1.000 Min. :1.000
## 1st Qu.:4.000 1st Qu.:5.000 1st Qu.:4.000 1st Qu.:1.000
## Median :5.000 Median :6.000 Median :5.000 Median :1.000
## Mean :5.114 Mean :5.256 Mean :4.923 Mean :1.474
## 3rd Qu.:6.000 3rd Qu.:6.000 3rd Qu.:6.000 3rd Qu.:2.000
## Max. :7.000 Max. :7.000 Max. :7.000 Max. :5.000
## Pay_Meth Insur_Type Gender Age
## Min. :1.000 Length:1049 Length:1049 Min. :18.00
## 1st Qu.:1.000 Class :character Class :character 1st Qu.:23.00
## Median :2.000 Mode :character Mode :character Median :34.00
## Mean :2.153 Mean :35.22
## 3rd Qu.:3.000 3rd Qu.:48.00
## Max. :3.000 Max. :60.00
## Education X Region Model
## Min. :1.000 Mode:logical Length:1049 Length:1049
## 1st Qu.:2.000 NA's:1049 Class :character Class :character
## Median :2.000 Mode :character Mode :character
## Mean :1.989
## 3rd Qu.:2.000
## Max. :3.000
## MPG Cyl acc1 C_cost. H_Cost
## Min. :14.00 Min. :4.0 Min. :3.600 Min. : 7.00 Min. : 6.000
## 1st Qu.:17.00 1st Qu.:4.0 1st Qu.:5.100 1st Qu.:10.00 1st Qu.: 8.000
## Median :19.00 Median :6.0 Median :6.500 Median :12.00 Median :10.000
## Mean :19.58 Mean :5.8 Mean :6.202 Mean :11.35 Mean : 9.634
## 3rd Qu.:22.00 3rd Qu.:6.0 3rd Qu.:7.500 3rd Qu.:13.00 3rd Qu.:11.000
## Max. :26.00 Max. :8.0 Max. :8.500 Max. :16.00 Max. :14.000
## Post.Satis
## Min. :2.00
## 1st Qu.:5.00
## Median :6.00
## Mean :5.28
## 3rd Qu.:6.00
## Max. :7.00Separating Age groups into 3 age segments
Car_Total$AgeGrp<-cut(Car_Total$Age,
breaks = c(0, 30, 50, Inf),
labels = c("Young Adults", "Adults", "Mature Adults"),
right = FALSE)
names(Car_Total)## [1] "Resp" "Att_1" "Att_2" "Enj_1" "Enj_2"
## [6] "Perform_1" "Perform_2" "Perform_3" "WOM_1" "WOM_2"
## [11] "Futu_Pur_1" "Futu_Pur_2" "Valu_Percp_1" "Valu_Percp_2" "Pur_Proces_1"
## [16] "Pur_Proces_2" "Residence" "Pay_Meth" "Insur_Type" "Gender"
## [21] "Age" "Education" "X" "Region" "Model"
## [26] "MPG" "Cyl" "acc1" "C_cost." "H_Cost"
## [31] "Post.Satis" "AgeGrp"head(Car_Total$AgeGrp)## [1] Young Adults Young Adults Adults Young Adults Young Adults
## [6] Young Adults
## Levels: Young Adults Adults Mature AdultsSplit Car into Make and Model
Car_Total[c("Make","Model_v1")] <-str_split_fixed(Car_Total$Model," ", 2)
View(Car_Total)Grouping by Parent Company
Car_Total <- Car_Total %>%
mutate(Parent = case_when(Make == "Buick" ~ "General Motors",
Make == "Chevrolet" ~ "General Motors",
Make == "Chrysler" ~ "Chrysler",
Make == "Dodge" ~ "Chrysler",
Make == "Fiat" ~ "Chrysler",
Make == "Ford" ~ "Ford",
Make == "Honda" ~ "Honda",
Make == "Kia" ~ "Kia",
Make == "Lincoln" ~ "Ford",
Make == "Toyota" ~ "Toyota",
TRUE ~ "Check"))
Car_Total$Parent<-as.factor(Car_Total$Parent)Assign a new data frame to preserve original data frame
ct <- Car_TotalDescriptive Statistics on Age Group
stat_table <- group_by(ct, ct$AgeGrp) %>% summarise(count = n(),
mean=mean(Post.Satis, na.rm = TRUE),
var=var(Post.Satis, na.rm = TRUE),
sd=sd(Post.Satis, na.rm = TRUE))
print(stat_table)## # A tibble: 3 × 5
## `ct$AgeGrp` count mean var sd
## <fct> <int> <dbl> <dbl> <dbl>
## 1 Young Adults 443 5.15 1.70 1.30
## 2 Adults 375 5.40 1.25 1.12
## 3 Mature Adults 231 5.34 1.59 1.26Make a Subset for Brand Toyota
head(stringr::str_detect(ct$Model, "Toyota"), 10)## [1] FALSE FALSE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUEhead(ct[str_detect(ct$Model,"Toyota"),], 10)## Resp Att_1 Att_2 Enj_1 Enj_2 Perform_1 Perform_2 Perform_3 WOM_1 WOM_2
