Question

Please answer using R studio

(1 point) Alzheimer's disease is a progressive disease of the brain. An article published in 2006 was critical of the methodology used in studies of Alzheimer's treatments conducted prior to 2006. In a random sample of n = 31 Alzheimer's studies, the quality of the methodology was measured on the Wong scale with scores ranging from 9 (low quality) to 27 (high quality) for each study. The data is provided below. data can be copied into R Studio using the following command: x=c(10.24,19,27,10,23,18,27,24,15,25,25,17,10,10,14,13,22,25,23,22,20,18,14,21,12,20,23,12,25,22); (a) Find a 87% confidence interval for the mean quality of studies on the treatment of Alzheimer's disease, Assuming the data is Normally distributed !!! <u< use as many digits as possible (b) Correctly complete the interpretation of the 87% confidence interval you computed in (a): the mean quality of ? is somewhere between is somewhere between 1 and and on the Wong scale. (c) If you were to decrease the level of confidence used to estimate the , what would happen to the confidence interval you found in (a)? A. The confidence interval would become narrower. B. Nothing. The confidence interval would be the same. C. The confidence interval would become wider. (d) Copy and paste the following code into R Studio to bootstrap the sample: library(mosaic) RNGkind(sample.kind="Rejection"); set.seed(33507); B=do(1000) * mean(resample(c(10,24,19,27,10,23,18,27,24,15,25,25,17,10,10,14,13,22,25,23,22,20,18,14,21,12,20,23,12,25,22), 31)); The 87% confidence interval for u is: <u< . use as many digits as possible

Answer 1

a)

> x<- c(10,24,19,27,10,23,18,27,24,15,25,25,17,10,10,14,13,22,25,23,22,20,18,14,21,12,20,23,12,25,22)
> n<-length(x) # size of variable
> n
[1] 31
>
> M <- mean(x) # mean of x
> M
[1] 19.03226
>
> S <- sqrt(var(x)) # std dev of x
> S
[1] 5.576641
>
> alpha = 0.13 # level of significance( 87% confidence interval)
> z <-qnorm(1-(0.13/2)) # z score for l.o.s.
> z
[1] 1.514102
>
> se= S / sqrt(n) # standard error
> se
[1] 1.001594
>
> ME <- z*se # Margin of Error
> ME
[1] 1.516516
>
>
> left <- M-ME # Lower confidene limit
> right <- M+ ME # upper confidence limit
>
> CI <- c( left,right)
> CI
[1] 17.51574 20.54877

b)

we are 87 % confident that the mean quality of studies on the treatment on Alzheimer's disease is somewhere between 17.52 and 20.55.

c) a lower confidence level is that you get a narrower, more precise confidence interval.

Answer:- The confidence interval would become narrower.

d)

> library(mosaic)
> RNGkind(sample.kind = "Rejection");
> set.seed(33507);
> B =do(1000)*mean(resample(x,31));
>
>
> a=B[,1] # subsetting to a vector
>
> n<-length(a) # size of variable
> n
[1] 1000
>
> M <- mean(a) # mean of x
> M
[1] 19.04826
>
> S <- sqrt(var(a)) # std dev of x
> S
[1] 1.026724
> se= S / sqrt(n) # standard error
> se
[1] 0.03246785
>
> ME <- z*se # Margin of Error
> ME
[1] 0.04915963
>
>
> left <- M-ME # Lower confidene limit
> right <- M+ ME # upper confidence limit
>
> CI2 <- c( left,right)
> CI2
[1] 18.99910 19.09742
>

19 <= u <= 19.10

***************** only code***************

x<- c(10,24,19,27,10,23,18,27,24,15,25,25,17,10,10,14,13,22,25,23,22,20,18,14,21,12,20,23,12,25,22)
n<-length(x) # size of variable
n

M <- mean(x) # mean of x
M

S <- sqrt(var(x)) # std dev of x
S

alpha = 0.13 # level of significance( 87% confidence interval)
z <-qnorm(1-(0.13/2)) # z score for l.o.s.
z

se= S / sqrt(n) # standard error
se

ME <- z*se # Margin of Error
ME

left <- M-ME # Lower confidene limit
right <- M+ ME # upper confidence limit

CI <- c( left,right)
CI

library(mosaic)
RNGkind(sample.kind = "Rejection");
set.seed(33507);
B =do(1000)*mean(resample(x,31));

a=B[,1] # subsetting to a vector

n<-length(a) # size of variable
n

M <- mean(a) # mean of x
M

S <- sqrt(var(a)) # std dev of x
S
se= S / sqrt(n) # standard error
se

ME <- z*se # Margin of Error
ME

left <- M-ME # Lower confidene limit
right <- M+ ME # upper confidence limit

CI2 <- c( left,right)
CI2