A genetic experiment involving peas yielded one sample of offspring consisting of 435 green peas and 133 yellow peas. Use a 0.01 significance level to test the claim that under the same circumstances, 27 % 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? A. Upper H 0 : p not equals 0.27 Upper H 1 : p less than 0.27 B. Upper H 0 : p not equals 0.27 Upper H 1 : p greater than 0.27 C. Upper H 0 : p equals 0.27 Upper H 1 : p greater than 0.27 D. Upper H 0 : p equals 0.27 Upper H 1 : p not equals 0.27 E. Upper H 0 : p not equals 0.27 Upper H 1 : p equals 0.27 F. Upper H 0 : p equals 0.27 Upper H 1 : p less than 0.27 What is the test statistic? zequals nothing (Round to two decimal places as needed.) What is the P-value? P-valueequals nothing (Round to four decimal places as needed.) What is the conclusion about the null hypothesis? A. Fail to reject the null hypothesis because the P-value is greater than the significance level, alpha
Answer)
Ho : P = 0.27
Ha : P is not equal to 0.27
N = 435
P = 0.27
First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not
N*p = 117.45
N*(1-p) = 317.55
Both the conditions are met so we can use standard normal z table to estimate the P-Value
Test statistics z = (oberved p - claimed p)/standard error
Standard error = √{claimed p*(1-claimed p)/√n
Observed p = 133/435
Claimed p = 0.27
N = 435
Z = 1.68
From z table,p(z>1.68) = 0.0465
But our test is two tailed
So, P-Value is = 2*0.0465 = 0.093
As the obtained P-Value is > 0.01 (given significance level)
We fail to reject Ho
Fail to reject the null hypothesis because the P-value is greater than the significance level, alpha
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