Suppose that we have two samples from two separate normally distributed populati

Question

Suppose that we have two samples from two separate normally distributed populations, both of size 200: X1, , x200 ~iid N(, 2) and Yi, ,Y200 ~iid N(2, 2). From these samples, we compute x-52.1, y = 49.7, sx6.5, sy7.1. (a) Construct a 99% confidence interval for the true difference between the means of these two populations (b) Use a t test to test Ho : -2 = 0 against the alternative Hi : -12 > 0. Let = 0.01. (c) Suppose that the true values of the population parameters are -52, 2-50, and 7. Compute the power of the t test from part (b)

Explanation / Answer

a.
TRADITIONAL METHOD
given that,
mean(x)=52.1
standard deviation , s.d1=6.5
number(n1)=200
y(mean)=49.7
standard deviation, s.d2 =7.1
number(n2)=200
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((42.25/200)+(50.41/200))
= 0.681
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, = 0.01
from standard normal table, two tailed and
value of |t | with min (n1-1, n2-1) i.e 199 d.f is 2.601
margin of error = 2.601 * 0.681
= 1.77
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (52.1-49.7) ± 1.77 ]
= [0.63 , 4.17]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=52.1
standard deviation , s.d1=6.5
sample size, n1=200
y(mean)=49.7
standard deviation, s.d2 =7.1
sample size,n2 =200
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 52.1-49.7) ± t a/2 * sqrt((42.25/200)+(50.41/200)]
= [ (2.4) ± t a/2 * 0.681]
= [0.63 , 4.17]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [0.63 , 4.17] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population proportion

b.
Given that,
mean(x)=52.1
standard deviation , s.d1=6.5
number(n1)=200
y(mean)=49.7
standard deviation, s.d2 =7.1
number(n2)=200
null, Ho: u1 - u2 = 0
alternate, H1: u1 - u2 > 0
level of significance, = 0.01
from standard normal table,right tailed t /2 =2.345
since our test is right-tailed
reject Ho, if to > 2.345
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =52.1-49.7/sqrt((42.25/200)+(50.41/200))
to =3.526
| to | =3.526
critical value
the value of |t | with min (n1-1, n2-1) i.e 199 d.f is 2.345
we got |to| = 3.52598 & | t | = 2.345
make decision
hence value of | to | > | t | and here we reject Ho
p-value:right tail - Ha : ( p > 3.526 ) = 0.00026
hence value of p0.01 > 0.00026,here we reject Ho
ANSWERS
---------------
null, Ho: u1 - u2 = 0
alternate, H1: u1 - u2 > 0
test statistic: 3.526
critical value: 2.345
decision: reject Ho
p-value: 0.00026

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