## 3 Res100 6 7 7 3 5 6 6 3 5
## 4 Res1000 6 6 7 6 6 6 6 6 6
## 5 Res1001 6 6 7 6 6 6 6 4 4
## 6 Res1002 3 1 4 3 5 6 6 2 6
## 7 Res1003 2 2 1 2 2 2 1 6 7
## 8 Res1004 7 7 7 6 5 6 5 6 6
## 9 Res1005 2 1 2 1 2 2 2 7 7
## 10 Res1006 6 6 6 5 5 5 5 3 3
## 11 Res1007 4 4 4 2 3 5 3 7 7
## 14 Res101 6 6 7 6 5 6 3 5 6
## Futu_Pur_1 Futu_Pur_2 Valu_Percp_1 Valu_Percp_2 Pur_Proces_1 Pur_Proces_2
## 3 6 6 7 6 5 5
## 4 6 6 4 6 6 3
## 5 4 6 5 6 6 7
## 6 6 6 5 4 5 5
## 7 6 5 4 4 4 5
## 8 6 7 6 5 5 5
## 9 7 7 4 6 6 7
## 10 6 6 5 6 6 5
## 11 5 6 6 2 2 5
## 14 4 3 3 2 2 2
## Residence Pay_Meth Insur_Type Gender Age Education X Region
## 3 2 1 Collision Female 32 1 NA American
## 4 2 3 Liability Female 24 2 NA Asian
## 5 1 3 Liability Female 24 2 NA Asian
## 6 1 3 Liability Female 25 2 NA Asian
## 7 1 3 Liability Female 26 2 NA Asian
## 8 2 3 Liability Female 26 2 NA Asian
## 9 1 3 Liability Female 27 2 NA Asian
## 10 2 3 Liability Female 27 2 NA Asian
## 11 2 3 Liability Male 27 1 NA Asian
## 14 2 2 Collision Female 32 2 NA American
## Model MPG Cyl acc1 C_cost. H_Cost Post.Satis AgeGrp Make
## 3 Toyota Rav4 24 4 8.2 10 8 4 Adults Toyota
## 4 Toyota Corolla 26 4 8.0 7 6 6 Young Adults Toyota
## 5 Toyota Corolla 26 4 8.0 7 6 5 Young Adults Toyota
## 6 Toyota Corolla 26 4 8.0 7 6 6 Young Adults Toyota
## 7 Toyota Corolla 26 4 8.0 7 6 5 Young Adults Toyota
## 8 Toyota Corolla 26 4 8.0 7 6 6 Young Adults Toyota
## 9 Toyota Corolla 26 4 8.0 7 6 7 Young Adults Toyota
## 10 Toyota Corolla 26 4 8.0 7 6 6 Young Adults Toyota
## 11 Toyota Corolla 26 4 8.0 7 6 6 Young Adults Toyota
## 14 Toyota Rav4 24 4 8.2 10 8 5 Adults Toyota
## Model_v1 Parent
## 3 Rav4 Toyota
## 4 Corolla Toyota
## 5 Corolla Toyota
## 6 Corolla Toyota
## 7 Corolla Toyota
## 8 Corolla Toyota
## 9 Corolla Toyota
## 10 Corolla Toyota
## 11 Corolla Toyota
## 14 Rav4 Toyotact_Toyota <- head(str_detect(ct$Model, "Toyota"), 10)
subtoyota<-ct[ct_Toyota,]
table(subtoyota$Make)##
## Buick Chevrolet Chrysler Dodge Fiat Ford Honda Kia
## 23 50 134 35 14 161 124 27
## Lincoln Toyota
## 33 238From Toyota Subset Normality Test
tapply(subtoyota$Post.Satis,subtoyota$AgeGrp,shapiro.test)## $`Young Adults`
##
## Shapiro-Wilk normality test
##
## data: X[[i]]
## W = 0.87974, p-value = 5.208e-16
##
##
## $Adults
##
## Shapiro-Wilk normality test
##
## data: X[[i]]
## W = 0.90027, p-value = 3.431e-13
##
##
## $`Mature Adults`
##
## Shapiro-Wilk normality test
##
## data: X[[i]]
## W = 0.88798, p-value = 1.627e-10Visual Inspection of Normality
res_aov <- aov(subtoyota$Post.Satis ~ as.factor(subtoyota$AgeGrp), data = ct)
hist(res_aov$residuals)
qqnorm(res_aov$residuals)
qqline(res_aov$residuals, col = "blue")
Perform a Shapiro-Wilk test on residuals
shapiro.test(res_aov$residuals)##
## Shapiro-Wilk normality test
##
## data: res_aov$residuals
## W = 0.92886, p-value < 2.2e-16Performing a Bartlett’s test homogeneity of variance
bartlett.test(subtoyota$Post.Satis ~ subtoyota$AgeGrp)##
## Bartlett test of homogeneity of variances
##
## data: subtoyota$Post.Satis by subtoyota$AgeGrp
## Bartlett's K-squared = 9.8668, df = 2, p-value = 0.007202One-Way ANOVA
H0: There is no significant difference in the Age Groups toward Post Purchase Satisfaction for Toyota.
H1: There is a significant difference in the Age Groups toward Post Purchase Satisfaction for Toyota.
