Please!!1 I need your help

#------------------1------------------------------
s=read.table("C:\\Users\\User\\Desktop\\P1.txt")
attach(s)
View(s)
#---------------(i)----------------------------------
a=mean(y)
n=nrow(s)
test_statistic=(a*sqrt(n))/sqrt(a*(1-a))
qnorm(0.975,0,1)
#reject the null hypothesis.
#---------------(ii)---------------------------------
model1=lm(y~x)
AIC(model1)
model2=lm(y~1)
AIC(model2)
# a model with x is better than without x as AIC value is less in that case.
#------------------(iii)---------------------------------
BIC(model1)
BIC(model2)
# a model with x is better than a model without x as the BIC value is less in that case.
#-------------------(iv)------------------------------------
# yes, in this case the AIC and BIC values give the same answer.
#Actually BIC is somewhat proportional to AIC in terms of its formula . BIC penalises a conmplex model more that is a complex model is more likely to have high BIC values.
#Since this is a simple linear model AIC and BIC hold similar...