Question

a genetic experiment involving peas yielded one sample of offspring consisting of 418 green peas and 180 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. ( h_0: p = 0.27 )
( h_1: plt0.27 )
b. ( h_0: p
eq0.27 )
( h_1: plt0.27 )
c. ( h_0: p = 0.27 )
( h_1: pgt0.27 )
d. ( h_0: p
eq0.27 )
( h_1: pgt0.27 )
e. ( h_0: p = 0.27 )
( h_1: p
eq0.27 )
f. ( h_0: p
eq0.27 )
( h_1: p = 0.27 )
what is the test statistic?
( z = 1.71 )
(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 sample size \(n = 418+180=598\). The number of yellow peas \(x = 180\). The sample proportion \(\hat{p}=\frac{x}{n}=\frac{180}{598}\approx0.301\).

Step2: Calculate \(P - value\)

Since the test - statistic \(z = 1.71\) and the test is two - tailed (because \(H_1:p
eq0.27\)), the \(P - value=2\times(1 - P(Z\leq|z|))\).
Using the standard normal table, \(P(Z\leq1.71)=0.9564\).
So \(P - value=2\times(1 - 0.9564)=2\times0.0436 = 0.087\)

Answer:

\(P - value = 0.087\)