Conduct a One-Way ANOVA Equal Variance not assumed P>0.05
welch_test<- oneway.test(Post.Satis ~ as.factor(AgeGrp), var.equal = FALSE, data = subtoyota)
print(welch_test)##
## One-way analysis of means (not assuming equal variances)
##
## data: Post.Satis and as.factor(AgeGrp)
## F = 4.1148, num df = 2.00, denom df = 468.87, p-value = 0.01692Results:
Since the p- value is greater than 0.05, we fail to reject the null hypothesis
ANOVA REGION and CAR PURCHASE PROCESS
Calculate the average of Pur_Proces_1 and Pur_Proces_2 to create a new dependent variable
ct$Avg_Purchase_Sat <- rowMeans(ct[, c("Pur_Proces_1", "Pur_Proces_2")], na.rm = TRUE)Visualize the mean difference of the average purchase satisfaction across regions
subtoyota <- ct[ct_Toyota,]
library(ggplot2)
ggplot(subtoyota, aes(x = Region, y = Avg_Purchase_Sat, fill = Region)) +
geom_boxplot(outlier.color = "red", outlier.shape = 16) +
stat_summary(fun = mean, geom = "point", shape = 4, size = 3, color = "black") +
stat_summary(fun = mean, geom = "text", vjust = -0.5, aes(label = round(after_stat(y), 2)), color = "black") +
theme_minimal() +
labs(title = "Average Purchase Satisfaction Across Regions for Toyota",
x = "Region",
y = "Average Purchase Satisfaction") +
scale_fill_brewer(palette = "Set3") +
theme(axis.text.x = element_text(angle = 45, hjust = 1),
legend.position = "none")
ANOVA test with TOYOTA subset dataset
Check normality assumption
tapply(subtoyota$Avg_Purchase_Sat, subtoyota$Region, shapiro.test)## $American
##
## Shapiro-Wilk normality test
##
## data: X[[i]]
## W = 0.9549, p-value = 8.373e-08
##
##
## $Asian
##
## Shapiro-Wilk normality test
##
## data: X[[i]]
## W = 0.94111, p-value = 2.937e-06
##
##
## $European
##
## Shapiro-Wilk normality test
##
## data: X[[i]]
## W = 0.94939, p-value = 9.963e-06
##
##
## $`Middle Eastern`
##
## Shapiro-Wilk normality test
##
## data: X[[i]]
## W = 0.92133, p-value = 2.109e-09NONE OF THE REGIONS FOLLOW A NORMAL DISTRIBUTION
Check homogeneity of variances in the dependent variable across conditions
bartlett.test(subtoyota$Avg_Purchase_Sat, subtoyota$Region)##
## Bartlett test of homogeneity of variances
##
## data: subtoyota$Avg_Purchase_Sat and subtoyota$Region
## Bartlett's K-squared = 14.01, df = 3, p-value = 0.002892ANOVA TEST: TOYOTA AVG PURCHASE SATISFACTION ACROSS REGIONS
aov_test_avg <- aov(Avg_Purchase_Sat ~ as.factor(Region), data = subtoyota)
print(aov_test_avg)## Call:
## aov(formula = Avg_Purchase_Sat ~ as.factor(Region), data = subtoyota)
##
## Terms:
## as.factor(Region) Residuals
## Sum of Squares 35.9247 1166.4424
## Deg. of Freedom 3 835
##
## Residual standard error: 1.181921
## Estimated effects may be unbalancedPerforming Post-Hoc Test (Tukey HSD Test)
TukeyHSD(aov_test_avg)## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = Avg_Purchase_Sat ~ as.factor(Region), data = subtoyota)
##
## $`as.factor(Region)`
## diff lwr upr p adj
## Asian-American 0.1263815 -0.17205054 0.42481353 0.6956795
## European-American -0.1497654 -0.44475846 0.14522770 0.5587499
## Middle Eastern-American 0.4201430 0.14776592 0.69252009 0.0004514
## European-Asian -0.2761469 -0.61117213 0.05887838 0.1469822
## Middle Eastern-Asian 0.2937615 -0.02153245 0.60905547 0.0781441
## Middle Eastern-European 0.5699084 0.25786749 0.88194928 0.0000179plot(TukeyHSD(aov_test_avg)) 
ANOVA test without the assumption of equal variances
oneway_result <- oneway.test(Avg_Purchase_Sat ~ as.factor(Region), data = subtoyota)
print(oneway_result)##
## One-way analysis of means (not assuming equal variances)
##
## data: Avg_Purchase_Sat and as.factor(Region)
## F = 10.048, num df = 3.00, denom df = 420.52, p-value = 2.087e-06Pairwise t-tests without the assumption of equal variances
pairwise.t.test(subtoyota$Avg_Purchase_Sat, subtoyota$Region, p.adjust.method = "BH", pool.sd = FALSE)##
## Pairwise comparisons using t tests with non-pooled SD
##
## data: subtoyota$Avg_Purchase_Sat and subtoyota$Region
##
## American Asian European
## Asian 0.2971 - -
## European 0.2536 0.0596 -
## Middle Eastern 0.0001 0.0252 7.9e-06
##
## P value adjustment method: BHT-TEST: MPG comparing toyota and honda
Independent sample t-test (miles per gallon by Toyota and Honda)
Filter data to include only Toyota and Honda
Car_Total_filtered <- subset(Car_Total, Make %in% c("Toyota", "Honda"))Visualize the mean difference
boxplot(Car_Total_filtered$MPG ~ Car_Total_filtered$Make,
col=c("lightblue", "lightgreen"),
main="Comparison of MPG Between Toyota and Honda",
xlab="Car brand",
ylab="Miles Per Gallon (MPG)")
Interpretation:the distribution of miles per gallon (MPG) for both Toyota and Honda. Visually, the median and range appear similar for both makes, but 4 MPG difference between Honda and Toyota.
