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}
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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.
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