7. Agronomist have identified seven different geographical areas with respect to

Question

7. Agronomist have identified seven different geographical areas with respect to raising corn in Louisiana and have managed to obtain an experimental farm in each area. To see if a single variety of corn can be recommended for the entire state, the two leading varieties are compared for yield all seven localities. The following yield in bushels per acre are obtained: test at = 0.05. Geographical Location 45 41 58 60 42 32 57 47 4462 63 46 35 59 Hypothesis: Ho: Ha: Test Statistic Graph: Conclusion: Construct a 95% CI for the mean difference:

Explanation / Answer

PART A.
Given that,
mean(x)=47.8571
standard deviation , s.d1=10.6055
number(n1)=7
y(mean)=50.8571
standard deviation, s.d2 =10.6055
number(n2)=7
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.447
since our test is two-tailed
reject Ho, if to < -2.447 OR if to > 2.447
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =47.8571-50.8571/sqrt((112.47663/7)+(112.47663/7))
to =-0.5292
| to | =0.5292
critical value
the value of |t | with min (n1-1, n2-1) i.e 6 d.f is 2.447
we got |to| = 0.52921 & | t | = 2.447
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -0.5292 ) = 0.616
hence value of p0.05 < 0.616,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -0.5292
critical value: -2.447 , 2.447
decision: do not reject Ho
p-value: 0.616
no significance diffrence between them

PART B>
TRADITIONAL METHOD
given that,
mean(x)=47.8571
standard deviation , s.d1=10.6055
number(n1)=7
y(mean)=50.8571
standard deviation, s.d2 =10.6055
number(n2)=7
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((112.477/7)+(112.477/7))
= 5.669
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, = 0.05
from standard normal table, two tailed and
value of |t | with min (n1-1, n2-1) i.e 6 d.f is 2.447
margin of error = 2.447 * 5.669
= 13.872
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (47.8571-50.8571) ± 13.872 ]
= [-16.872 , 10.872]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=47.8571
standard deviation , s.d1=10.6055
sample size, n1=7
y(mean)=50.8571
standard deviation, s.d2 =10.6055
sample size,n2 =7
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 = [( 47.8571-50.8571) ± t a/2 * sqrt((112.477/7)+(112.477/7)]
= [ (-3) ± t a/2 * 5.669]
= [-16.872 , 10.872]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-16.872 , 10.872] 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

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