Check normality assumption on the dependent variable
shapiro.test(Car_Total_filtered$MPG) ##
## Shapiro-Wilk normality test
##
## data: Car_Total_filtered$MPG
## W = 0.70381, p-value < 2.2e-16Hypotheses:
H0: The data follows a normal distribution.
H1: The data does not follow a normal distribution.
Interpretation: For both the MPG data and the residuals from the ANOVA, the p-values are significantly below 0.05, leading to the rejection of the null hypothesis
ANOVA to check normality of residuals
res_aov <- aov(MPG ~ Make, data=Car_Total_filtered)
summary(res_aov)## Df Sum Sq Mean Sq F value Pr(>F)
## Make 1 0 0.286 0.036 0.849
## Residuals 449 3536 7.876Histogram of residuals
hist(res_aov$residuals, main="Histogram of Residuals", xlab="Residuals", col="lightcoral", border="black")
Interpretation: The histogram shows the distribution of residuals from the ANOVA test. The plot reveals that the residuals are not symmetrically distributed, confirming the results of the Shapiro-Wilk test that the normality assumption is violated.
QQ plot to visually check normality of residuals
qqnorm(res_aov$residuals, pch=1, frame=FALSE)
qqline(res_aov$residuals, col="blue", lwd=2)
#### Interpretation: The strong deviation from the diagonal line,
coupled with the clustering of points and presence of outliers, confirms
that the residuals are not normally distributed. This supports the
findings from the Shapiro-Wilk test, where the p-value was significantly
below 0.05, indicating that the assumption of normality is violated.
Shapiro-Wilk test for normality of residuals
shapiro.test(res_aov$residuals) ##
## Shapiro-Wilk normality test
##
## data: res_aov$residuals
## W = 0.70945, p-value < 2.2e-16Bartlett test for homogeneity of variances
bartlett.test(Car_Total_filtered$MPG, Car_Total_filtered$Make) ##
## Bartlett test of homogeneity of variances
##
## data: Car_Total_filtered$MPG and Car_Total_filtered$Make
## Bartlett's K-squared = 2.4429, df = 1, p-value = 0.1181Hypotheses:
H0: The variances are equal between the groups (Toyota and Honda).
H1: The variances are not equal between the groups.
Interpretation: The p-value (0.1181) is above 0.05, so we fail to reject the null hypothesis. This indicates that the assumption of equal variances holds.
Research Question:
H0: There is no significant difference in fuel consumption (miles per gallon) between Toyota and Honda.
H1: There is a significant difference in the mean MPG between Toyota and Honda.
Independent sample t-test (two-sided, equal variances assumed)
t.test(MPG ~ Make, data=Car_Total_filtered, var.eq=TRUE)##
## Two Sample t-test
##
## data: MPG by Make
## t = 0.19062, df = 449, p-value = 0.8489
## alternative hypothesis: true difference in means between group Honda and group Toyota is not equal to 0
## 95 percent confidence interval:
## -0.4908668 0.5963204
## sample estimates:
## mean in group Honda mean in group Toyota
## 22.79245 22.73973Interpretation: The p-value is 0.8489, which is far above the 0.05 significance level. This means we fail to reject the null hypothesis, confirming that there is no significant difference in MPG between Toyota and Honda. The confidence interval also includes 0, further supporting this conclusion.
Value Perception and Gender for brand Toyota
Merge together Valu_Percp_1 and Valu_Percp_2
Car_Total$Valu_Percp_Mean = (Car_Total$Valu_Percp_1 +
Car_Total$Valu_Percp_2) / 2
View(Car_Total[c("Valu_Percp_1", "Valu_Percp_2", "Valu_Percp_Mean")])Conduct independent sample t-test for Toyota between gender and value perception
Choose the target car brand Toyota for statistical analyses
select subset data on the target brand Toyota.
