Question

Consider the following Two-(Independent) Sample setting: A farmer who own 200 cows wants to compare two different feeds 1 and 2. The farmer first did simple random selection of 11 cows from his herd. Then, farmer randomly assigned 6 of the 11 cows to get Trt A, leaving the other 5 to get Trt B. For the 6 Trt A cows, the average weight gain was +2.1, while the Std dev of their(6) weight gains was 0.8. For the 5 Trt B cows, the average weight gain was +1.7 , while the Std dev of the (5) weight gains was 0.4.   Utilizing Chpt 21's Option 2 (having dof for 2-sample t = min of n₁ - 1 , n₂ - 1), conduct a 95% C.I for the mean Trt A - Trt B weight gain difference over the herd (i.e. population of 200 cows). Also, perform a 2-sided test of hypothesis to see if one can declare a significant difference in the two feeds, at the critical p-value level (alpha) of 0.05.

Answer 1

a.
TRADITIONAL METHOD
given that,
mean(x)=2.1
standard deviation , s.d1=0.8
number(n1)=6
y(mean)=1.7
standard deviation, s.d2 =0.4
number(n2)=5
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((0.64/6)+(0.16/5))
= 0.372
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 4 d.f is 2.776
margin of error = 2.776 * 0.372
= 1.034
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (2.1-1.7) ± 1.034 ]
= [-0.634 , 1.434]
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DIRECT METHOD
given that,
mean(x)=2.1
standard deviation , s.d1=0.8
sample size, n1=6
y(mean)=1.7
standard deviation, s.d2 =0.4
sample size,n2 =5
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 = [( 2.1-1.7) ± t a/2 * sqrt((0.64/6)+(0.16/5)]
= [ (0.4) ± t a/2 * 0.372]
= [-0.634 , 1.434]
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interpretations:
1. we are 95% sure that the interval [-0.634 , 1.434] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion

b.
Given that,
mean(x)=2.1
standard deviation , s.d1=0.8
number(n1)=6
y(mean)=1.7
standard deviation, s.d2 =0.4
number(n2)=5
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.776
since our test is two-tailed
reject Ho, if to < -2.776 OR if to > 2.776
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =2.1-1.7/sqrt((0.64/6)+(0.16/5))
to =1.0742
| to | =1.0742
critical value
the value of |t α| with min (n1-1, n2-1) i.e 4 d.f is 2.776
we got |to| = 1.07417 & | t α | = 2.776
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.0742 ) = 0.343
hence value of p0.05 < 0.343,here we do not reject Ho
ANSWERS
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null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 1.0742
critical value: -2.776 , 2.776
decision: do not reject Ho
p-value: 0.343
we do not have enough evidence to support the claim that if one can declare a significant difference in the two feeds mean Trt A - Trt B weight gain difference over the herd.