Preliminary data collected on the abundance of Staghorn Lichens on the trunks of
ID: 2929285 • Letter: P
Question
Preliminary data collected on the abundance of Staghorn Lichens on the trunks of White Fir trees indicated that the average difference in percent cover of lichen between the west and east side of a tree was 18% with a standard deviation of 13.8 (n = 2 trees). Data on percent cover of Staghorn Lichen was also obtained by sampling four trees, two trees on their west side and two trees on their east side yielded an average percent cover of 43% on the east side and 25% on the west side, with standard deviations of 22.5 and 14.9 respectively. What sample size would be needed to test the null hypotheses at = 0.05 that the percent cover of lichens does not differ between the east and west sides of fir trees against the alternative hypothesis that lichen cover on the east side of fir trees is double that of the west side of trees? Which design, a within-subjects or a between-subjects design, would be more efficient for answering this question?
Explanation / Answer
Given that,
mean(x)=43
standard deviation , 1 =22.5
number(n1)=2
y(mean)=25
standard deviation, 2 =14.9
number(n2)=2
null, Ho: u1 = u2
alternate, H1: 1 > u2
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=43-25/sqrt((506.25/2)+(222.01/2))
zo =0.94
| zo | =0.94
critical value
the value of |z | at los 0.05% is 1.645
we got |zo | =0.943 & | z | =1.645
make decision
hence value of | zo | < | z | and here we do not reject Ho
p-value: right tail -Ha : ( p > 0.94 ) = 0.17277
hence value of p0.05 < 0.17277,here we do not reject Ho
ANSWERS
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null, Ho: u1 = u2
alternate, H1: 1 > u2
test statistic: 0.94
critical value: 1.645
decision: do not reject Ho
p-value: 0.17277
we do not have enough evidence to support the claim