A genetic expenment involing peas yielded one sample of offspring consisting of
ID: 3356883 • Letter: A
Question
A genetic expenment involing peas yielded one sample of offspring consisting of 443 green peas and 173 yelow peas Use a 0 05 signácance level to test the clam bhat under the same circumstances, 26ss of osprng peas be yellow Identily the mull hypothesis, altemative hypothesis, test statistic, Pvalue, conclusion about the null hypothesis, and tinal conclusion that addresses the orgnal clam Use the Pa mehod and the norml stion as approximation to the binomial distnbubion What are the null and atemative hypotheses? A. Hop:026 H, p 0 26 m, p0 26 HP0 26 What in the test statistic? Round to tee decimal places as needed) What is the Pvalue? Round to four decimal places as needed )Explanation / Answer
Given that,
possibile chances (x)=173
sample size(n)=443
success rate ( p )= x/n = 0.3905
success probability,( po )=0.26
failure probability,( qo) = 0.74
null, Ho:p=0.26
alternate, H1: p!=0.26
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.39052-0.26/(sqrt(0.1924)/443)
zo =6.2629
| zo | =6.2629
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =6.263 & | z | =1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 6.26287 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
a.
null, Ho:p=0.26
alternate, H1: p!=0.26
b.
test statistic: 6.26
c.
p-value: 0
d.
reject the null hypothesis because p value is less than or equal to the
significance level
e.
Option A
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