Problem 5

library(readxl)
data <- read_excel("Que 5.xlsx")
data2<-read_excel("Que 5.xlsx")

head(data)

#a) Compute the covariance
covariance<-cov(data$Husband_Height_cm,data$Wife_Height_cm)

cat("The covariance between the height husband and wives",covariance)

#b)THe covariance if height is measured in inches 
data2$Husband_Height_cm<-data$Husband_Height_cm/2.54
data2$Wife_Height_cm<-data$Wife_Height_cm/2.54

covarianceinches<-cov(data$Husband_Height_cm,data$Wife_Height_cm)
cat("The covariance between the height husband and wives in inches",covarianceinches)

#c)correlation coefficient between the heights of the husband and wife.
correlation<-cor(data2$Husband_Height_cm,data2$Wife_Height_cm)
correlation

#d)THe correlation if height is measured in inches 


correlation2<-cor(data2$Husband_Height_cm,data2$Wife_Height_cm)
correlation2

#e) If every man married a women exactly 5 cm shorter than him , the correlation would be likely a perfect positive correlation.

#f) Fit a linear regression 
modelli<-lm(Wife_Height_cm~Husband_Height_cm,data=data)

summary(modelli)

#F)
#As the link between the heights of husbands and wives is often symmetrical and not causal in character, it is common to regard the choice of the response variable as random when building a regression model comparing the heights of husbands and wives. Because the results and interpretations are comparable, you can use either the height of husbands or the height of women as the response variable.In conclusion, it is not essential to use the height of the wife or the husband as the response variable for correlation analysis; instead, you can choose depending on convenience or custom.

#g)
coefficient<-coef(modelli)

coefficient
model_summary<-summary(modelli)
standarderror<-model_summary$coefficients["Husband_Height_cm", "Std. Error"]
standarderror

tstatistics<-coefficient["Husband_Height_cm"]/standarderror
tstatistics

degreeof<-nrow(data)-2

pvalue2<- 2 * (1 - pt(abs(tstatistics),df=degreeof))
pvalue2
cat("tstatistics",tstatistics)
 #We can reject the null hypothesis because p value is 0              
               

#h)

coefficient<-coef(modelli)
coefficient
model_summary<-summary(modelli)
standarderrorint<-model_summary$coefficients["(Intercept)", "Std. Error"]
standarderrorint

tstatisticsint<-coefficient["(Intercept)"]/standarderrorint
tstatisticsint

degreeoff<-nrow(data)-2

pvalue3<- 2 * (1 - pt(abs(tstatisticsint),df=degreeoff))
pvalue3
cat("tstatistics",tstatisticsint)

#From a small p value , we would reject the null hypothesis that the intercept is zero. From this we can see there is a strong evidence that the intercept is different from the zero