Question

a genetic experiment involving peas yielded one sample of offspring consisting of 435 green peas and 175 yellow peas. use a 0.01 significance level to test the claim that under the same circumstances, 26% 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.

e. ( h_0:p
eq0.26 )
( h_1:p = 0.26 )

f. ( h_0:p = 0.26 )
( h_1:p>0.26 )

what is the test statistic?
( z = 1.52 )
(round to two decimal places as needed.)

what is the p - value?
( p - value=square )
(round to three decimal places as needed.)

Explanation:

Step1: Calculate sample proportion

The total number of peas \(n = 435+175=610\). The sample proportion of yellow peas \(\hat{p}=\frac{175}{610}\approx0.287\). The hypothesized proportion \(p = 0.26\), and \(q=1 - p=0.74\).

Step2: Calculate the P - value

Since this is a right - tailed test (because \(H_1:p>0.26\)), the P - value is \(P(Z>z)\). Given \(z = 1.52\). Using the standard normal distribution table or a calculator, \(P(Z>1.52)=1 - P(Z\leq1.52)\). From the standard normal table, \(P(Z\leq1.52)=0.9357\). So \(P - value=1 - 0.9357 = 0.064\)

Answer:

\(P - value = 0.064\)