Question

A genetic experiment involving peas yielded one sample of offspring consisting of

428

green peas and

136

yellow peas. Use a

0.05

significance level to test the claim that under the same​ circumstances,

24​%

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?

The test statistic is z=

The​ P-value is

Identify the conclusion about the null hypothesis and the final conclusion that addresses the original claim.

Answer 1

Solution :

Given that,

= 0.24

1 - = 0.76

n = 428

x = 136

Level of significance = = 0.05

Point estimate = sample proportion = = x / n = 0.318

This a left (One) tailed test.

The null and alternative hypothesis is,

Ho: p = 0.24

Ha: p < 0.24

Test statistics

z = ( - ) / *(1-) / n

= ( 0.318 - 0.24) / (0.24*0.76) / 428

= 3.730

P-value = P(Z < z)

= P(Z < 3.730 )

= 0.9999

The p-value is p = 0.9999, and since p = 0.9999 > 0.05, it is concluded that fail to reject the null hypothesis.

There is no sufficient evidence to claim that under the same​circumstances,24​% of offspring peas will be yellow, at 0.05 significance level.