Question

4.

A genetic experiment involving peas yielded one sample of offspring consisting of 408 green peas and 124 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?

A.

H0: p≠0.24

H1: p<0.24

B.

H0: p≠0.24

H1: p>0.24

C.

H0: p=0.24

H1: p≠0.24

D.

H0: p=0.24

H1: p<0.24

E.

H0: p=0.24

H1: p>0.24

F.

H0: p≠0.24

____________________________________

What is the test​ statistic?

z= "?" ​(Round to two decimal places as​ needed.)

What is the​ P-value?

P-value= ​(Round to four decimal places as​ needed.)

______________________________

What is the conclusion about the null​ hypothesis?

A.

Reject

the null hypothesis because the​ P-value is

less than or equal to

the significance​ level,

α.

B.

Reject

the null hypothesis because the​ P-value is

greater than

the significance​ level,

α.

C.

Fail to reject

the null hypothesis because the​ P-value is

greater than

the significance​ level,

α.

D.

Fail to reject

the null hypothesis because the​ P-value is

less than or equal to

the significance​ level,

α.

_________________________________

What is the final​ conclusion?

A.There

is not

sufficient evidence to warrant rejection of the claim that

24​%

of offspring peas will be yellow.

B.There

is

sufficient evidence to support the claim that less than

24​%

of offspring peas will be yellow.

C.There

is

sufficient evidence to warrant rejection of the claim that

24​%

of offspring peas will be yellow.

D.There

is not

sufficient evidence to support the claim that less than

24​%

of offspring peas will be yellow.

Answer 1

Solution :

Given that,

= 0.24

1 - = 0.76

n = 408

x = 124

Level of significance = = 0.05

Point estimate = sample proportion = = x / n = 0.304

This a left (One) tailed test.

The null and alternative hypothesis is,

D)

Ho: p = 0.24

Ha: p < 0.24

Test statistics

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

= ( 0.304 - 0.24) / (0.24*0.76) / 408

= 3.02

P-value = P(Z < z )

= P(Z < 3.02)

= 0.9987

P-value > 0.05, Fail to reject the H0.

C)

Fail to reject the null hypothesis because the​ P-value is greater than the significance​ level,

D)

There is not sufficient evidence to support the claim that less than 24​% of offspring peas will be yellow.