1. The following data resulted from an experiment studying the effect of nitroge
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Question
1. The following data resulted from an experiment studying the effect of nitrogen fertilizer on lettuce Data are taken from Kuehl (2000) Amount of fertilizer Heads of lettuce per plot 0 lbs/acre 200 lbs/acre 104 114 90 140 131 148 154 163 Assume that the number of heads of lettuce in the plots is approximately Normally distributed (a) Let o? represent the variance of the number of heads of lettuce under 0 lbs/acre of fertilizer and let represent that under 200 lbs/acre. For each of the following sets of hypotheses get the p-value based on the data in the table. i. Ho: versus H1: > (b) Do you believe that the variances and are different? Why/why not? (c) Let 1 represent the mean of number of heads of lettuce under 0 lbs/acre of fertilizer and let 2 represent that under 200 lbs/acre. For each of the following sets of hypotheses get the p-value based on the data in the table, assuming that - (d) Do you believe the means 1 and 2 are different? Why/why not?Explanation / Answer
TEST FOR VARIANCE
Given that,
sample 1
s1^2=445.333, n1 =4
sample 2
s2^2 =182, n2 =4
null, Ho: sigma^2 = sigma^2
alternate, H1: sigma^2 != sigma^2
level of significance, alpha = 0.05
from standard normal table, two tailed f alpha/2 =15.439
since our test is two-tailed
reject Ho, if F o < -15.439 OR if F o > 15.439
we use test statistic fo = s1^1/ s2^2 =445.333/182 = 2.45
| fo | =2.45
critical value
the value of |f alpha| at los 0.05 with d.f f(n1-1,n2-1)=f(3,3) is 15.439
we got |fo| =2.447 & | f alpha | =15.439
make decision
hence value of |fo | < | f alpha | and here we do not reject Ho
ANSWERS
---------------
null, Ho: sigma^2 = sigma^2
alternate, H1: sigma^2 != sigma^2
test statistic: 2.45
critical value: -15.439 , 15.439
decision: do not reject Ho
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TEST FOR INDEPENDENT MEAN
Given that,
mean(x)=112
standard deviation , s.d1=21.1029
number(n1)=4
y(mean)=149
standard deviation, s.d2 =13.4907
number(n2)=4
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.45
since our test is two-tailed
reject Ho, if to < -2.45 OR if to > 2.45
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (3*445.3324 + 3*181.999) / (8- 2 )
s^2 = 313.6657
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=112-149/sqrt((313.6657( 1 /4+ 1/4 ))
to=-37/12.5233
to=-2.9545
| to | =2.9545
critical value
the value of |t | with (n1+n2-2) i.e 6 d.f is 2.45
we got |to| = 2.9545 & | t | = 2.45
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != -2.9545 ) = 0.0213
hence value of p0.05 > 0.0213,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -2.9545
critical value: -2.45 , 2.45
decision: reject Ho
p-value: 0.0213
FINAL ANASWERS
a.
null, Ho: sigma^2 = sigma^2
alternate, H1: sigma^2 != sigma^2
b.
we do not reject Ho and believe that variances are equal since we do n't reject Ho
c.
null, Ho: u1 = u2
alternate, H1: u1 != u2
d.
we reject Ho and believe that mean are not equal since we reject Ho
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