Question

PLEASE INSERT R SCRIPT

We wish to compare the expected values, μΧ andHy of two independent normal populations, say X and Y, with known standard deviations σχ-8.3 and Ơy-10.1 . we take a random sample of size 13 from X ( X1M, ,x13 ) and a random sample of size 7 from Y ( Yi,½, , ) as follows: X: 0.54, -3.67, 1.41, 23.51, 20.77, 15.90, 16.70, 2.48, 12.27, 2.64, 22.12, 14.03, 17.80 Y: 21.22, 12.61, 15.04, 14.51, 10.91, -4.90, 22.71 We are interested in examining Hx -Hy. Call the sample means of X and Y, Xbar and Ybar respectivelyxbar and ybar realized values). Assume that all distributions are normal. Use R for computations. a)Calculate xbar| b) Calculate the variance of Xbar c) Calculate ybar d) Calculate the variance of Ybar. e) Calculate the variance of Xbar - Ybar f) what is the critical value used for a 96% confidence interval for Ah-uy? g) Create a 96% confidence interval for Ax- h) What is the length of your 96% confidence interval for,h-ly? Enter a number i) What would the p value have been if we used this data to test Ho:x -Hy 0 against the alternative Hay0? k) Copy your R script for the above into the text box here

Answer 1

R-Output

  
> X = c(0.51,-3.67,1.41,23.51,20.77,15.90,16.70,2.48,12.27,2.64,22.12,14.03,17.80)
> Y = c(21.22,12.61,15.04,14.51,10.91,-4.90,22.71)
>
>
> ##given
>
> sig_x = 8.3
> sig_y = 10.1
>
> #a)
> mean(X)
[1] 11.26692
>
>
> #b)
> var_of_xbar = sig_x/length(X)
> var_of_xbar
[1] 0.6384615
>
>
> #c)
> mean(Y)
[1] 13.15714
>
>
> #d)
> var_of_ybar = sig_y/length(Y)
> var_of_ybar
[1] 1.442857
>
>
> #f)
> qt(1-0.04/2, 13-7)
[1] 2.612242
>
>
> #g)
> fit = t.test(X,Y, conf.level = 0.96)
> fit$conf.int
[1] -11.713954 7.933515
attr(,"conf.level")
[1] 0.96
>
> #h)
>
> lenthofCI = 7.93351 - (-11.713954)
> lenthofCI
[1] 19.64746
>
> #I) p-value
> fit$p.value
[1] 0.6673191
>
>
>
>
>
>
>
>

#########################################################################

R-Script -

X = c(0.51,-3.67,1.41,23.51,20.77,15.90,16.70,2.48,12.27,2.64,22.12,14.03,17.80)
Y = c(21.22,12.61,15.04,14.51,10.91,-4.90,22.71)

##given

sig_x = 8.3
sig_y = 10.1

#a)
mean(X)

#b)
var_of_xbar = sig_x/length(X)
var_of_xbar

#c)
mean(Y)

#d)
var_of_ybar = sig_y/length(Y)
var_of_ybar

#f)
qt(1-0.04/2, 13-7)

#g)
fit = t.test(X,Y, conf.level = 0.96)
fit$conf.int

#h)

lenthofCI = 7.93351 - (-11.713954)
lenthofCI

#I) p-value
fit$p.value