head(stringr::str_detect(Car_Total$Model, "Toyota"), 10)## [1] FALSE FALSE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUEhead(Car_Total[str_detect(Car_Total$Model,"Toyota"),], 10)## Resp Att_1 Att_2 Enj_1 Enj_2 Perform_1 Perform_2 Perform_3 WOM_1 WOM_2
## 3 Res100 6 7 7 3 5 6 6 3 5
## 4 Res1000 6 6 7 6 6 6 6 6 6
## 5 Res1001 6 6 7 6 6 6 6 4 4
## 6 Res1002 3 1 4 3 5 6 6 2 6
## 7 Res1003 2 2 1 2 2 2 1 6 7
## 8 Res1004 7 7 7 6 5 6 5 6 6
## 9 Res1005 2 1 2 1 2 2 2 7 7
## 10 Res1006 6 6 6 5 5 5 5 3 3
## 11 Res1007 4 4 4 2 3 5 3 7 7
## 14 Res101 6 6 7 6 5 6 3 5 6
## Futu_Pur_1 Futu_Pur_2 Valu_Percp_1 Valu_Percp_2 Pur_Proces_1 Pur_Proces_2
## 3 6 6 7 6 5 5
## 4 6 6 4 6 6 3
## 5 4 6 5 6 6 7
## 6 6 6 5 4 5 5
## 7 6 5 4 4 4 5
## 8 6 7 6 5 5 5
## 9 7 7 4 6 6 7
## 10 6 6 5 6 6 5
## 11 5 6 6 2 2 5
## 14 4 3 3 2 2 2
## Residence Pay_Meth Insur_Type Gender Age Education X Region
## 3 2 1 Collision Female 32 1 NA American
## 4 2 3 Liability Female 24 2 NA Asian
## 5 1 3 Liability Female 24 2 NA Asian
## 6 1 3 Liability Female 25 2 NA Asian
## 7 1 3 Liability Female 26 2 NA Asian
## 8 2 3 Liability Female 26 2 NA Asian
## 9 1 3 Liability Female 27 2 NA Asian
## 10 2 3 Liability Female 27 2 NA Asian
## 11 2 3 Liability Male 27 1 NA Asian
## 14 2 2 Collision Female 32 2 NA American
## Model MPG Cyl acc1 C_cost. H_Cost Post.Satis AgeGrp Make
## 3 Toyota Rav4 24 4 8.2 10 8 4 Adults Toyota
## 4 Toyota Corolla 26 4 8.0 7 6 6 Young Adults Toyota
## 5 Toyota Corolla 26 4 8.0 7 6 5 Young Adults Toyota
## 6 Toyota Corolla 26 4 8.0 7 6 6 Young Adults Toyota
## 7 Toyota Corolla 26 4 8.0 7 6 5 Young Adults Toyota
## 8 Toyota Corolla 26 4 8.0 7 6 6 Young Adults Toyota
## 9 Toyota Corolla 26 4 8.0 7 6 7 Young Adults Toyota
## 10 Toyota Corolla 26 4 8.0 7 6 6 Young Adults Toyota
## 11 Toyota Corolla 26 4 8.0 7 6 6 Young Adults Toyota
## 14 Toyota Rav4 24 4 8.2 10 8 5 Adults Toyota
## Model_v1 Parent Valu_Percp_Mean
## 3 Rav4 Toyota 6.5
## 4 Corolla Toyota 5.0
## 5 Corolla Toyota 5.5
## 6 Corolla Toyota 4.5
## 7 Corolla Toyota 4.0
## 8 Corolla Toyota 5.5
## 9 Corolla Toyota 5.0
## 10 Corolla Toyota 5.5
## 11 Corolla Toyota 4.0
## 14 Rav4 Toyota 2.5Assign into a new dataframe
head(Car_Total_Toyota<-str_detect(Car_Total$Model,"Toyota"), 5)## [1] FALSE FALSE TRUE TRUE TRUEhead(Car_Total_Toyota,5)## [1] FALSE FALSE TRUE TRUE TRUEAssign subset data into a dataframe
head(subtoyota<-Car_Total[Car_Total_Toyota,],5)## Resp Att_1 Att_2 Enj_1 Enj_2 Perform_1 Perform_2 Perform_3 WOM_1 WOM_2
## 3 Res100 6 7 7 3 5 6 6 3 5
## 4 Res1000 6 6 7 6 6 6 6 6 6
## 5 Res1001 6 6 7 6 6 6 6 4 4
## 6 Res1002 3 1 4 3 5 6 6 2 6
## 7 Res1003 2 2 1 2 2 2 1 6 7
## Futu_Pur_1 Futu_Pur_2 Valu_Percp_1 Valu_Percp_2 Pur_Proces_1 Pur_Proces_2
## 3 6 6 7 6 5 5
## 4 6 6 4 6 6 3
## 5 4 6 5 6 6 7
## 6 6 6 5 4 5 5
## 7 6 5 4 4 4 5
## Residence Pay_Meth Insur_Type Gender Age Education X Region Model
## 3 2 1 Collision Female 32 1 NA American Toyota Rav4
## 4 2 3 Liability Female 24 2 NA Asian Toyota Corolla
## 5 1 3 Liability Female 24 2 NA Asian Toyota Corolla
## 6 1 3 Liability Female 25 2 NA Asian Toyota Corolla
## 7 1 3 Liability Female 26 2 NA Asian Toyota Corolla
## MPG Cyl acc1 C_cost. H_Cost Post.Satis AgeGrp Make Model_v1 Parent
## 3 24 4 8.2 10 8 4 Adults Toyota Rav4 Toyota
## 4 26 4 8.0 7 6 6 Young Adults Toyota Corolla Toyota
## 5 26 4 8.0 7 6 5 Young Adults Toyota Corolla Toyota
## 6 26 4 8.0 7 6 6 Young Adults Toyota Corolla Toyota
## 7 26 4 8.0 7 6 5 Young Adults Toyota Corolla Toyota
## Valu_Percp_Mean
## 3 6.5
## 4 5.0
## 5 5.5
## 6 4.5
## 7 4.0head(subtoyota,5)## Resp Att_1 Att_2 Enj_1 Enj_2 Perform_1 Perform_2 Perform_3 WOM_1 WOM_2
## 3 Res100 6 7 7 3 5 6 6 3 5
## 4 Res1000 6 6 7 6 6 6 6 6 6
## 5 Res1001 6 6 7 6 6 6 6 4 4
## 6 Res1002 3 1 4 3 5 6 6 2 6
## 7 Res1003 