SOLUTION: A genetic experiment involving peas yielded one sample of offspring consisting of 440 green peas and 154 yellow peas. Use
a 0.05 significance level to test the claim that under th
Algebra.Com
Question 1193687: A genetic experiment involving peas yielded one sample of offspring consisting of 440 green peas and 154 yellow peas. Use
a 0.05 significance level to test the claim that under the same circumstances, 23% of offspring peas will be yellow. Identify the
null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method and the normal distribution as an approximation to the binomial
distribution.
What are the null and alternative hypotheses?
What is the test statistic?
What is the P value?
What is the conclusion about the null hypothesis?
A:Reject the hypothesis because the P-value is less than or equal to the significance level, a .
B: Fail to reject the null hypothesis because the P-value is greater than the significance level, a
C:Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, a . D:Reject the null hypothesis because the P-value is greater than the significance level, a
What is the final conclusion?
A:There is not sufficient evidence to warrant rejection of the claim that 23% of offspring peas will be yellow
B: There is sufficient evidence to warrant rejection of the claim that 23% of offspring peas will be yellow C: There is sufficient evidence to support the claim that less than 23% of offspring peas will be yellow D:There is not sufficient evidence to support the claim that less than 23% of offspring peas will be yellow
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Ho: <=23% of peas are
Ha: >23% of peas are yellow
alpha=0.05 for this one way test prob {reject Ho|Ho true}
-
normal approximation has mean np=136.62
variance is np(1-p)=105.20
sd is sqrt (V)=10.26
found 154
so z=(153.5-136.62)/10.256 using the continuity correction factor.
=1.645
probability z is >1.645 is 0.050. This is a one way test, so will be half the p-value of a two way.
The answers to the multiple choice questions are A and B. One rejects Ho and concludes that there is sufficient evidence to warrant the claim that 23% of the peas are not yellow.
The way I would do this is with a one-sample two way proportion test
Ho: p=0.23
Ha: P NE 0.23
alpha-0.05 p{reject Ho|Ho true}
test statistic is a z=(p hat-p)/sqrt(0.23*0.77/594)
critical value is |z|>1.96
p hat=0.02593
z= 0.0293/0.0173=1.693
fail to reject Ho, p-value is 0.090. Two-way tests double the p-value of one way, since both sides have areas in the rejection region. This is more exact.
RELATED QUESTIONS
A genetic experiment involving peas yielded one sample of offspring consisting of 433... (answered by Theo)
A genetic experiment involving peas yielded one sample of offspring consisting of 441... (answered by CPhill)
genetic experiment with peas resulted in one sample of offspring that consisted of 447... (answered by math_tutor2020)
A genetic experiment with peas resulted in one sample of offspring that consisted of 439... (answered by Boreal)
A genetic experiment with peas resulted in one sample of offspring that consisted of 431... (answered by ewatrrr)
A genetic experiment involving peas yielded one sample of offspring consisting of 449... (answered by ikleyn)
A genetic experiment with peas resulted in one sample of offspring that consisted of 429... (answered by math_tutor2020)
A genetic experiment with peas resulted in one sample of offspring that consisted of 409... (answered by Boreal)
A genetic experiment with peas resulted in one sample of offspring that consisted of 403... (answered by Boreal)