Data Analysis Using R (Statistics Project Sample)

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Data analysis using R
2. In an experiment to compare two diets for fattening beef steers, nine pairs of animals were chosen from the heard; members of each pair were matched as closely as possible with respect to hereditary factors. The members of each pair were randomly allocated, one to teach diet. The following table shows the weight gains (lb) of the animals over a 140-day test period on Diet 1 (X ) and on Diet 2 ( Y ).
(a) Construct a 90% confidence interval for the difference between the two diets.
#value of X and Y
#value of X and Y
X<-c(596,422,524,454,538,522,478,564,556)
Y<-c(498,460,468,458,530,582,528,598,456)
t.test(Y,X,alternative = c("two.sided", "less", "greater"),
mu = 0, paired = TRUE, var.equal = FALSE,
conf.level = 0.90)
Confidence intervals -46.95389 ≤ µ ≥30.06500
(b) Conduct a hypothesis test at the 5% significance level that tests whether the diets are equal.
Which one is better?
# T-tes for hypothesis.
t.test(X,Y,alternative = c("two.sided", "less", "greater"),
mu = 0, paired = TRUE, var.equal = FALSE,
conf.level = 0.95)
Results of above test.
Paired t-test
data: X and Y
t = 0.40777, df = 8, p-value = 0.6941
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-39.31068 56.19957
sample estimates:
mean of the differences
8.444444
6. Compute the following intervals the long way, feel free to check your work with test() .
(a) Produce a 92% confidence interval for the mean ToothGrowth for Guinea pigs who received
Vitamin C.
#92% confidence interval for the mean ToothGrowth for Guinea pigs who received Vitamin C.
# n sample size
n<-length(tooth_VC)
# mean for tooth_OJ
m<-mean(tooth_Vc)
#confidence level
conf.level<-0.92
# Z score
z<-qt((1+conf.level)/2,df=n-1)
se<-sd(tooth_VC)/sqrt(n)
se<-sd(tooth_VC)/sqrt(n)
# confidence interval
ci<-z*se
# upper level
upper<-m+ci
#lower level
lower<-m-c
Confidence interfeval 14.2253≤ µ ≥ 19.703
(b) Produce a 97% confidence interval for the mean ToothGrowth for Guinea pigs who received
Orange Juice.
# 97% confidence interval for the mean ToothGrowth for Guinea pigs who received Orange Juice.
# n sample size
n<-length(tooth_OJ)
# mean for tooth_OJ
m<-mean(tooth_OJ)
#confidence level
conf.level<-0.95
# Z score
z<-qt((1+conf.level)/2,df=n-1)
se<-sd(tooth_OJ)/sqrt(n)
se<-sd(tooth_OJ)/sqrt(n)
# confidence interval
ci<-z*se
# upper level
upper<-m+ci
#lower level
lower<-m-c
Confidence intervals 17.21916≤ µ ≥ 23.41565
(c) Produce a 95% confidence interval for the difference of the means VC-OJ
#95% confidence interval for the difference of the means VC-OJ
# n sample size
n<-length(tooth_OJ)
# mean for tooth_OJ
m1<-mean(tooth_OJ)
# mean for tooth_VC
m2<-mean(tooth_VC)
# difference in the Mean
m<-m2-m1
#confidence level
conf.level<-0.95
# Z score
z<-qt((1+conf.level)/2,df=n-1)
se<-sd(tooth_VC)/sqrt(n)
se<-sd(tooth_VC)/sqrt(n)
# confidence interval
ci<-z*se
# upper level
upper<-m+ci
#lower level
lower<-m-c
Confidence intervals -6.786586≤ µ ≥- 0.6134142
(d) Use your answer in part (c)
Is there a statistical difference in the mean ToothGrowth Lengths of the two populations (the guinea pigs who get Vitamin C and those who get Orange Juice)? This is not a yes or no question!! State you conclusion properly.
7. The dataset found in rats’ zinc.csv contains measurements of zinc concentrations (in mg/ml) in the blood of two groups of rats. Group A received a dietary supplement of calcium, and group B did not. Researchers are interested in variations in zinc level as a side effect of dietary supplementation of calcium.
Investigate the claim that the zinc concentration for rats who take calcium supplements is less than the zinc concentration for rats that do not. Justify your analysis.
R code:
# reading the rats_zinc.csv file
zinccon= read.csv("rats_zinc.csv", Header="T")
#Grouping the the two categories A and B
Groupz<-split(zinccon$Zinc,zinccon$Gr...