2 2 1 2 2 2 1 6 7
## Futu_Pur_1 Futu_Pur_2 Valu_Percp_1 Valu_Percp_2 Pur_Proces_1 Pur_Proces_2
## 3 6 6 7 6 5 5
## 4 6 6 4 6 6 3
## 5 4 6 5 6 6 7
## 6 6 6 5 4 5 5
## 7 6 5 4 4 4 5
## Residence Pay_Meth Insur_Type Gender Age Education X Region Model
## 3 2 1 Collision Female 32 1 NA American Toyota Rav4
## 4 2 3 Liability Female 24 2 NA Asian Toyota Corolla
## 5 1 3 Liability Female 24 2 NA Asian Toyota Corolla
## 6 1 3 Liability Female 25 2 NA Asian Toyota Corolla
## 7 1 3 Liability Female 26 2 NA Asian Toyota Corolla
## MPG Cyl acc1 C_cost. H_Cost Post.Satis AgeGrp Make Model_v1 Parent
## 3 24 4 8.2 10 8 4 Adults Toyota Rav4 Toyota
## 4 26 4 8.0 7 6 6 Young Adults Toyota Corolla Toyota
## 5 26 4 8.0 7 6 5 Young Adults Toyota Corolla Toyota
## 6 26 4 8.0 7 6 6 Young Adults Toyota Corolla Toyota
## 7 26 4 8.0 7 6 5 Young Adults Toyota Corolla Toyota
## Valu_Percp_Mean
## 3 6.5
## 4 5.0
## 5 5.5
## 6 4.5
## 7 4.0Check
table(subtoyota$Make)##
## Toyota
## 292Convert Gender into numerical values
subtoyota <-subtoyota %>%
mutate(Gender_numeric = case_when(Gender == "Female" ~ 1,
Gender == "Male" ~ 0))head(subtoyota, 5)## Resp Att_1 Att_2 Enj_1 Enj_2 Perform_1 Perform_2 Perform_3 WOM_1 WOM_2
## 3 Res100 6 7 7 3 5 6 6 3 5
## 4 Res1000 6 6 7 6 6 6 6 6 6
## 5 Res1001 6 6 7 6 6 6 6 4 4
## 6 Res1002 3 1 4 3 5 6 6 2 6
## 7 Res1003 2 2 1 2 2 2 1 6 7
## Futu_Pur_1 Futu_Pur_2 Valu_Percp_1 Valu_Percp_2 Pur_Proces_1 Pur_Proces_2
## 3 6 6 7 6 5 5
## 4 6 6 4 6 6 3
## 5 4 6 5 6 6 7
## 6 6 6 5 4 5 5
## 7 6 5 4 4 4 5
## Residence Pay_Meth Insur_Type Gender Age Education X Region Model
## 3 2 1 Collision Female 32 1 NA American Toyota Rav4
## 4 2 3 Liability Female 24 2 NA Asian Toyota Corolla
## 5 1 3 Liability Female 24 2 NA Asian Toyota Corolla
## 6 1 3 Liability Female 25 2 NA Asian Toyota Corolla
## 7 1 3 Liability Female 26 2 NA Asian Toyota Corolla
## MPG Cyl acc1 C_cost. H_Cost Post.Satis AgeGrp Make Model_v1 Parent
## 3 24 4 8.2 10 8 4 Adults Toyota Rav4 Toyota
## 4 26 4 8.0 7 6 6 Young Adults Toyota Corolla Toyota
## 5 26 4 8.0 7 6 5 Young Adults Toyota Corolla Toyota
## 6 26 4 8.0 7 6 6 Young Adults Toyota Corolla Toyota
## 7 26 4 8.0 7 6 5 Young Adults Toyota Corolla Toyota
## Valu_Percp_Mean Gender_numeric
## 3 6.5 1
## 4 5.0 1
## 5 5.5 1
## 6 4.5 1
## 7 4.0 1Visually see the mean difference
boxplot(subtoyota$Valu_Percp_Mean ~ subtoyota$Gender_numeric, col=c(5,7))
Normality assumption on the dependent variable
shapiro.test(subtoyota$Valu_Percp_Mean)##
## Shapiro-Wilk normality test
##
## data: subtoyota$Valu_Percp_Mean
## W = 0.94228, p-value = 2.831e-09Chcek normality assumption on residuals
res_aov<-aov(Valu_Percp_Mean~Gender_numeric, data=subtoyota)
res_aov## Call:
## aov(formula = Valu_Percp_Mean ~ Gender_numeric, data = subtoyota)
##
## Terms:
## Gender_numeric Residuals
## Sum of Squares 0.3812 318.7387
## Deg. of Freedom 1 287
##
## Residual standard error: 1.053844
## Estimated effects may be unbalanced
## 3 observations deleted due to missingnessHistogram
hist(res_aov$residuals)
Create QQ plot (visually check normality of the data)
qqnorm(res_aov$residuals, pch=1, frame=FALSE)
qqline(res_aov$residuals, col="red", lwd=4)
Test normality of the residuals
shapiro.test(res_aov$residuals)##
## Shapiro-Wilk normality test
##
## data: res_aov$residuals
## W = 0.94881, p-value = 1.687e-08Bartlett test of homogeneity of variances
bartlett.test(subtoyota$Valu_Percp_Mean, subtoyota$Gender_numeric) ##
## Bartlett test of homogeneity of variances
##
## data: subtoyota$Valu_Percp_Mean and subtoyota$Gender_numeric
## Bartlett's K-squared = 2.4132, df = 1, p-value = 0.1203Research Question: independent sample t-test
H0: There is no significant difference in the average value perception for Toyota brand cars between males and females.
