statistics homework 8.2

Question

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

448 green peas and 140 yellow peas. Use a 0.01 significance level to test the claim that under the same​ circumstances, 23​% 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.

Answer 1

The null hypothesis (H₀) is that 23% of offspring peas will be yellow, while the alternative hypothesis (H₁) is that the proportion of yellow peas is not equal to 23%.

H₀: p = 0.23

H₁: p ≠ 0.23

The significance level is at α = 0.01.

There is a total of 588 peas (448 green peas and 140 yellow peas).

p_{yellow peas} = 140/588 = 0.2381

z_statistic = (0.2381 - 0.23) / √[(0.23)*(0.77)/588] = 0.0081 / 0.017 ≈ 4.76

Based on the z-score table, the p-value is well below 0.01. Therefore, the null hypothesis is rejected. There is sufficient evidence that the proportion of yellow peas in the offspring is different from 23%.