A genetic experiment involving peas yielded one sample of offspring consisting of 446446 green peas and 155155 yellow peas. Use a 0.010.01 significance level to test the claim that under the same circumstances, 2424% 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.
H0: Null Hypothesis: P = 0.24
HA: Alternative Hypothesis: P 0.24
n = Sample Size = 446 + 155 = 601
p = Sample Proportion = 155/601 = 0.2579
P = Population proportion = 0.24
Q = 1- P = 0.76
SE =
Test statistic is:
Z = (p - P)/SE
= (0.2579 - 0.24)/0.0174 = 1.0287
Table of Area Under Standard Normal Curve gives area = 0.3485
So,
P - Value = (0.5 - 0.3485) X 2 = 0.3030
= 0.01
Since P - Value = 0.3030 is greater than = 0.01, the difference is not significant. Fail to reject null hypothesis.
Conclusion:
The data support the claim that under the same circumstances, 24% of offspring peas will be yellow.
A genetic experiment involving peas yielded one sample of offspring consisting of 446446 green peas and...
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