Question

Question Help * A genetic experiment involving peas yielded one sample of offspring consisting of 415 green peas and 139 yellow peas. Use a 0.05 significance level to test the claim that under the same crumstances, 25% of offspring peas will be yellow. Ide distribution. What are the null and alternative hypotheses? tify the nul hypothesis ate ative hypothesis test statistic, p ate·conciso about he null hypotheais, and final conclusion that addresses the original daim. Use the P.value method and the normal distibution as an approximation to the binomial A. Ho:p#025 H: p>025 О с. но: p.025 H: p0.25 O E. Ho'p#025 H: p 025 ов, но: р-О25 H: p 025 H P>025 Ho' p#025 H:pc025 OF,

Answer 1

n = 415

x = 139

$\hat{p}=\frac{x}{n}=\frac{139}{415}=0.3349$

Claim:25 % of offspring peas will be yellow.

Null and alternative hypothesis is

option B is correct

$H1 : p \neq 0.25$

Level of significance = 0.05

Test statistic is

$z =\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$z =\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}} =\frac{0.0849}{0.02126}=3.99$

$\alpha =0.05$

p-value = 0.00006 ( using z-table )

p-value = 0.0001

p-value $< \alpha$ , Reject Ho

conclusion : At $\alpha =0.05$ ,There is insufficient evidence to conclude that 25 % of offspring peas will be yellow