Question

A genetic experiment involving peas yielded one sample of offspring consisting of 409 green peas and 140 yellow peas. Use a 0.05 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

What are the null and alternative hypothesis?   

What is the test statistic?   

What is the P-value?   

What is the conclusion about the null hypothesis?   

What is the final conclusion?

Answer 1

This is the two tailed test .

The null and alternative hypothesis is

H0 : p =0.23

Ha : p 0.23

n = 409 green peas + 140 yellow peas =549 total peas

x = 140 yellow peas

= x / n = 140 / 549 =0.25

P0 = 0.23

1 - P0 = 1 - 23 =0.77

Test statistic = z

= - P0 / [P0 * (1 - P0 ) / n]

= 0.25 -0.23/ [(0.23*0.77) / 549]

= 1.11

Test statistic = z =1.11

P(z >1.11 ) = 1 - P(z < 1.11) = 0.8588

P-value = 2*0.1412 =0.2824

= 0.05

P-value >

0.2824 > 0.05

Do not reject the null hypothesis .

There is insufficient evidence to suggest that