# H1: There is a significant difference in the average value perception for Toyota brand cars between males and females.
Conduct t-test
t.test(Valu_Percp_Mean~Gender_numeric, data=subtoyota, var.eq=TRUE)##
## Two Sample t-test
##
## data: Valu_Percp_Mean by Gender_numeric
## t = -0.58585, df = 287, p-value = 0.5584
## alternative hypothesis: true difference in means between group 0 and group 1 is not equal to 0
## 95 percent confidence interval:
## -0.3596413 0.1946552
## sample estimates:
## mean in group 0 mean in group 1
## 5.118003 5.200496Results:
Since the p- value is greater than 0.05, we fail to reject the null hypothesis
Conduct independent sample t-test for attitude between Toyota and Honda
Filter rows by test values (choose “Toyota” and “Honda” brands for analysis)
head(filtered_car_total <- Car_Total %>%
filter(str_detect(Parent, "Toyota") | str_detect(Parent, "Honda")),5)## Resp Att_1 Att_2 Enj_1 Enj_2 Perform_1 Perform_2 Perform_3 WOM_1 WOM_2
## 1 Res100 6 7 7 3 5 6 6 3 5
## 2 Res1000 6 6 7 6 6 6 6 6 6
## 3 Res1001 6 6 7 6 6 6 6 4 4
## 4 Res1002 3 1 4 3 5 6 6 2 6
## 5 Res1003 2 2 1 2 2 2 1 6 7
## Futu_Pur_1 Futu_Pur_2 Valu_Percp_1 Valu_Percp_2 Pur_Proces_1 Pur_Proces_2
## 1 6 6 7 6 5 5
## 2 6 6 4 6 6 3
## 3 4 6 5 6 6 7
## 4 6 6 5 4 5 5
## 5 6 5 4 4 4 5
## Residence Pay_Meth Insur_Type Gender Age Education X Region Model
## 1 2 1 Collision Female 32 1 NA American Toyota Rav4
## 2 2 3 Liability Female 24 2 NA Asian Toyota Corolla
## 3 1 3 Liability Female 24 2 NA Asian Toyota Corolla
## 4 1 3 Liability Female 25 2 NA Asian Toyota Corolla
## 5 1 3 Liability Female 26 2 NA Asian Toyota Corolla
## MPG Cyl acc1 C_cost. H_Cost Post.Satis AgeGrp Make Model_v1 Parent
## 1 24 4 8.2 10 8 4 Adults Toyota Rav4 Toyota
## 2 26 4 8.0 7 6 6 Young Adults Toyota Corolla Toyota
## 3 26 4 8.0 7 6 5 Young Adults Toyota Corolla Toyota
## 4 26 4 8.0 7 6 6 Young Adults Toyota Corolla Toyota
## 5 26 4 8.0 7 6 5 Young Adults Toyota Corolla Toyota
## Valu_Percp_Mean
## 1 6.5
## 2 5.0
## 3 5.5
## 4 4.5
## 5 4.0head(filtered_car_total,5)## Resp Att_1 Att_2 Enj_1 Enj_2 Perform_1 Perform_2 Perform_3 WOM_1 WOM_2
## 1 Res100 6 7 7 3 5 6 6 3 5
## 2 Res1000 6 6 7 6 6 6 6 6 6
## 3 Res1001 6 6 7 6 6 6 6 4 4
## 4 Res1002 3 1 4 3 5 6 6 2 6
## 5 Res1003 2 2 1 2 2 2 1 6 7
## Futu_Pur_1 Futu_Pur_2 Valu_Percp_1 Valu_Percp_2 Pur_Proces_1 Pur_Proces_2
## 1 6 6 7 6 5 5
## 2 6 6 4 6 6 3
## 3 4 6 5 6 6 7
## 4 6 6 5 4 5 5
## 5 6 5 4 4 4 5
## Residence Pay_Meth Insur_Type Gender Age Education X Region Model
## 1 2 1 Collision Female 32 1 NA American Toyota Rav4
## 2 2 3 Liability Female 24 2 NA Asian Toyota Corolla
## 3 1 3 Liability Female 24 2 NA Asian Toyota Corolla
## 4 1 3 Liability Female 25 2 NA Asian Toyota Corolla
## 5 1 3 Liability Female 26 2 NA Asian Toyota Corolla
## MPG Cyl acc1 C_cost. H_Cost Post.Satis AgeGrp Make Model_v1 Parent
## 1 24 4 8.2 10 8 4 Adults Toyota Rav4 Toyota
## 2 26 4 8.0 7 6 6 Young Adults Toyota Corolla Toyota
## 3 26 4 8.0 7 6 5 Young Adults Toyota Corolla Toyota
## 4 26 4 8.0 7 6 6 Young Adults Toyota Corolla Toyota
## 5 26 4 8.0 7 6 5 Young Adults Toyota Corolla Toyota
## Valu_Percp_Mean
## 1 6.5
## 2 5.0
## 3 5.5
## 4 4.5
## 5 4.0Merge together Att_1 and Att_2
filtered_car_total$Att_Mean = (filtered_car_total$Att_1 +
filtered_car_total$Att_2) / 2
View(filtered_car_total[c("Att_1", "Att_2", "Att_Mean")])Convert parent into numeric values
