Problem 3

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library(readxl)
data <- read_excel("problem3-oxygenpurity.xls")

modellr <- lm(purity ~ hydro, data = data)

coefficients <- coef(modellr)

print(coefficients)

(Intercept)       hydro 
   78.90306    10.71324 

#Answer 1
#The coefficient of beta0=78.90306 and beta1=10.71324 after fitting the simple linear regression model.

#Answer 2 
summary(modellr)

Call:
lm(formula = purity ~ hydro, data = data)

Residuals:
    Min      1Q  Median      3Q     Max 
-4.5048 -3.5605  0.0294  2.6550  7.3408 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   78.903      3.843  20.532 2.73e-16 ***
hydro         10.713      3.204   3.344  0.00281 ** 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 3.494 on 23 degrees of freedom
Multiple R-squared:  0.3272,    Adjusted R-squared:  0.2979 
F-statistic: 11.18 on 1 and 23 DF,  p-value: 0.002814

#From statistics,we can see that it rejects Null hypothesis because it is a significant linear regression.

#Answer 3 
#The coefficient of determination value is r^2 is 0.3272.This value indicates that there is 32.2% of the variation in the dependent variable is explained by the independent variable.

#Answer 4 
# Calculating  the 95% confidence interval for a specific coefficient ( beta1)
ci <- confint(modellr, "hydro", level = 0.95)
ci

        2.5 %   97.5 %
hydro 4.08618 17.34031

#Answer4 
ci1 <- confint(modellr, level = 0.95)
ci1

               2.5 %   97.5 %
(Intercept) 70.95322 86.85289
hydro        4.08618 17.34031

#The confidence interval on slope Beta 0 (Lower bound is 70.95322 and upper bound is 86.85289).
#The confidence interval on slope Beta 1 (Lower bound is  4.08618 and upper bound is 17.34031).

#Answer 5 
# For finding the mean purity , we are defining the value of hydrocarbon percentage 
valuehydro<-1.05
data2<-data.frame(hydro=valuehydro)

predic<-predict(modellr,newdata=data2,interval = "confidence",level=0.95)

cat("predicted mean purity:",predic[1],"\n")

predicted mean purity: 90.15196 

cat("Lower bound:",predic[2],"\n")

Lower bound: 88.47041 

cat("upper bound:",predic[3],"\n")

upper bound: 91.83351 

#Answer 6
#correlation coefficient r between Y and X
correlation<- cor(data$purity,data$hydro)
cat("The correlation coefficient:",correlation,"\n")

The correlation coefficient: 0.5719793 

#Answer 7
#Test the hypothesis:

no<-length(data$purity)

tstats<-correlation*sqrt((no-2)/1-correlation^2)

degrfre<-no-2

# Perform a two-sample t-test
t_test_result <- t.test(data)

# Get the two-tailed p-value
pvalue <- t_test_result$p.value

#significant level
a<-0.05

#comparing p value to a 
if(pvalue<a) {
  cat("It reject the null hypothesis,\n")
  cat("There is a significant correlation between purity and hydrocarbons,\n")
} else {
  cat("It fails to  reject the null hypothesis,\n")
  cat("There is no significant correlation between purity and hydrocarbons,\n")
}

It reject the null hypothesis,
There is a significant correlation between purity and hydrocarbons,

#print t stats 
cat("t statistics",tstats)

t statistics 2.723537

#print p value
cat("pvalue",pvalue)

pvalue 3.64393e-09