The data below are yields for two different types of corn seed that were used on

Question

The data below are yields for two different types of corn seed that were used on adjacent plots of land. Assume that the data are simple random samples and that the differences have a distribution that is approximately normal. Construct a 95% confidence interval estimate of the difference between type 1 and type 2 yields. What does the confidence interval suggest about farmer Joe's claim that type 1 seed is better than type 2 seed? Type 1 2060 1932 2009 2516 2122 2076 2255 1434 Type 2 2006 1942 2008 2442 2103 1986 216 1448 In this example, mu Subscript dd is the mean value of the differences d fo for the population of all pairs of data, where each individual difference d is defined as the type 1 seed yield minus the type 2 seed yield. The 95% confidence interval is nothingless than

Explanation / Answer

TRADITIONAL METHOD
given that,
mean(x)=2050.5
standard deviation , s.d1=307.1063
number(n1)=8
y(mean)=1768.875
standard deviation, s.d2 =683.4412
number(n2)=8
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((94314.2795/8)+(467091.8739/8))
= 264.9071
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, = 0.1
from standard normal table, two tailed and
value of |t | with min (n1-1, n2-1) i.e 7 d.f is 1.8946
margin of error = 1.895 * 264.9071
= 501.9989
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (2050.5-1768.875) ± 501.9989 ]
= [-220.3739 , 783.6239]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=2050.5
standard deviation , s.d1=307.1063
sample size, n1=8
y(mean)=1768.875
standard deviation, s.d2 =683.4412
sample size,n2 =8
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 = [( 2050.5-1768.875) ± t a/2 * sqrt((94314.2795/8)+(467091.8739/8)]
= [ (281.625) ± t a/2 * 264.9071]
= [-220.3739 , 783.6239]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [-220.3739 , 783.6239] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.