filtered_car_total <-filtered_car_total %>%
mutate(Parent_numeric = case_when(Parent == "Toyota" ~ 1,
Parent == "Honda" ~ 0))Visually see the mean difference
boxplot(filtered_car_total$Att_Mean ~ filtered_car_total$Parent_numeric, col=c(5,7))
Normality assumption on the dependent variable
shapiro.test(filtered_car_total$Att_Mean)##
## Shapiro-Wilk normality test
##
## data: filtered_car_total$Att_Mean
## W = 0.92712, p-value = 5.313e-14Chcek normality assumption on residuals
res_aov<-aov(Att_Mean~Parent_numeric, data=filtered_car_total)
res_aov## Call:
## aov(formula = Att_Mean ~ Parent_numeric, data = filtered_car_total)
##
## Terms:
## Parent_numeric Residuals
## Sum of Squares 6.3422 726.1419
## Deg. of Freedom 1 449
##
## Residual standard error: 1.271709
## Estimated effects may be unbalancedHistogram
hist(res_aov$residuals)
Create QQ plot (visually check normality of the data)
qqnorm(res_aov$residuals, pch=1, frame=FALSE)
qqline(res_aov$residuals, col="red", lwd=4)
Test normality of the residuals
shapiro.test(res_aov$residuals)##
## Shapiro-Wilk normality test
##
## data: res_aov$residuals
## W = 0.94609, p-value = 9.63e-12Bartlett test of homogeneity of variances
bartlett.test(filtered_car_total$Att_Mean, filtered_car_total$Parent_numeric) ##
## Bartlett test of homogeneity of variances
##
## data: filtered_car_total$Att_Mean and filtered_car_total$Parent_numeric
## Bartlett's K-squared = 7.0355, df = 1, p-value = 0.007991Research Question: independent sample t-test
H0: There is no significant difference in the average attitude toward the vehicle between the brands Honda and Toyota.
H1: There is a significant difference in the average attitude toward the vehicle between the brands Honda and Toyota.
Conduct t-test
t.test(Att_Mean~Parent_numeric, data=filtered_car_total, var.eq=TRUE)##
## Two Sample t-test
##
## data: Att_Mean by Parent_numeric
## t = 1.9803, df = 449, p-value = 0.04828
## alternative hypothesis: true difference in means between group 0 and group 1 is not equal to 0
## 95 percent confidence interval:
## 0.001885513 0.494532916
## sample estimates:
## mean in group 0 mean in group 1
## 5.503145 5.254935Results:
Since the p- value is smaller than 0.05, we reject the null hypothesis
Multiple Linear Regression
Is there a linear relationship between Value Perception, together with Region, Acceleration and MPG
visualizing their relationships
ggplot(subtoyota)+aes(x=Valu_Percp_Mean, y=Region, colour=acc1, size=MPG)+
geom_point()+ scale_color_gradient() +
labs (y = "Region", x = "Value", color = "Acceleration",
size = "MPG") +
theme_minimal()
model2<-lm(Valu_Percp_Mean ~ Region+acc1+MPG, data=subtoyota)
summary(model2)##
## Call:
## lm(formula = Valu_Percp_Mean ~ Region + acc1 + MPG, data = subtoyota)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3.1515 -0.6034 0.2487 0.7956 1.9849
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 2.91205 1.61289 1.805 0.0721 .
## RegionAsian 0.08833 0.15871 0.557 0.5783
## RegionEuropean 0.18931 0.16178 1.170 0.2429
## RegionMiddle Eastern 0.73495 0.39572 1.857 0.0643 .
## acc1 0.42850 0.34923 1.227 0.2208
## MPG -0.04911 0.05763 -0.852 0.3948
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.048 on 286 degrees of freedom
## Multiple R-squared: 0.01927, Adjusted R-squared: 0.002123
## F-statistic: 1.124 on 5 and 286 DF, p-value: 0.3478